How to derive the conductivity titration curve which accounts for salt formation I derived an equation that gave me the volume of base required to get a certain $\mathrm{pH}$:

In a titration between a weak acid $\ce{HA}$ and weak base $\ce{B}$ (adding base into acid solution) the following equations apply:

$$
\begin{align}
\ce{HA &<=> H+ + A- & K_{a}&=\frac{[H+][A-]}{[HA]}}\\
\ce{B + H2O & <=> HB+ + OH- & K_{b}&=\frac{[HB+][OH-]}{[B]}}\\
\ce{H2O & <=> H+ + OH- &K_{w}&=[H+][OH-]=10^{-14}} \\
\ce{Deg[A-] &= \frac{K_{a}}{K_{a} +[H+]} &Deg[HB+]&=\frac{K_{b}}{K_{b} + [OH-]}}
\end{align}
$$

These can be substituted into a charge balance and solved for $V_{b}$ (volume of base added):

$$
\begin{align}
\ce{[A-] + [OH-] &= [H+] + [HB+]}\\
\ce{\frac{K_{a}}{K_{a} + [H+]}\times\frac{C_{a}V_{a}}{V_{a} +V_{b}} + \frac{K_{w}}{[H+]} &=[H+] + \frac{K_{b}}{K_{b} + \frac{K_{w}}{[H+]}}\times\frac{C_{a}V_{a}}{V_{a} +V_{b}}}\\
\cdots&\cdots\\
\ce{V_{b} &=\frac{\left([H+] - \frac{K_{w}}{[H+]}-\frac{C_{a}K_{a}}{K_{a} + [H+]}\right)\times V_{a}}{\frac{K_{w}}{[H+]} -[H+] - \frac{K_{b}C_{b}}{K_{b} + \frac{K_{w}}{[H+]}}}}
\end{align}
$$

I used this to generate a pH curve and conductivity curve for the titration using the concentration of the various chemical species in solution:

For calculating the conductivity of the solution:

$$
\kappa_{\rm{soln}}=\lambda_{\ce{A-}}\times\left[\ce{A-}\right] + \lambda_{\ce{HB+}}\times\left[\ce{HB+}\right]+\lambda_{\ce{H+}}\times\left[\ce{H+}\right] + \lambda_{\ce{OH-}}\times\left[\ce{OH-}\right]
$$

Where $\lambda\chi$ is the conductivity constant for chemical species $\chi$

I was told (by a person we shall call $X$) that this model is incorrect as the weak acid/weak base titration does not have a sharp gradient change at the equivalence point but is rather sloped/gently curved due to the following equilibrium:

$$
\ce{HBA <=> HB+ + A-}\tag{$K_{\rm{sp}}=[\ce{HB+}][\ce{A-}]$}
$$

Is there a more mathematical explanation and is it possible to incorporate this equilibrium expression into the titration curve.

Please note that I know that a strong acid/base-weak base/acid conductivity curve does not have a well defined equivalence point (being rather curved), and upon asking $X$, she said that the weak acid/weak base conductivity curve was also curved at equivalence and not as distinctly sharp as my model predicts

I believe have found a answer.
  It's valid mathematically, make sense physically but I don't know if chemically is true.
  I posted to community appreciation.
  There we go!

The reactions

ionization of weak acid:
$$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}
\label{eq:KWeakAcid}$$
ionization of weak base:
$$\ce{B + H2O <=> BH+ + OH-}\qquad K_\ce{b}=\frac{\ce{[HB+][OH-]}}{\ce{[B]}}
\tag{2}
\label{eq:KWeakBase}$$
self-ionization of water:
$$\ce{2 H2O <=> H3O + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]}
\tag{3}
\label{eq:KWater}$$
equilibrium of poorly soluble salt:
$$\ce{HBA <=> HB+ + A-}\qquad K_\ce{sp}=\ce{[HB+][A-]}
\tag{4}
\label{eq:KSalt}$$

Charge balance

$$\ce{[H3O+] + [HB+] = [OH-] + [A-]}\tag{5}\label{eq:ChangeBalance}$$

Mass balance

To mass balance we can use an expression based on amount of substance:

Weak acid:

\begin{align}
{n_\ce{HA}}_\text{total}&={n_\ce{HA}}_\text{solution}+{n_\ce{A-}}_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
C_\ce{HA}V_\ce{HA}&=\left(\ce{[HA] + [A-]}\right)V_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
\\
\ce{[HA] + [A-]}&=\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{6}
\label{eq:AcidMassBalance}\\
\end{align}

Weak base:

\begin{align}
{n_\ce{B}}_\text{total}&={n_\ce{B}}_\text{solution}+{n_\ce{HB+}}_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
C_\ce{B}V_\ce{B}&=\left(\ce{[B] + [HB+]}\right)V_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
\\
\ce{[B] + [HB+]}&=\frac{C_\ce{B}V_\ce{B}-{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{7}
\label{eq:BaseMassBalance}\\
\end{align}

The general equation

Combining ($\ref{eq:KWeakAcid}$) and ($\ref{eq:AcidMassBalance}$) equations we have:

$$\ce{[A-]}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{8}
\label{eq:Anion}$$

Combining ($\ref{eq:KWeakBase}$), ($\ref{eq:KWater}$) and ($\ref{eq:BaseMassBalance}$) equations we have:

$$\ce{[HB+]}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-
{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{9}
\label{eq:Cation}$$

From ($\ref{eq:KSalt}$) equation we have ${n_\ce{HB+}}_\text{ppt}={n_\ce{A-}}_\text{ppt}=
n_\ce{HBA}=n_\text{ppt}=n$ and replacing ($\ref{eq:KWater}$), ($\ref{eq:Anion}$) and
($\ref{eq:Cation}$) equations in ($\ref{eq:ChangeBalance}$) equation we have:

$$\ce{[H3O+]} + \frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}
\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
= \frac{K_\ce{w}}{\ce{[H3O+]}} + \frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}
\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
\tag{10}
\label{eq:TitrationCurve}$$

Besides that combining the ($\ref{eq:KSalt}$), ($\ref{eq:Anion}$) and ($\ref{eq:Cation}$) equations we have:

$$K_\ce{sp}=\overbrace{
\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[HB+]}}
\overbrace{
\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[A-]}}
\tag{11}
\label{eq:Kppt}$$

Reproducing your data

If we ignore the poorly soluble salt formation, the ($\ref{eq:TitrationCurve}$)
equation is exactly the same which you deduced.

From your image we have
$V_\ce{HA} = \pu{1.00 L}$,
$C_{A} = \pu{0.01 mol L^{-1}}$,
$K_\ce{a} = \pu{2.0E-4}$,
$C_\ce{B} = \pu{0.20 mol L^{-1}}$,
$K_\ce{b} = \pu{2.0E-4}$, and
$K_\ce{w} = \pu{1.0E-14}$
for which the titration curve would be:

The plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$ through titration would be:

Using the conductivity equation of the solution

$$\kappa_\ce{solution} = \lambda_\ce{A-}\ce{[A-]} + \lambda_\ce{HB+}\ce{[HB+]}
+ \lambda_\ce{H3O+}\ce{[H3O+]} + \lambda_\ce{OH-}\ce{[OH-]}
\tag{12}
\label{eq:ConductivitySolution}$$

and
$\lambda_\ce{A-} = \pu{ 90.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{HB+} = \pu{172.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{H3O+} = \pu{849.6E-4 m^2 S mol^{-1}}$, and
$\lambda_\ce{OH-} = \pu{398.0E-4 m^2 S mol^{-1}}$
from your image is possible to plot the conductivity curve:

If we look to contribution of each species to conductivity curve
we can understand its final shape.

Some considerations

To take account the formation of $\ce{HBA}$, suppose that $K_\ce{sp} = \pu{7.5E-5}$.
The $Q_\ce{sp}$ plot through titration would be:

While $Q_\ce{sp}<K_\ce{sp}$ we have $n=0$, because $\ce{[HB+][A-]}$ is less than
critical concentrations to start precipitation and the titration curve follows as usual.
When $Q_\ce{sp}=K_\ce{sp}$ we have the precipitation eminence and $n=0$.
When $Q_\ce{sp}>K_\ce{sp}$ I'm not sure what will happen.

The $V_\text{critical}$ in which $Q_\ce{sp}=K_\ce{sp}$ can be found graphical or analytical way.
The last way in not so easy, but I believe then can be done.
Personally I will use the graphical way.

Solving ($\ref{eq:TitrationCurve}$) equation to $V_\ce{B}$ (with $n=0$) we can write:

$$V_\ce{B}=\frac{V_\ce{HA}\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+] +
C_\ce{HA}\alpha_\ce{A-}}\right)}{C_\ce{B}\alpha_\ce{HB+} -
\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+]}\right)}
\tag{13}
\label{eq:Vtit}$$

Again, the ($\ref{eq:Vtit}$) equation is exactly the same which you deduced.

Solving ($\ref{eq:Kppt}$) to $V_\ce{B}$ (with $n=0$) we have:

$$V_\ce{B}=\displaystyle\frac{-\left[V_\ce{A}\left(2 -
\displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}{K_\ce{sp}}\right)\right]-
\sqrt{\left[V_\ce{A}\left(2 - \displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}
{K_\ce{sp}}\right)\right]^2 - 4V_\ce{HA}^2}}{2}
\tag{14}
\label{eq:Vcritical}$$
where
$${\alpha}_\ce{HB+}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}$$
and
$${\alpha}_\ce{A-}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}$$

The $V_\ce{B}\ vs.\ \ce{pH}$ plot to ($\ref{eq:Vtit}$) and ($\ref{eq:Vcritical}$) equations
allows to find $V$ and $\ce{pH}$ for which the curves cut across each other.

Another (graphical) way could be the $Q_\ce{sp}$ plot.
Any way the values found are $\pu{45.20 mL}$ and $4.685$ respectively.
What happens with $\ce{pH}$ when $V_\ce{B}$ is greater than $V_\text{critical}$?
Some hypothesis can been proposed.

Hypothesis #1

  $\ce{HBA}$ will precipitate and the titration curve don't be affected.

In this case the ($\ref{eq:Kppt}$) equation can be solved to $n$:

$$n=\displaystyle\frac{\left(n_\ce{HA}+n_\ce{B}\right)-
\sqrt{\left(-\left(n_\ce{HA}+n_\ce{B}\right)\right)^2 -
4\left(n_\ce{HA}n_\ce{B}-
\displaystyle\frac{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2}
{{\alpha}_\ce{HB+}{\alpha}_\ce{A-}}\right)}}{2}
\tag{15}
\label{eq:nppt}$$

where $n_\ce{HA} = C_\ce{HA}V_\ce{HA}$ and $n_\ce{B} = C_\ce{B}V_\ce{B}$.

The ($\ref{eq:nppt}$) equation allows calculate the amount of precipitate is formed
given a $V_\ce{B}$ and $\ce{pH}$ value. But given a $V_\ce{B}$, what is $\ce{pH}$ value?
I believe that, if this hypothesis is true, the $\ce{pH}$ value can be calculated
from ($\ref{eq:TitrationCurve}$) equation and the titration curve don't be affected.

The the $n\ vs.\ V_\ce{B}$ plot of ($\ref{eq:nppt}$) equation is

and the $n\ vs.\ \ce{pH}$ plot of ($\ref{eq:nppt}$) equation is

The effect of precipitation can be seen in plots below.
The $\ce{[HB+]}$ and $\ce{[A-]}$ plot

The $Q_\ce{sp}$ plot

The plot of contribution of each species to conductivity

The total conductivity plot

Hypothesis #2

  $\ce{HBA}$ don't precipitates and the $\ce{pH}$ rises.

In this case the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.

The ($\ref{eq:Kppt}$) equation can be rearranged to give:

$$\ce{[H3O+]}^2 - \ce{[H3O+]}\left(
\frac{K_\ce{a}C_\ce{HA}V_\ce{HA}C_\ce{B}V_\ce{B}}
{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2} - K_\ce{a} -
\frac{K_\ce{w}}{K_\ce{b}}\right) +
\frac{K_\ce{a}K_\ce{w}}{K_\ce{b}}=0
\tag{16}
\label{eq:TitrationCurvePpt}$$

Note that, using one of the roots from ($\ref{eq:TitrationCurvePpt}$) equation,
the $\ce{pH}$ jumps from $4.685$ to $9.315$ (see titration curve below) when
$V_\ce{B}=V_\text{critical}$ and, while $V_\ce{B}$ increases, the titration curve
is look like as usual.

Mathematically in this hypothesis no precipitate is formed, but the
solution cross the region where there should be precipitate.
The consequence this is the concentrations of $\ce{[A-]}$ and $\ce{[HB+]}$,
which depends on $\ce{pH}$ values, jumps too.

Hypothesis #3

  $\ce{HBA}$ don't precipitates and the $\ce{pH}$ falls.

Again the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.

Using the other root from ($\ref{eq:TitrationCurvePpt}$) equation, the titration curve
have a discontinuity at $V_\text{critical}$ and then the $\ce{pH}$ values falls
while $V_\ce{B}$ increases. No jumps, neither on concentrations or pH, are observed.

If hypothesis #3 is true, the plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$
through titration is

the $Q_\ce{sp}$ plot is

the conductivity curve to each species

and finally the conductivity curve is

Addendum

To those interested follows the gnuplot code to produce these graphs. All equations needed to reproduce this answer are available.

The present code creates several graphs used in this answer but no all. Some of them are combination of two or more data sets or a graph from an equation available.

#
#  Implements a lot of equations and create a couple of graphs showing what
#  happens with pH, concentrations, conductivity, and Qsp when a poorly soluble
#  salt is formed during a titration.
#
#  To more details see the original question/answer
#
#  Question: https://chemistry.stackexchange.com/q/117984/80491
#  Answer:   https://chemistry.stackexchange.com/a/118760/80491
#
#  Author: GRSousaJr (https://chemistry.stackexchange.com/users/80491/grsousajr)
#  gnuplot 5.2 patchlevel 7
#

# ========== INITIAL CONFIGURATION ===============================================
reset               # return all graph-related options their default values
set encoding utf8   # selects a character encoding

# Save each graph as pdf file
set terminal pdfcairo size 8 in, 6 in font "Times New Roman,18" enhanced

# Colors of lines
set style line 1 lc "#e41a1c"   # red
set style line 2 lc "#377eb8"   # blue
set style line 3 lc "#4daf4a"   # green
set style line 4 lc "#984ea3"   # purple

# Margins
set lmargin screen 0.110    # left margin
set tmargin screen 0.940    # top margin
set bmargin screen 0.130    # bottom margin

set grid ls -1 lc "gray"    # style for grid
set tics out nomirror       # tics marks on axis
set samples 1000            # number of points to each curve
set dummy pH                # "pH" as "x" for all plot-commands
unset key                   # turn-off the legend (key) exhibition

# ========== INPUT PARAMETERS ====================================================
# ---------- Weak acid (HA) ------------------------------------------------------
VA    = 1.0     # L, HA volume (mandatory)
CA0   = 0.01    # mol/L, HA concentration (mandatory)
pKa   = 3.69897 # Ka = 0.00020 (mandatory)

# ----------- Weak Base (B) ------------------------------------------------------
CB0   = 0.2     # mol/L, B concentration (mandatory)
pKb   = 3.69897 # Kb = 0.00020 (mandatory)
VBmax = 120.0   # mL, B volume (optional)

# ----------- Poor soluble salt (HBA) --------------------------------------------
pptn  = "True"  # "True/False" (mandatory) indicates if occurs precipitation
Ksp   = 7.5E-5  # HBA poor soluble salt (mandatory if pptn = "True")

# ----------- Solvent ------------------------------------------------------------
pKw   = 14.0    # H2O self-ionization

# ----------- Conductivity constants ---------------------------------------------
array lambda[4]
lambda[1] =  90.00 # (E-4) A-
lambda[2] = 172.00 # (E-4) HB+
lambda[3] = 849.60 # (E-4) H+
lambda[4] = 398.00 # (E-4) OH-

# ========== DERIVATE PARAMETERS =================================================
# ---------- Equilibrium constants -----------------------------------------------
Ka     = 10**(-pKa)     # Ka from pKa
Kb     = 10**(-pKb)     # Kb from pKa
Kw     = 10**(-pKw)     # Kw from pKw
# ---------- Logical test to VBmax------------------------------------------------
if (!exists("VBmax")){
    VBmax = 2.0*VA*CA0/CB0  # Double that required for neutralization
} else{
    VBmax = VBmax/1000.0    # Converting mL to L
}

# ========== FUNCTIONS ===========================================================
# ---------- H+ and OH- concentrations (as a function of pH) ---------------------
H(pH)  = 10**(-pH)      # H+ from pH
OH(pH) = 10**(pH-14.0)  # OH- from pH

# ---------- Ionization fractions (as a function of pH) --------------------------
aA(pH)  = Ka/(H(pH) + Ka)       # A-
aB(pH)  = Kb/(Kw/H(pH) + Kb)    # HB+

# ---------- B volume (as a function of pH) --------------------------------------
VB(pH) = (VA*(Kw/H(pH) - H(pH) + aA(pH)*CA0))/(aB(pH)*CB0 - (Kw/H(pH) - H(pH)))

# ---------- Amount of substance (as a function of pH) ---------------------------
nA       = CA0*VA         # HA
nB(pH)   = CB0*VB(pH)  # B

# Amount of precipitate formed
nppt(pH) = ((nA + nB(pH)) - sqrt((-(nA + nB(pH)))**2 - \
    4*(nA*nB(pH) - (Ksp*(VA + VB(pH))**2/(aB(pH)*aA(pH))))))/2.0

# Logical test to know if precipitation must be taken account
if (pptn eq "True" ){
    n(pH) = nppt(pH) <= 0 ? 0 : nppt(pH)  # Disregards negative values of 'n'.
    sufix = "_with_ppt"
} else {
    n(pH) = 0   # No occurs precipitation
    sufix = "_without_ppt"
}

# ---------- A- and HB+ concentrations (as a function of pH) ---------------------
CA(pH) = aA(pH)*(nA - n(pH))/(VA + VB(pH))       # [A-]
CB(pH) = aB(pH)*(nB(pH) - n(pH))/(VA + VB(pH))   # [HB+]

# ---------- Qsp (as a function of pH) -------------------------------------------
Qsp(pH) = CB(pH)*CA(pH)

# ---------- Individual and Total conductivities (as a function of pH) -----------
kappaA(pH)   = lambda[1]*CA(pH)
kappaB(pH)   = lambda[2]*CB(pH)
kappaH(pH)   = lambda[3]*H(pH)
kappaOH(pH)  = lambda[4]*OH(pH)
kappaSol(pH) = kappaA(pH) + kappaB(pH) + kappaH(pH) + kappaOH(pH)

# ---------- Initial and final pH ------------------------------------------------
pH0   = -log10((-Ka + sqrt(Ka**2 + 4*CA0*Ka))/2.0)    # Initial pH
pHmax = pKw + log10((CA0*2*VA - CA0*VA)/(VA + VBmax)) # Estimate for pHmax at VBmax

#---------- Finding pHmax by iteration -------------------------------------------
while (VB(pHmax) > VBmax || VB(pHmax) < 0){
    pHmax = pHmax - 0.0001
}

# ========== CREATING THE GRAPHS =================================================
# ---------- Titration curve -----------------------------------------------------
filename = "Titration_Curve" . sufix # Define a filename to pdf file

# Plotting using a data block (like a temp file)
set table $Titration_Data
    plot [pH0:pHmax] 1000*VB(pH)
unset table

set xlabel "Volume of titrant / mL" # name for x-axis (the same to all plot-commands)
set ylabel "pH"                     # name for y-axis
set xrange [0:VBmax*1000]           # VBmax in mL
set yrange [0:14]                   # pH 0 to 14
set xtics 10                        # tics at each 10 mL
set ytics 1                         # tics at each 1 pH unit

set output filename . ".pdf"        # Create the pdf file

plot $Titration_Data using 2:1 with lines linestyle 1 linewidth 2

# ---------- Concentrations curves ------------------------------------------------
filename = "Concentrations_Curve" . sufix

set table $ConcB_Data
    plot [pH0:pHmax] CB(pH)
unset table

set table $ConcA_Data
    plot [pH0:pHmax] CA(pH)
unset table

set ylabel "Concentration / mol L^{–1}"
set yrange [0:CA0]
set ytics .001
set format y "%.3f" # number with 3 decimal places

# Labels
set label 1 "{/:Bold [HB^{+}]}" at 20.0, 0.003 tc ls 1 font ',20' offset 0,0.75
set label 2 "{/:Bold [A^{–}]}"  at 05.0, 0.003 tc ls 2 font ',20' offset 0,0.75

set output filename . ".pdf"

plot \
    $ConcB_Data u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConcA_Data u (1000*VB($1)):2 w l ls 2 lw 2

unset for [i=1:2] label i   # Clean the labels
unset format                # Clean the format

# ---------- Individual conductivity curves --------------------------------------
filename = "Individual_Conductivity_Curve" . sufix

set table $ConducA_Data
    plot [pH0:pHmax] kappaA(pH)
unset table

set table $ConducB_Data
    plot [pH0:pHmax] kappaB(pH)
unset table

set table $ConducH_Data
    plot [pH0:pHmax] kappaH(pH)
unset table

set table $ConducOH_Data
    plot [pH0:pHmax] kappaOH(pH)
unset table

set ylabel "{/:Italic κ} / S m^{–1}"
set yrange [0:2]
set ytics .2
set format y "%.2f"

set label 1 "{/=24 {/:Italic κ}_{/=12 A^{–}}}"      at 95, 0.70 center tc ls 1
set label 2 "{/=24 {/:Italic κ}_{/=12 HB^{+}}}"     at 95, 1.50 center tc ls 2
set label 3 "{/=24 {/:Italic κ}_{/=12 H_{3}O^{+}}}" at 07, 0.90 center tc ls 3
set label 4 "{/=24 {/:Italic κ}_{/=12 OH^{–}}}"     at 95, 0.15 center tc ls 4

set output filename . ".pdf"

plot \
    $ConducA_Data  u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConducB_Data  u (1000*VB($1)):2 w l ls 2 lw 2 ,\
    $ConducH_Data  u (1000*VB($1)):2 w l ls 3 lw 2 ,\
    $ConducOH_Data u (1000*VB($1)):2 w l ls 4 lw 2

unset for [i=1:4] label i
unset format

# ---------- Total conductivity curves -------------------------------------------
filename = "Total_Conductivity_Curve" . sufix

set table $ConducSolution_Data
    plot [pH0:pHmax] kappaSol(pH)
unset table

set yrange [0.75:3]
set ytics .25
set format y "%.2f"

set output filename . ".pdf"

plot $ConducSolution_Data u (1000*VB($1)):2 w l ls 1 lw 2
unset format

# ---------- Qsp curves ----------------------------------------------------------
filename = "Qsp_Curve" . sufix

set table $Qsp_Data
    plot [pH0:pHmax] Qsp(pH)
unset table

set yrange [0:0.000100]
set ytics 0.00001
set ylabel "{/:Italic Q}_{sp}"
set format y "%h" # scientific notation format

set label 1 "{/:Bold {/:Italic Q}_{sp}}"\
    at 32.0, 3.0E-5 tc ls 1 font ",20" offset 0,0.75
set label 2 "{/:Italic Q}_{sp} = {/:Italic K}_{sp}" \
    at 70.0, Ksp    tc ls 1 font ",20" offset 0,0.75

set output filename . ".pdf"

plot $Qsp_Data u (1000*VB($1)):2 w l ls 1 lw 2


Do amphoteric cations exist? I know that anions such as $\ce{CH3COO-}, \ce{OH-}, \ce{CN-}$ are basic in nature, since they are conjugate bases of weak acids. Similarly, cations such as $\ce{NH4+}, \ce{H3O+}$ are acidic in nature , since they conjugate acids of weak bases.

However, we also have another case whereby, anions such as $\ce{HSO3-}, \ce{HS-}$ are amphoteric in nature. This is due to the fact that their conjugate acids are polyprotic acids. Thus, one can have the two reactions:

$\ce{HS- + H2O <=> S^2- + H3O+}$ and $\ce{HS- + H2O <=> H2S + OH-}$

Since, for this case, we got amphoteric anions from polyprotic acids, is there a possibility that we can do the same for polyprotic bases so as to get amphoteric cations? i.e. Can we have a conjugate acid of a polyprotic base, which can be regarded as an amphoteric cation? I haven't seen any such examples of amphoteric cations yet, hence I am asking this question.

Yes, for example lysine.

Lysine has two amino groups and one carboxyl group. When dissolved in aprotic solvents, all of these will be uncharged so the carboxyl group will be $\ce{-COOH}$ and the two amino groups $\ce{-NH2}$. When dissolved in water, these three groups have distinct $\mathrm pK_\mathrm a$ values of $2.15, 9.16$ and $10.67$. Depending on the acidity of the solution, the following (acid-sided) equilibriums occur:

When dissolved in water, the carboxyl group’s proton will immediately be removed and used to immediately protonate the ε-amino group (the far end). However, the α-amino group is still more basic than water and thus will be partially protonated; meaning that the right-hand equilibrium will be happening predominantly if lysine is dissolved in water. Dissolved lysine will be a mixture of uncharged lysine and lysine cation.

When an acid such as hydrochloric acid is added, at first the α-amino group is completely protonated and then the left-hand equilibrium occurs predominantly as more and more lysine cation is converted to lysine dication.

If a base is added (such as sodium hydroxide), the ε-amino group’s proton can also be removed yielding lysine anion; I have not shown that equilibrium in the scheme above.

Lysine cation thus fits your question as it is an amphoteric cation, the conjugate acid of a dibasic compound.


What is the correct way to calculate the rate constant for the reaction of NO and O2? The following data are given for the reaction of $\ce{NO}$ and $\ce{O2}$:

$$
\ce{2NO + O2  ->  2NO2}
$$

The the reaction is second  order  in $\ce{[NO]}$ and first order in $\ce{[O2]}$, and the rate of disappearance of $\ce{NO}$ is $2.5 \times 10^{-5}~\mathrm{mol\over L\,s}$ at the instant when $\ce{[NO] = [O2]} = 0.01~\mathrm{mol\over L}$.

The question asks me to calculate the rate constant.

I've thought of two ways of approaching the calculation—which of these solutions is correct?

1) Take the rate of the reaction as one-half the rate of disappearance of $\ce{NO}$:

$$
\begin{align}
R &= {1\over 2} * 2.5 \times 10^{-5} = k \ce{[NO]^2[O2]}=k(0.01)^3 \\
k &= 12.5~\mathrm{L^2\over mol^2\,s}
\end{align}
$$

2) Take the rate of the reaction as equal to the rate of disappearance of $\ce{NO}$:

$$
\begin{align}
R &=  2.5 \times 10^{-5} = k \ce{[NO]^2[O2]} = k(0.01)^3 \\
k &= 25 ~\mathrm{L^2\over mol^2\,s}
\end{align}
$$

Either definition is acceptable, as long as you clearly identify relationship between the calculated rate value $R$ and the rates-of-change for each of the species in the system.

In case (1), you've effectively calculated the rate constant defined as:

$$
R_1 = k_1\ce{[NO]^2[O2]}
    = -{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{1\over 2}{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {1\over 2}{\mathrm d \ce{[NO2]}\over \mathrm dt}
$$

In case (2), you've instead calculated the rate constant defined as:

$$
R_2 = k_2 \ce{[NO^2][O2]}
    = -2{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {\mathrm d \ce{[NO2]}\over \mathrm dt}
$$

Both cases accurately describe the kinetics, but you have to be careful to identify which definition of the rate you've used in calculating $k$.


How to calculate energy gap when determining electron configuration? The weird electron configuration of Niobium was well explained on this post:

How can one explain Niobium’s weird electronic configuration?

It was pointed out that electrons fill orbitals with higher exchange energy and lower repulsion energy, only if the sum of the positive exchange energy and negative repulsion energy overcomes the energy gap between orbitals.

The only problem is that I do not know how to quantify the energy gap.

How can I calculate the energy gap between orbitals?

You could calculate the energy of your initial electron configuration, then do the same for your final state and work out the difference. The Los Alamos Flexible Atomic Code (https://github.com/flexible-atomic-code/fac) can do that for you. There should be no state where you fill shells from the ground up that has a lower energy than what we now call the ground state of nobium.


Lithography: Is isopropanol a developer and/or a stopper? In the ebeam lithography with PMMA resists isopropanol is often used in a mixture 3:1 (IPA:MIBK) as a developer solution. To my knowledge the developing process is usually stopped by putting the sample in distilled water and drying it with a nitrogen airgun. But I also see collegeaus who use the 3:1 mixture for developing and then "stop" the development in pure IPA. I actually think they develop further, as I know some people also use pure IPA instead of 3:1 mixture for developing.

What am I missing, I know there many different recipes in lithography, but this IPA thing doesn't make sense to me, using IPA as a stopper means to me the developing process is strengthened further and the resist isotropically resolved, which you normally want to avoid with kind of written sharp edged rectangular structures written by ebeam

Methyl isobutyl ketone is the solvent for PMMA - the alcohol is just the carrier. MIBK has low solubility in water, so if you stopped with water, you would probably leave more MIBK residue in your PMMA.


Why is the reaction quotient Q always equal to one at standard state? I understand that at equilibrium, $Q$ is equal to $K$. This should not depend how I express concentrations (as amount of substance concentrations $c$ or as partial pressures $p$ or as molality $b$ or as mole fraction $x$) and which units I use (e.g. partial pressure in Pa or in atm) as long as I am consistent.

Various expressions rely on $Q$ being dimensionless (because we take its logarithm) and on being equal to one when all species are at standard state. For example, the equation for calculating the change in Gibbs free energy for a reaction should result in the standard Gibbs free energy for the standard state:

$$\Delta_r G = \Delta_r G^\circ + R T \ln Q = \Delta_r G^\circ \text{(for standard state)}$$

This requires that $Q = 1$ at standard state. The Nernst equation is another (related) example.

What I am wondering about is why $Q$ is always equal to one at standard state, no matter how I express concentrations when calculating reaction quotients.

For example, if I have a reaction like $\ce{A(g) + B(g) -> C(g)}$, doesn't the value of $Q$ change if I switch from concentration $c$ to partial pressure $p$, or use units of atm instead of Pa when measuring partial pressures? Come to think of it, doesn't the value of the equilibrium constant $K$ change as well when I change measures or units of concentration?

As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases).
The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it does not need to be practically reachable by the chemical species. By definition, solutes in standard state have an activity of 1, and gas have a fugacity of 1. Therefore, whatever you reaction, if all the species are assumed to be in standard state, the reaction quotient will be a product of 1s, that is, 1.


How to separate a racemic mixture of 1,2-diols [upjohn dihydroxylation]2I want to ask that i synthesise R-substituted-1,2-Diol with (ee=1:1) ratio of R-, S- with the help of Upjohn Dihydroxylation. i try to seprate it by different mobile phasese but failed to separate, then i try -OTBS at primary alcohol by wish that may it work and make two seprate points at TLC but i failed, i need both R- and S- enantiomers of Diol.
my question is that what possible method can i use to separate two diols completely or make enantio-selective diol by using different conditions like by Sharpless method.

The standard method for ketalization is illustrated in the reaction of cyclohexanone 1 with ethylene glycol to form the ketal 2. The reaction is reversible which requires irreversible removal of water shifting the reaction to the right. In this reaction the toluene/water azeotrope may be separated by a Dean-Stark trap. Because of an unfavorable equilibrium, hindered ketones, such as camphor 3, are more difficult to ketalize. Standard conditions may be employed in this instance using more efficient removal of water.1
Dauben, et al.,2 have demonstrated an alternative technique for ketalizing camphor and other hindered ketones by employing high pressure at ambient temperature as illustrated in 3 --> 4.3. To illustrate the resolution technique I have chosen (R)-camphor because it is hindered, inexpensive and recoverable. The full structure of your racemic 1,2-diol was not provided so I will employ racemic butan-1,2-diol to this end. There are four possible diastereomers that can be formed in the ketalization--namely-- 5, 6, 7 and 8. The (R)-diol can form ketals 5 and 6 while the (S)-diol can afford ketals 7 and 8. Because each pair is formed under equilibrating conditions, ideally one would hope that the less hindered of the 5/6 pair and 7/8 pair dominate such that a single ketal would arise for each of the enantiomers of the diol. The success of such a technique requires separation of diastereomers, presumably via chromatography, or, ideally, by selective crystallization. If the latter method were viable, an enantiopure diol can be liberated by acid hydrolysis. The enantiomeric diol would arise by utilizing (S)-camphor, which, unfortunately, is much more expensive than (R)-camphor. continued...

One need not be limited to camphor. Other hindered ketones may be considered. The methodology described here is not a guarantee of success. Conceptually, it is no different from Ladenburg's resolution of racemic coniine with tartaric acid in the 19th-century.
1) Estrone methyl ether was successfully ketalized by user55119 using a spinning band fractionating column to remove water. The Dean-Stark method had failed.
2) W. G. Dauben, J. M. Gerdes and G. C. Cook, J. Org. Chem., 1986, 51, 4964.
3) The role of ethyl orthoformate is to scavenge water thereby forming ethanol and ethyl formate. Ketalization was also accomplished at high pressure using TMSOCH2CH2OTMS/TMSOTf.


The most stable resonance structure I came across this question:

  Which is the most stable resonating structure of p-nitrosobenzene
  among the following?

I have been taught about the priorities for resonance structure stability as follows:

  Complete octet $\rightarrow$ more stable
  Uncharged $\rightarrow$ more stable
  Lower formal charge $\rightarrow$ more stable
  Less separation between opposite charges $\rightarrow$ more stable
  Negative charge near more electronegative atoms and positive charge near less electronegative atoms $\rightarrow$ more stable

So according to the above rules, (b) and (c) are the among the most stable. Bu then applying rule #4 above, (b) should be more stable, because here each positive charge is in vicinity of negative charge.

However the answer given is (c). Why is this?

Because on C; the negative charge should be on oxygen (the more electronegative element) instead of on carbon


What is heat of vaporization? How can it be used at temperature as low as 25 °C? When I first learned this term. I was introduced to this graph:

The heat in this case is at 100 °C, which I had ever since believed to be the only temperature of vaporization, assuming vaporization = boiling.

Then I saw a table with all the heat of vaporization values for different compounds. Most importantly, all these values apply to reactions at 25  °C.

Different heat of vaporization values apply to different temperature. but water, for ex., cannot boil, with vapor bubbles, unless at 100 °C, thus all the energy used will just be used to simply raise the temperature. wouldn't it?

Think of vaporization as the removal of the fastest molecules. The separation of the gaseous form from the more closely-bound liquid therefore removes energy from the liquid. The heat of vaporization can apply at much lower temperatures than the BP of water at atmospheric pressure. Even on a day when the temperature is well blow freezing, laundry will dry on a clothesline because ice evaporates (well, sublimes). Also, water at skin temperature soon feels cold as it loses heat through evaporation.

BTW, the boiling point of water is 100 °C only at 1 atmosphere; see this video of water boiling at room temperature, ~25 °C.


What methods could be used for determining the concentration of Zn2+ in a solution? I know that complexometric titration with EDTA is one method. What other methods could be used?

Mass spectroscopy(MS)
HPLC-MS, if separation is required
Atomic emission spectroscopy(AES)
(Probably)
Polarography/voltammetry

The list is not exhaustive.


What temperature is required to burn pencil lead graphite? I tried to burn a bunch of 8B pencil lead in a gas flame the other day. None of it actually caught fire, but being made of carbon (in the form of graphite) should cause it to burn, as graphite is essentially purified coal. What temperature is required to set it burning, and is extra oxygen required?

You have two problems.
1.) pencil lead graphite is actually a graphite-clay composite.
2.) carbon does not sustain burning easily unless it is held at very high temperatures.

Pencil lead will not burn effectively for the first reason as clay is non-combustible and smothers any fire load. If you did have pure graphite to burn, a simple flame wouldn't be hot enough to combust it.  You need to sustain temperatures of around $1000\mathrm{-}2000~^\circ\mathrm{C}$ in order for pure carbon to burn.


Demountable simple fume hood I found this build-plan recycling on the internet (I believe it was originally published in a book written by the infamous clandestine chemist Hobart Huson) which seemed interesting to me. I do not have the money to pay for a professional fume hood, nor do I have the space to fit such thing. I have to switch the function of my work workplace sometimes because of lack of space (it is also used for woodworking, another hobby of me), so a simple demountable fume hood would be ideal. As said, I found this one with a simple build description, but I was wondering about its effectiveness and safety.

As shown on the picture, the fume hood consists of a demountable frame, a thick plastic sheet covering the frame that is attached to a tube, and a simple exhaust system connected to an other tube that is placed outside the building, so that the fumes can diffuse in the atmosphere. The following four points concern me:

Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?
Is this plastic sheet a potential danger in case of fire? Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?
Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? Or is reading the MSDS (i.e. the environment section) for the chemicals being used enough?
There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?

Take note of the fact that we are talking about amateur chemistry here. There are some nasty fumes that need to be removed, like chlorine for example, but no dense fogs of hydrogen cyanide or phosgene.

Since the chemicals will not be in contact with the plastic for very long, very little will diffuse through the plastic.

If you are concerned about fire, you could wrap the outside (not the inside) with aluminum foil; this should keep the fire contained. Have a container of sand handy in case the fire is extremely hot and the aluminum starts to burn. In fact, sand would be handy for most fires since it will decrease the amount of oxygen getting to the flames.

As long as you have a high air flow and the volatiles are coming from a small area, there should be no problem. Added: The concentration of the volatiles will not be high enough for a spark to ignite.

I can't really comment about the environmental effect unless you provide more detail. If the MSDS doesn't mention a hazardous exposure from the fumes, then you probably don't need a fume hood. You can get some activated charcoal (from a pet store) and pass the fumes through the charcoal, but I don't know how effective that would be.


Why Lysine is abbreviated as 'K'? It does not has a 'K' in its name. It does not have a potassium in its chemical composition.

Would be nice to know why Glutamine is Q too, some of these abbreviation seems simply random.

From here

Back in 1960's Dr.Dayhoff was shortening the names of amino acids so that ancient computers could save computing time.

  By the time Dr. Dayhoff got to lysine, there were not too many letters left, so she used the letter K, explaining that K is at least near L in the alphabet.

As for Glutamine.

  For the remaining 5 amino acids, Dr. Dayhoff was reaching somewhat to find an easy-to-remember connection between the single letter and the amino acid.  She assigned aspartic acid, asparagine, glutamic acid and glutamine the letters D, N, E and Q, respectively, noting that D and N are nearer the beginning of the alphabet than E and Q, and that Asp is smaller than Glu, while Asn is smaller than Gln.


Calculation of Gibbs free energy of a chemical reaction So, I'm trying to derive the equation: $$\Delta G = \Delta G^o + RT \ln Q\tag{1}$$
And trying to follow the steps detailed on this website here. I am having trouble deriving equation 7, which tries to sum chemical potentials. Previously, it derives the change in chemical potential, $\Delta \mu _i$ when the reaction is poised away from equilibrium, and for the reaction $A + B \rightarrow C + D$, four equations, each for $A, B, C$ and $D$ respectively.
$$\Delta \mu _A = \mu_A' -\mu_A = RT \ln([A']/[A])\tag{2}$$
Where $[A']$ is the activity away from equilibrium. Afterwards, it proceeds by:
$$\Sigma\Delta\mu_i = RT \ln\left(\frac{[C'][D'][A][B]}{[C][D][A'][B']}\right)\tag{3}$$
My confusion is regarding the arrangement of the activities in the $\ln()$ function. I assumed that:
$$\Sigma\Delta\mu_i = \Delta\mu_A + \Delta\mu_B + \Delta\mu_C + \Delta\mu_D = RT \ln\left(\frac{[A'][B'][C'][D']}{[A][B][C][D]}\right)\tag{4}$$
It says that we assume the concentration of the products increases by convention, but I don't get it. Even if the concentration of the reactants decrease and the products increase, the change in sign will be due to the $\ln()$ function. If you flip the activities of the reactants around, it doesn't give the net change in Gibb's free energy anymore.
If anyone can help explain this, I will be very grateful.
P.S. Sorry if the question seemed low effort, but the context is mostly given in the website, and I can't find an explanation anywhere :(


Hybridization of carbon atoms present in the smallest ester? I found this question in the Resonance DLPD Organic Chemistry book on Page 55:

  What is the hybridization of carbon atoms present in the smallest ester?

The smallest ester as far as I know is methyl formate ($\ce{HCOOCH3}$):

The first carbon atom (on left) has 3 σ bonds and the one on the right has 4 σ bonds. Hence, the hybridization is $\mathrm{sp}^2$ and $\mathrm{sp}^3$, respectively.

However, according to the answer given in the book, all carbon atoms in the smallest ester are $\mathrm{sp}^2$ hybridized.

What did I do wrong? Is there a smaller ester? Or the hybridization of the one I have above not correctly determined?

You are right. Methyl formate has one $\mathrm{sp}^2$ and one $\mathrm{sp}^3$ hybridized orbital.

Your book probably has a misprint.


Why is imidazole considered an electron deficient heterocycle? A paper I'm looking at1 mentions a few electron deficient heterocycles:

  electron deficient heterocyclic compounds such as di(azines), benzothiazole and imidazole...

What are the criteria to determine whether a heterocycle is electron deficient or not? (Why is imidazole considered electron deficient, for example?)

Reference

(1) Bureš, F. Fundamental aspects of property tuning in push–pull molecules. RSC Adv. 2014, 4 (102), 58826–58851. DOI: 10.1039/C4RA11264D.

As indicated in the comment below, a rule of thumb for these particular compounds, is that aromatic nitrogens with double bonds (e.g., not pyrrole) are "electron deficient."

The term is a bit ill-defined. It does not mean the aromatic system (which usually still has 4n+2 electrons) actually has a different electron count.

More broadly, there can be several ways of declaring "electron rich" or "electron deficient," for organic compounds, but in the context of a "push-pull" conjugated or aromatic system (in the paper you link), these refer to rings or functional groups that are electron donors or electron acceptors.

So:

Electron deficient - tends to gain negative charge as an electron accepting group or ring
Electron rich - tends to gain positive charge as a donor group or ring

The charge transfer need not be a full electron - it is almost always much less (e.g., a small fraction).

The second part of your question is how you can determine this.

Conjugated systems are typically imagined as a particle-in-a-box.. that the electrons are highly delocalized. The "electron donor" is a region with higher potential energy and the electron acceptor is a region with lower potential energy. Thus, the electrons have slightly greater probability on the electron acceptor.

Therefore, you determine "electron rich" or "electron deficient" in comparison with the rest of the conjugated system.

Typically, electron deficient rings have more negative HOMO and LUMO energies - making them more difficult to oxidize (i.e., they are "deficient" in electrons and it's hard to give up another) and easier to reduce (i.e., adding an electron makes them less "deficient").

Electron rich rings have more positive HOMO and LUMO energies - making them easier to oxidize and give up an electron.

Again, this is a rule of thumb and it's almost always partial charge transfer.


Why is dimethyloxonium cation stronger acid than hydronium? $\ce{(CH_3)_2O^+ H}$ has $\rm pK_a \; -3.8$ while $\ce{H_3O^+}$ has $\rm pK_a \; 1.74\;.$ So, the former is stronger acid than latter hydronium ion.

But why is it so?

Is it due to the $\rm +I$-effect of $\ce{-(CH_3)}$? But it would rather decrease the attraction of $\ce O$ on the electron pair of $\ce H$ and that would make it indifferent to release $\ce {H^+}$ isn't it?

What is actually going that makes $\ce{(CH_3)_2O^+ H}$ stronger acid than $\ce{H_3O^+}$ ?

A powerful but often overlooked factor in determining a substance's acidity and basicity is the effect of the solvent (or lack thereof) in which it is dissolved.

You are not wrong; in the gas phase, $\ce{H_3O^{+}}$ is indeed a stronger acid than $\ce{(CH_3)_2OH^{+}}$, by a factor of about $5\times 10^{18}$. This can be seen by comparing the proton affinities of the conjugated bases, dimethyl ether ($\mathrm{804\ kJ\ mol^{-1}}$) and water ($\mathrm{697\ kJ\ mol^{-1}}$), respectively.

In other words, a mole of gaseous dimethyl ether reacting with a mole of gaseous protons releases $\mathrm{804\ kJ}$, while a mole of gaseous water releases only $\mathrm{697\ kJ}$. This means that in the gas phase, water is a weaker base than dimethyl ether, which implies that protonated water is a stronger acid than protonated dimethyl ether.

Why does the tendency invert in aqueous solution? Water can form hydrogen bonds, a very strong type of molecular interaction. $\ce{H_3O^{+}}$ has three protons available for hydrogen bond donation, whereas $\ce{(CH_3)_2OH^{+}}$ only has one. This means that the solvating water stabilises $\ce{H_3O^{+}}$ more than $\ce{(CH_3)_2OH^{+}}$. Higher stability means reduced acidity, and the difference happens to be enough that the relative order of acidity in water is the opposite of that in the gas phase.


What is meant by the numbers on reagents above and below arrows? I was taught that molecules on top of arrows are reactants and below arrow are solvent.

But I've come across a series of molecules written above and below arrows with numbers and commas. Do these numbers mean that all these reagents above and below arrows are in separate containers and are added one by one?

The numbers indicate steps. The only problem is that the definition of a "step" can be slightly different. Sometimes, separate steps mean that you do something, wait some time, and then do the next thing. Other times, you need to work-up or destroy the excess reagents that are left and isolate the desired product before you can proceed with the next step.

In your example, you need to keep the mercury and the borohydride separate. If you used any excess in the first step, you might need to destroy it before you add the borohydride.


How do I know crystal structures from formula? Give an explanation why $\ce{MgF2}$ and $\ce{CaF2}$ adopt different structure types.

For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae,  on how I determine this?

I'll just point out the direction in which you could look for an answer.

In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.

You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.

For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.

For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.


Calculation of equilibrium constant for dissolution of iron(II) fluoride in acid I am reviewing for the AP Chemistry test, and the question is this

$$
\ce{FeF2(s)  <=> Fe^2+(aq) + 2F^- (aq) } \quad  K_1 = 2\times 10^{-6}
$$

$$
\ce{F- (aq) + H+ (aq) <=> HF(aq)} \quad K_2 = 1 \times 10^3
$$
$$
 \ce{FeF2 (s) + 2 H+ <=> Fe^2+ (aq) + 2 HF (aq)} \quad K_3 = ?
$$

I solved for $K_3$ by multiplying $K_1$ and $K_2$, and then dividing by 2. I got $K_3 = 1 \times 10^{-3}$.  I would think that this would be not thermodynamically favorable since $K_3 < 1$, but the answer was the opposite: $K_3$ is thermodynamically favorable, because $K_3 > 1$.  Why is this the answer?

Let’s approach the entire problem from the other direction. Consider your final reaction:

$$\ce{FeF2(s) + 2H+ (aq) <=> Fe^2+ (aq) + 2 HF (aq)}\\
K_\mathrm{tot} = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2}$$

We have our initial dissociation with the following equation:

$$\ce{FeF2(s) <=> Fe^2+ (aq) + 2 F- (aq)}\\
K_1 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]}$$

And of course we have the second equilibrium which equates to:

$$\ce{F- (aq) + H+ (aq) <=> HF (aq)}\\
K_2 = \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]}$$

Now we want to express $K_\mathrm{tot}$ in terms of $K_1$ and $K_2$. Since we are dealing with products all the way through, it will somehow be a multiplication. So let’s try $K_1 \times K_2$:

$$K_1 \times K_2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} = \frac{[\ce{Fe^2+}][\ce{F-}][\ce{HF}]}{[\ce{FeF2}][\ce{H+}]}$$

We’re not quite there yet. We still have an extraneous fluoride concentration in the numerator and we are missing an $\ce{HF}$ concentration and a proton concentration. Luckily, that is exactly what an additional $K_2$ will give us:

$$K_1 \times \left( K_2 \right )^2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \left ( \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} \right )^2 = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2} = K_\mathrm{tot}$$

This is the mathematical way of saying ‘we need the second equation twice’. Due to the way equilibrium constants work, you always have to multiply or divide, so ‘doing something twice’ is equivalent to squaring.

Numerically, this means the answer is:

$$K_\mathrm{tot} = K_1 \times \left ( K_2 \right )^2 = 2 \times 10^{-6} \times \left ( 1 \times 10^3 \right )^2 = 2 \times 10^{-6} \times 1 \times 10^6 = 2$$

Yes, $\ce{FeF2}$ is a solid, so it has an activity of $1$ so we could ignore it when calculating the equilibrium constant. But we don’t actually use its value, so its being there doesn’t hurt either.


How does sulbutiamine work in the same way as thiamine, even if it doesn't have a thiazole ring? So sulbutiamine is said to be a thiamine substitute.

Yet - it doesn't contain the thiazole ring that's so important for many of thiamine's functions (highlighted in red).

How does it function as a thiamine substitute?

To get started, here are the structures of some compounds that are relevant.

Thiamine

Sulbutiamine

Allithiamine

Looking at these structures we see that you are of course correct, sulbutiamine does not contain a 5-membered thiazole ring like thiamine.

Thiamine (vitamin B1) was an early focus of interest because sailors at sea for long periods of time developed beriberi due to a nutritional thiamine deficiency.  Since those early days, thiamine substitiutes were of medicinal interest.  One of the early substitutes was allithiamine.  It is a natural product that can be extracted from garlic.

You can see the structural similarities between allithiamine and sulbutiamine.  Sulbutiamine has the advantage that both sides of the disulfide (S-S) bond contain the same active ingredient, so one would expect increased pharmacological properties from sulbutiamine.  The advantage that both allithiamine and sulbutiamine possess over thiamine is that they are more soluble in lipids (the positively charged thiazole ring in thiamine makes it very soluble in water, but not so soluble in lipids).  This enhanced solubility in lipids means that they can cross the blood-brain barrier much more effectively than thiamine, this improves their bioavalabilty.

With that background we can now answer your question, "how does sulbutiamine work in the same way as thiamine."

Once in the brain, sulbutiamine can be easily metabolized to thiamine.  Look at the structure of sulbutiamine and here is what happens:

reductively break the disulfide bond and form two identical mercaptans
$$\ce{RS-SR ->C[{[H]}]\   2 ~RS-H}$$
protonate the acetyl (that's on the nitrogen that will become part of our thiazole ring) oxygen to make it more attractive to nucleophilic substitution
Let the mercaptan sulfur nucleophilically attack the acetyl carbonyl carbon
eliminate water and form the 5-membered thiazole ring
and we now have thiamine

But a much larger dosage in the brain than had we simply ingested thiamine.


Electrostatic potential map and local ionization map What's the difference between electrostatic potential map and local ionization map?

I think that the main difference is that the electrostatic potential map doesn't take into account resonance (delocalization), but only electronegativity, therefore nucleophilicity or electrophilicity of a species is much better estimated by using the local ionization map. Am I wrong?

The excellent MultiWFN is able to produce both functions and luckily, a reference for it has explanations for both as well:

Electrostatic potential

According to the reference above,

  $$V_{tot}(r)
    = V_{nu}(r) + V_{ele}(r)
      = \sum_A \frac{Z_A}{|r - R_A|} - \int \frac{\rho(r')}{|r - r'|} dr'$$
  This function measures the electrostatic interaction between a unit point charge placed at $r$ and the system of interest.
  A positive (negative) value implies that current position is dominated
  by nuclear (electronic) charges.
  Molecular ESP has been widely used for prediction of nucleophilic and electrophilic sites for a long time.
  It is also valuable in studying hydrogen bonds, halogen bonds, molecular recognitions, and the inter-molecular interaction of aromatics.

Average local ionisation energy

Again, according to the reference above,

  $$\bar{I}(r)
    = \frac{\sum_i \rho_i(r) |\epsilon_i|}{\rho(r)}$$
  where $\rho_i(r)$ and $\epsilon_i$ are the electron density function and orbital energy of the $i$th molecular orbital, respectively.
  This function has many uses, for example, reproducing atomic shell structure, measuring electronegativity, quantifying local polarizability and hardness, and predicting sites for electrophilic or radical attack.

Thus, while the electrostatic potential represents the hypothetical electrostatic interaction energy between the molecule and a positive unit point charge at $r$, the average local ionisation energy is the orbital density-weighted absolute orbital energy average.

Finally, resonance/delocalization/electronegativity are built in the method you use to produce the electron density $\rho(r)$ (edit: a nice discussion about it can be found here).


Why are so many different organophosphates able to deactivate acetylcholinesterase? From sarin to tricresyl phosphate, there are organophosphates that are able to deactivate acetylcholinesterase.

What's so special about the simple organophosphate group that makes them able to do it?

Acetylcholine's normal function is to cleave acetylcholine's ester bond to convert it into acetic acid and choline, but certain compounds can interfere. The short answer is that these nerve agents all share similar structural components allowing them to be bound by acetylcholinesterase, where they then either react to alter the enzyme's binding site, preventing activity, or are just very slow to hydrolyze, temporarily inhibiting the enzyme until it can remove the compound.

Not all of these related acetylcholinesterase inhibitors work in exactly the same way, but if you examine their structures and compare them with acetylcholine, the normal substrate of the enzyme, you'll notice some similarities:

Acetylcholine:

Sarin:

Tricresyl phosphate:

You can see that they all have this central double-bonded oxygen (a few have sulfur instead, but are converted to oxygen metabolically), which is important for fitting into the enzyme. The ester bond holding the choline to the carbonyl in acetylcholine is relatively easily hydrolized, but phosphoesters are much slower to hydrolyze. From what I can tell, this seems to be one mechanism behind trecresyl phosphate's toxicity and probably why it's not as acutely toxic as something like sarin or VX. Sarin, on the other hand, has a pretty good fluoride leaving group. This lets it form a strong covalent bond with a serine's sidechain in ACE's binding pocket. This reaction is effectively irreversible, making sarin very dangerous as it totally disables any ACE it reacts with, meaning it takes very little to cause a significant effect.


Why doesn't baking soda react with the solid form of an acid? Baking soda (powder) will react when it comes into contact with a solution of citric acid crystals dissolved in a liquid medium such as water. But why doesn't baking soda react when mixed with a solid form of an acid (citric acid crystals)? Why is water needed for a reaction to occur?

Baking Soda (Solid) + Citric Acid Solution (Liquid) = Fizzy Reaction

Baking Soda (Solid) + Citric Acid Crystals (Solid) = No Reaction (??)

How do you know a reaction is not occurring when you mix the two solids? The only cue we have that a reaction occurs in the solution state is the formation of bubbles of escaping carbon dioxide. If there is no liquid to displace, would we be able to tell if carbon dioxide was being produced? Yes, but not with the naked eye.

  The real issue is that the solid state reaction is much, much slower.

To produce carbon dioxide, the bicarbonate ion needs to come in contact with an acidic proton on citric acid. This is unlikely in the solid state since there is very little kinetic energy.

In solution, there is more kinetic energy, so the molecules are more likely to encounter each other. Additionally, the proton transfer can propagate quickly through the hydrogen bond network in water.

$$\ce{HCO3- \bond{...} H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-Cit}$$


How to determine if an unknown is potassium chloride or potassium bromate? In class, we were given unknown substances. I've narrowed the possible substances down to either potassium chloride or potassium bromate.
We were given several acids, indicators, and other chemicals (silver nitrate, calcium hydroxide, ammonium hydroxide) to test with. How could one tell them apart? Maybe with a precipitate reaction?

Noting the solubilities of potassium chloride and potassium bromate gives a nice point in the identification but apart from it, there are certain chemical test specific for certain anion that gives pinpoint information about the salt.

First is the chromyl chloride test for chloride ions. Perform the test and you will identify potassium chloride (potassium bromate will give -ve result to the test). Second is conc. $\ce{H2SO4}$ test. Potassium chloride will gives vapors of hydrogen chloride. Potassium bromate will give bromine gas.

$$\ce{KCl + H2SO4 -> HCl ^ + KHSO4}$$
$$\ce{4KBrO3 + 2H2SO4 -> 2Br2 ^ + 2O2 ^ + 4K+ + 2SO4^{2-} + 2H2O}$$

Third test is the heat test. Potassium chloride is stable to heat while potassium bromate is prone to decomposition $\ce{\approx 370°C}$.

$$\ce{2KBrO3 ->[\Delta] 2KBr + 3O2 ^}$$

About the silver nitrate test pointed by @Dallon Penney, precipitation will occur in both cases albeit depending on the concentration of respective ions, it seems that the precipitate forming has different properties. Silver chloride exist as a white, curdy precipitate, insoluble in water and in dilute nitric acid while silver bromate exist as a white crystalline precipitate soluble in hot water and sparingly soluble in nitric acid. Both precipitates however are soluble in ammonia solution.


Mechanism for hydrolysis of enol ether to aldehyde I'm fairly confident in my answers for steps 1-3, but I am not certain how to have the methyl group bonded that is bonded to the alkene carbon leave in step 5 without forming methanol.

Here is what I have worked out so far.

Your issue is hardly with the Wittig reaction, although one could very well nitpick. The addition of phosphonium ylid to carbonyl is generally thought to be concerted: see Which is the currently accepted mechanism of a Wittig reaction? for more details.

From what I can tell, you are having more trouble with the acidic hydrolysis of an enol ether to an aldehyde. As has been pointed out by some comments, your proposed mechanism is not really possible. Vinyl $\mathrm{sp^2}$ cations are very unstable and an $\mathrm{S_N1}$-type dissociation of MeOH is very unlikely.1

Instead, you need to use the fact that an enol ether is very much like an enol. It is nucleophilic on the α-carbon, and you can protonate it on that carbon, almost like you are tautomerising an enol back to a ketone. You could also protonate it on oxygen, but that is an unproductive route: nothing really happens apart from the proton just falling off again.

The oxonium ion thus formed is electrophilic, and water can attack it leading to a hemiacetal.

From here on it is standard carbonyl chemistry.

I have deliberately left some stuff quite vague here.

Although it's acceptable shorthand to write $\ce{H+}$ in acidic media, you should know that naked $\ce{H+}$ doesn't exist in water. It's usually bonded to something else. Is it $\ce{HCl}$, or $\ce{H3O+}$?
The identity of the base is left unclear. Is it $\ce{Cl-}$, or $\ce{H2O}$? Hints: Which one of them is a stronger base? Which one is available in greater quantities?

Notes

(1) Just for interest, $\mathrm{S_N1}$-type chemistry leading to vinyl cations has been done before, but with much better leaving groups than MeOH. See: Okuyama, T.; Takino, T.; Sueda, T.; Ochiai, M. J. Am. Chem. Soc. 1995, 117 (12), 3360–3367. DOI: 10.1021/ja00117a006.


What causes photovoltaic (solar) cells to degrade? Photovoltaic panels can have 20 or 25 year underwritten warranties with a guaranteed remaining efficiency of 80 % of the new panel. That means, that photovoltaic panels seem to degrade somehow.

Why do they degrade? What exactly is reducing their efficiency?
How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?

Why do they degrade? What exactly is reducing their efficiency?

This link outlines several modes of solar panel degradation, and this report by the National Renewable Energy lab is a very detailed review of studies on solar panel degradation worldwide.

To summarize:

Internal resistance of the cell can increase due to infiltration of contaminants (usually water vapor) when the encapsulating material cracks due to long-term UV exposure or temperature cycling.
The shunt resistance (resistance of the path to ground) can decrease when metal ions migrate (diffuse) through the cell. This happens naturally, with the rate depending on temperature and the type of cell. In general, higher temperatures lead to faster diffusion, and amorphous silicon has higher diffusion rates.
The anti-reflective coating can deteriorate. Anti-reflective coatings can be made of lots of different types of materials, but in general, heat, UV exposure, and exposure to contaminants are what cause them to degrade.

These are all degradation mechanisms of the cells themselves - any of the other components of the panel can also degrade. In all cases, the mechanisms are related to thermal stress or mechanical damage due to things like wind or hail, or corrosion from humidity.

Given all of the above, the main causes of solar panel degradation that can be (sort of) controlled by you are:

Temperature
UV exposure
Mechanical damage

  How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?

Anything you can do to reduce these three factors would help. However, covering a solar panel with an opaque tarp would only reduce UV exposure. The temperature (and cyclical thermal stress) would likely increase.

Unfortunately, both 1 and 2 increase with exposure to sunlight - and exposure to sunlight is the whole point of solar panels. Mechanical damage can be reduced by installing panels in places and orientations that make it unlikely that they will be stressed by winds, or have things hitting them very often.

However, it looks like the only effective option would be to completely enclose the panels in a temperature-controlled low humidity environment when they are not in use. Whether or not the benefit of doing so is worth the cost will depend on the degradation rate and cost of the particular panels.


E2 Reactions in Cyclic Structures In my textbook, the percentages of anti-eliminations are given which are quite contrary to expectations. The reason is also not given quite clearly. Why does the preference for anti-periplanarity change and what are the reasons for the disparity between the different cyclic structures?

\begin{array}{lr}
\text{Ring Size} & \%\text{ SYN Elimination} \\
\hline
\text{Cyclobutyl} & 90 \\
\text{Cyclopentyl} & 46 \\
\text{Cyclohexyl} & 4 \\
\text{Cycloheptyl} & 37
\end{array}

The values are for $\ce{(CH2)_nC-HY}$.

To begin with, I want to present you approximate representations of the four cycloalkenes in question. Unfortunately though, these representations cannot, in this case, be used to fully replace building the molecules; I suggest you do so with the molecular modelling kit of your choice.

I’ll start off with the six-membered ring. This — as is often correctly pointed out — is essentially strainless and the chair conformation displayed is the most stable one. While frequent interconversion between different chair forms can occur, switching axial and equatorial substituents. We can thus expect that a reaction will, in principle, always take place in this chair form.

If we examine the dihedral angles, we notice that an axial substituent is anti-periplanar to the adjacent two carbons’ hydrogens. This means, it can easily anti-eliminate in both directions. No substituent whether equatorial or axial has any ecliptic hydrogen in either direction. This means, that a syn-specific elimination, which requires an ecliptic configuration, cannot occur out of the chair conformation. The $4~\%$ syn-elimination the table references must occur from other, minor configurations such as the boat.

Cyclopentane is also hardly strained. However, the configuration of the ring is much more volatile. Furthermore, even in the shown configuration there are possibilities of syn-eliminations; for example see the two carbons opposite to the upwards-pointing one. Still, anti-eliminations are also well possible, especially from the other three carbons that form the upwards arch, and account for most of the elimination.

A similar case can be made for cycloheptane, which is even more strained. There is not much for me to say here that I haven’t said for cyclopentane other than that you should build it.

And finally cyclobutane. This is the most strained of the systems considered and probably the one with the most fluctuating configuration. Here, it looks like an anti-elimination is possible; but this requires the substituent to occupy an axial position. This configuration is strongly sterically hindered. Syn-eliminations can occur from any position by only moderate shifts from the geometry shown. This may explain why anti-eliminations are strongly disfavoured here. I would, however, once again stress how helpful it is (especially in this case) to build the model.


What does this sentence mean? I was reading the book: " Nanotechnology: A Gentle Introduction to the Next Big Idea" and I came across this sentence:

  For these molecules to perform useful functions, they must be easy to
  assemble and easy to recognize and bind to by other molecules.

Who can explain this sentence? I have problem with "easy to assemble and easy to recognize" part?

It was in the paragraph below:-

  Generally, the molecules found in nature are complex and the source of
  much dismay to beginning organic chemistry students. For these
  molecules to perform useful functions, they must be easy to assemble
  and easy to recognize and bind to by other molecules. They must also
  be made by biological processes and have variable properties.

I think the sentence means that the molecule the nanotechnician is using has to be quite easy for other molecules to interact with, either by chemically bonding to it or acting like a catalyst (or otherwise).


nickel-copper fabric: shielding viability after being subjected to weathering I know the solution can be as simple as a well-sealed garbage can, but I am contemplating making a more malleable fabric based faraday-cage-esque contraption. The plan is to use nickle-copper fabric to achieve the shielding, I'll experiment with how many layers/folds create the best seal. It is to be placed outside, exposed to the elements, potentially incurring moderate to heavy rain. Before I get too heavily invested in this, I would like to really think about the shielding viability of the fabric in the long-term.

For reference, here is a generic picture of the nickel-copper fabric. Amazon has its fair share too, for those interested.

Question:

Unlike its ferrous counterparts, nickel/copper do not rust but rather have a patina or tarnish that develops after weathering. Will nickel-copper fabric tarnish like we might expect a penny to tarnish? If so, what compromises to shielding would be sustained? Also, if so, would that compromise the structural integrity of the fabric (more prone to tears, ect)?

Further Clarifications:
 - exact chemical makeup of the fabric is uncertain, the one I bought was simply marketed as "nickel-copper fabric" so I assume its mostly nickel and copper, but it seems to have a fabric-like feel, so perhaps its mixed with some kind of substrate or something. If the consensus is that this information is crucial I'll try to track down the manufacturer and get specifics, but at the moment I don't have a practical means of going about that(manufacturer is unknown). You are free to speculate about common implementations of nickel-copper fabric and state your assumptions accordingly

shielding: I'm concerned that if the fabric weathers poorly, the shielding will be affected
tear resistance: I'm concerned that if the fabric weathers poorly there may be more points of failure, leading to easier tears in the fabric
health: the fabric will occasionally come in to close contact with humans (me) and I'm not sure if tarnished nickel-copper has any adverse health risks
set up: my design is too hypothetical to put into words at the moment, but you can imagine a tarp-like thing, outside layer is cordura (water-resistant, not water-proof) and inside layer is nickel-copper fabric
placement: outside, exposed to the elements, incurring rain

Your alloy should be sufficient. For one, cupronickel alloys are very corrosion resistant and is used to make many different coins across the world. occasionally yes, a nickel or dime will tarnish but this is usually from exposure to soda, salt or some other corrosive liquid, other wise they can sit in a fountain for weeks without tarnishing. AS long as the acid rain isn't too severe, you should be okay.

As for the product you are considering, the description on amazon describes it as:

  Copper+Nickel+Polyester Plated Fabric

Now plated metals do not necessarily have the same properties as the pure casted alloy since to plate on polymer requires using electroless plating and does not guarantee the same microstructure as a casted alloy, but corrosion should be unaffected.

See also: Monel


Is there a readily available liquid that's more dense than water and insoluble? I am looking for any liquid that does not dissolve in water with higher density and does not react with water, one that is readily available.

Trichloroethylene is a common chemical solvent that has reasonably low toxicity, it has density ~1.5 g/ml. However, it slowly becomes restricted. Tetrachloroethylene also may be considered. Dichlormethan has reasonably low toxicity, but its BP is too low. I would not use the liquids mentioned without fume hood on regular basis. One-time exposure is probably OK, but repeating exposure may lead to cancer.  Chloroform may be considered, but I recommend to avoid it, it has accumulating liver toxicity. Tetrachloromethane should be avoided, it also has undesired toxicity.

Mercury MAY be considered, it is toxic if one is repeatedly exposed to vapours, but has very low vapour pressure and is reasonably cheap. But it has annoying tendency to produce small drops that are very hard to collect and move into smallest holes and accumulates in the system, so it clearly requires dedicated  workplace.

Liquids mentioned above are reasonably cheap, so if they are not restricted in your home country, you should be able to allow yourself one. Unfortunately, that's all ideas I got. Well, there are some more, but they either has unreasonable cost, or are too toxic. Bromine, for example, is toxic and, though not mixing with water, partially dissolves in it.


Nomenclature and chemical formulae for bridged dinuclear chromium complexes In theory, $\ce{[Cr(H2O)6]^3+}$ ions have a bluish violet colour. However, many times in the lab we end up with a dark green solution. This is due to hydrolysis, coordination with sulphate (or other) ions, and polymerisation, most often dimerisation.

Hydrolysis$$\ce{[Cr(H2O)6]^3+(aq) +H2O(l) <=> [Cr(OH)(H2O)5]^2+(aq) + H3O+(aq)}$$

Coordination and dimerisation

The degree of coordination depends on conditions, e.g., temperature. It is common to find the following equilibria.

$$\ce{2[Cr(OH)(H2O)5)]^2+(aq) + $\mathrm{SO}_4^{2-}$(aq) <=> Picture1(aq) + 4H2O(l)}$$

$\ce{Picture1}$:

$$\ce{2[Cr(OH)(H2O)5)]^2+(aq) + 3\ $\mathrm{SO}_4^{2-}$(aq) <=> Picture2(aq) + 6H2O(l)}$$

$\ce{Picture2}$:

Question

How can one write these ions in condensed structural formulae? I wish not to lose any information.
How can one unambiguously name these two ions?

I am looking for official IUPAC names and formulae using latest recommendations for complexes. For instance, $\ce{[Pt(NH3)4][Pt(Cl)4]}$ is named tetraammineplatinum(II) tetrachloroplatinate(II); $\ce{[Cr(OH)(H2O)5]^2+}$ is pentaaquahydroxochromium(III) ion.

Thank you in advance!

P.S. Pictures were drawn using the free online software eMolecules. Also, feel free to correct any formatting and/or grammar errors. As for chemistry-related mistakes, please comment rather than edit.

$\require{unicode}$The rules for formulating and naming coordination compounds are presented in Chapter IR-9 of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.

The line formula for a simple coordination entity is constructed within square brackets. The symbol for the central atom is placed first and is then followed by the symbols or abbreviations for the ligands (in alphabetical order according to the way they are presented in the formula).

  IR-9.2.3.1 Sequence of symbols within the coordination formula

  (i) The central atom symbol(s) is (are) listed first.

  (ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in alphabetical order (…). Thus, $\ce{CH3CN}$, $\ce{MeCN}$ and $\ce{NCMe}$ would be ordered under C, M and N respectively, and $\ce{CO}$ precedes $\ce{Cl}$ because single letter symbols precede two letter symbols. The placement of the ligand in the list does not depend on the charge of the ligand.

  (iii) More information is conveyed by formulae that show ligands with the donor atom nearest the central atom; this procedure is recommended wherever possible, even for coordinated water.

Note that according to the recommendations in Item (iii), where possible, the coordinating (ligating) atom should be placed nearer the central atom in order to provide more information about the structure of the complex. Therefore, the line formula of the hexaaquachromium(3+) complex ion should be written as $\ce{[Cr(OH2)6]^3+}$ rather than $\ce{[Cr(H2O)6]^3+}$.

For the same reason, if possible, bridging ligands should be placed between the central atom symbols. Bridging ligands are those bound to more than one central atom. They are differentiated in names by the addition of the prefix ‘μ’.

  IR-9.2.5.2 Bridging ligands

  Bridging ligands, as far as they can be specified, are indicated by the Greek letter μ appearing before the ligand symbol or name and separated from it by a hyphen; (…)

  (…) In formulae, bridging ligands are placed after terminal ligands of the same kind. (…)

  The bridging index $n$, the number of coordination centres connected by a bridging ligand, is placed as a right subscript. The bridging index $2$ is not normally indicated. (…)

Polynuclear complexes may be named differently according to whether only stoichiometry is to be specified or partial or complete structural information is to be included.

  IR-9.2.3 Formulae of coordination compounds

  (…) Different applications may require flexibility in the writing of formulae. Thus, on occasion it may be desirable to violate the (…) guidelines in order to provide more information about the structure of the compound that the formula represents. In particular, this is the case for dinuclear compounds where a great deal of structural information can be provided by relaxing the ordering principles outlined in Section IR-9.2.3.1. (…)

  IR-9.2.5.1 General

  (…)

  As a general principle, as much structural information as possible should be presented when writing the formula or name of a polynuclear complex. However, polynuclear complexes may have structures so large and extended as to make a rational structure-based nomenclature impractical. Furthermore, their structures may be undefined or not suitably elucidated. In such cases, the principal function of the name or formula is to convey the stoichiometric proportions of the various moieties present.

  (…)

  Note, however, that the rules for formula writing may be relaxed in various ways in order better to display particular features of the structures in question. (…)

As a first approximation, the systematic line formula for the complex ion shown in Picture 1 is $\ce{[(H2O)3Cr($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)Cr(OH2)3]^2+}$. Note: pay attention to the Greek letter; it is to be written upright.

However, since this complex ion is symmetric, its formula may be further simplified.

  IR-9.2.5.4 Symmetrical dinuclear entities

  For symmetrical dinuclear entities, the name may be simplified by employing multiplicative
  prefixes.

  (…)

The resulting formula is: $\ce{[\{Cr(OH2)3\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2+}$. The corresponding name is di-μ-hydroxido-μ-sulfato-bis(triaquachromium)(2+).

The corresponding formula for the complex ion shown in Picture 2 is $\ce{[\{Cr(OH2)2(SO4)\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2-}$. The corresponding name is
di-μ-hydroxido-μ-sulfato-bis(diaquasulfatochromate)(2−).

Furthermore, if necessary, you can indicate the connectivity of the sulfato ligand using the kappa convention (sulfato-2κO).


What is a triglyceride? I'm confused on what a triglyceride is, from what my text book it says its a type of gylercide, then from a website it said

  Glycerides can be subdivided into two categories. The first group, the neutral glycerides are nonionic and nonpolar. The second group, the phosphoglycerides contain a polar region, the phosphoryl group.

Which makes triglyceride not a directly under a glyceride but under one of these groups. Then it says

  Esterification may occur at one, two or all three positions producing monoglycerides (monoacylglycerols), diglycerides (diacylglycerols), or triglycerides (triacylglycerols). The most prevalent and most important are the triglycerides.

So does that mean that triglyceride can be produced by either of the two groups phosphoglycerides and neutral glycerides. And if that's correct then how are monoglycerides and diglycerides produced.

Source

There is some minor argument as far as nomenclature goes for these lipids. Most sources I can find, including IUPAC have glycerides as only including esters of glycerol with fatty acids. Monoglycerides have a single fatty acid ester with two free OH groups, diglycerides have two fatty acid esters, and triglycerides have three. Phosphoglycerides are defined separately as esterified derivatives of phosphatidic acids, so you can think of them like glycerides where for one of the substituents, instead of a fatty acid, some derivative of phosphoric acid is present. However, the way the two terms are defined, they don't encompass any common molecules, i.e. there are no molecules that are both glycerides and phosphoglicerides and neither group is a subset of the other.

So: mono-, di-, and triglycerides are subsets of glycerides while phosphoglycerides are a different group altogether, though structurally similar.


Why is europium used in Euro bank notes? Europium is apparently used as an anti-counterfeiting measure in Euro bank notes because of its fluorescence under UV light.

Is there any reason why it is specifically $\ce{Eu}$ and not any of the other lanthanides that fluoresce as well?

Is it because $\ce{Eu}$ is relatively rare and expensive so it's harder to fake? Is it because it has a very characteristic fluorescence? Is it because they just thought it's funny to use $\ce{Eu}$ in Euros?

As already noted by the OP, many rare earths fluoresce under ultraviolet light and there seems to be no particular reason why Europium is a better choice than any other element, purely based on its fluorescent properties. (See update at bottom)

The European Central Bank keeps the exact nature of the compounds used a secret. However, it was discovered in 2002 by Frank Suijver and Andries Meijerink [1] that europium, complexed with two $\beta$ diketone molecules provides the red light but the other colours were more difficult to identify.

However, the reason for the choice of Europium is unclear. According to Hugh Aldersey-Williams[2]:

  It [the European Central Bank] wilfully misunderstands my request to know who fought for Europium and drearily asks for my 'understanding that, for security reasons, we cannot comment on the chemical components of the euro banknote security features.'

Apparently, in the original paper by Suijver and Meijerink they say that they were visited several years beforehand by European Bank officials who were investigating potential luminescent materials so it may be that they inadvertently gave the bank the idea for using Europium in its banknotes.

Europium is one of the more expensive lanthanides and so this is probably a factor in its usage to prevent against forgeries, as the OP suggests.

I suspect there is also simply an element of irony in the fact that Europium is used in Euro bank notes and it may be that it was just some bank official tasked with finding a suitable compound to use who thought that it would be funny to use Europium.

Anyway, the bottom line appears to be that nobody knows the reason because the European Central Bank refuses to tell anyone anything about the compounds they use.

UPDATE: In response to @Nicolau Saker Neto's comment I did some more research into the fluorescence of Europium. Europium(III) (red) is regularly used alongside Europium(II) (blue) and Terbium(III) (green) in trichromatic lighting systems. It may therefore be the case that Europium was chosen because it is already well studied and in common use and that they created a $\beta$ diketone complex in order to fine tune the exact wavelength of the emissions and make it unique to their banknotes.

References:

[1] http://dspace.library.uu.nl/handle/1874/25831 (Paper in Dutch which I can't read. However, Williams refers to some of the content in his book and I have alluded to this above. If anyone can translate the original paper and add anything to the answer that would be very helpful.)

[2]Hugh Aldersey-Williams; 'Periodic Tales: The Curious Lives of the Elements'; ISBN 978-0-141-04145-2 http://www.amazon.co.uk/Periodic-Tales-Curious-Lives-Elements/dp/0141041455


How to calculate the equivalent weight of a metal in a metal sulphate? The question is:

$\pu{2g}$ of a metal is present in$\pu{4.6g}$ of its metal sulphate. Find the equivalent weight of the metal.

I understand the concept that equivalent weight of an element is the molecular weight divided by the n-factor (in this case it's the valency), but I couldn't arrive at a conclusion.
Could you also help me understand how equivalent weight of a metal is related to the equivalent weight of the sulphate? In my textbook, it's written like this:
$$\textrm{eq. wt. of metal} = \textrm{eq. wt. of metal sulphate(2-)}$$
and then they carried on with the calculations.
PS I know the answer, I need to know how one should conceptually arrive at it!

I assume that if you are dealing in equivalent weights then you must be well associated with the concepts of equivalents.
Now,
$$ \textrm{equivalent} = \textrm{weight} / \textrm{eq. weight}$$
and since compounds combine in equal amounts of equivalents we have
$$\frac{\textrm{weight of metal}}{ \textrm{eq weight}} = \frac{\textrm{weight of sulphate}}{ \textrm{its eq weight}}$$
$$\rightarrow 2/x = 2.6/48$$
Hence $x= 39.92$.


How to calculate the detection limit? I am having trouble figuring out how to calculate the detection limit. Here is the question:

You and your friend working in a different lab have been given an unknown water sample that contains potassium.  You are both asked to measure the potassium concentration in the solution (10 replicate measurements).  The methods that you and your friend are using are different.  The results that were obtained are listed below.  Determine:

(a) the concentration, standard deviation and relative standard deviation for the unknown as measured using the two methods (check for outliers!),

(b) calculate the detection limit (3sigma) for each method,

(c) compare the standard deviations and evaluate whether the two averages are significantly different (or not) at the 95% confidence level.

RESULTS:

Standard    Method 1    Method 2
Concentration   Intensity   Intensity
(mg/L)  (nA)    (mV)
0.000   0.624        1.955
0.000   0.488        2.490
0.000   0.522        2.166
0.000   0.355        1.500
5.000   9.245       15.644
10.000  17.069      31.220
15.000  26.200      44.266
20.000  33.881      62.394
25.000  43.826      75.611

Replicate
1       26.544      46.977
2       25.449      49.120
3       21.053      50.998
4       24.353      46.615
5       23.899      49.326
6       24.010      46.666
7       25.554      45.291
8       23.549      42.995
9       26.008      43.678
10      24.404      49.012

So my problem is with part (b). I know the detection limit is calculated with the formula DL = 3sigma/slope and that I can find the slope by calculating Sxy/Sxx but I'm not sure what points I'm supposed to use? What points do I use for the standard deviation (sigma) and for Sxy/Sxx?

I've tried looking at examples in the textbook but nothing was really similar. I also tried searching online, but I couldn't find anything that helped me.

Have a look at this: http://www.chem.utoronto.ca/coursenotes/analsci/stats/LimDetect.html

It is possible to calculate the limit of detection from the standard error of the regression:

$$C_{\mathrm{LOD}} = \frac{3s_{y/x}}{b}$$ These values are calculated from the regression of all the data points, not the standard deviation of any subset of data.

However, in this case we have replicate blank measurements (four measurements at C=0), so we can also calculate the limit of detection as simply 3 times the standard deviation of the blank measurements:

$$y_{\mathrm{LOD}} = y_{\mathrm{blank}}+3s_{\mathrm{blank}}$$

(in instrument response units—convert to concentration from a known standard or from the regression equation)


What does the ZZE tag in an Chemical Table file refer to? I have a number of Chemical Table files which include a "M  ZZE" tag in the properties block.

For example, in Galactobuxin, I have:
M  ZZE  1  31  23

In Rubrobrassicin there is:
M  ZZE  3  28  23  36  23  48  23

In R-hexanoyl CoA there is:
M  ZZE  1  55  23

The typical format for such tags is that it starts with an M, followed by two spaces. Then there is a three-letter code, in this case ZZE. Other codes include ISO (for isotope) or CHG (for charged atoms in the compound), then a number which indicates how many records on that line. This is followed by the actual data.
So
M  ZZE  3  28  23  36  23  48  23
would probably mean that there are three sets of ZZE data, being "28 23" "36 23" and "48 23". In every record I have the second number is 23.

The ZZE tags seem to occur only for compounds containing ribose rings, and the first number is the index of one of the bonds inside the ribose ring. Specifically, of three three bonds adjacent bonds with a stereochemistry marker, it always refers to the one in the middle:

I'm guessing the 23 is some property of the bond: so in rubrobrassicin the 28th, 36th and 48th bonds (there's one, and only one, ZZE record for each of the ribose rings in rubrobrassicin).

Does anyone know what the ZZE property in a chemical table file refers to, or what it indicates? I can't find any information on the format.

The current version of the CTFile spec can be downloaded for free from Accelrys. Based on the files you have, you'll want to look at the "V2000" section.

Short answer: I believe [M  ZZE] is a vendor-specific property key, possibly introduced by ACD Labs, given it's the package cited in the header on line 2. The CTFile format spec for V2000 files says nothing about an "M  ZZE" property key.

The CTFile spec says that property keys that are not recognized by parsing software are to be ignored. This makes vendor-specific extensions feasible.

You seem to be on the right track in your interpretation. The Galactobuxin entry has this line in the properties block:

  M  ZZE  1  31  23

Breaking this down:

{M  ZZE} is the mystery properties key.

{1} is the number of data entries on the line.

{31} is the 1-based index of either an atom or bond.

Atom 31 is one of the rather boring methoxy groups.

On the other hand, bond 31 is a real problem child:

This file is abusing the meaning of 'wedge bonds' by trying to use them to indicate chair conformation perspective, rather that what they're intended for - flat stereoprojection. My guess is that the file is trying to annotate this bond.

The only question left is - what does {23} refer to?

Bond 23 is unremarkable. Atom 23 is one of the origins of a wonky wedge bond. But it's not clear why it should be associated with bond 31.

My best guess is that the software that created this file knew it was abusing wedge bonds and wanted to include a flag on bond 31 so that software reading the file wouldn't misinterpret.

Given that you're seeing the same {23} in every file, this could be a code to specially process bond 31's stereo flag.


What is the mechanism of AChE inhibition by Onchidal? A naturally-occuring neurotoxin, called Onchidal, produced by a species of sea slug acts as an irreversible inhibitor of acetylcholinesterase. The structure of Onchidal is presented below:

How and why does Onchidal act as an AChE inhibitor? What does the adduct of Onchidal with AChE look like? Why is the inhibition of AChE by Onchidal irreversible?

Q.1: How and why does Onchidal act as an AChE inhibitor?
Onchidal is an antipredatory, feeding deterrent product secreted by marine organisms such as Onchidella binneyi and Elysia cripata, not having the protection of an external shell and thus relying on the production of a defensive secretion. [Gavignin et al., 2000, 1996, 1997; Ireland et al. 1978] To mammals, it is a neurotoxin that acts as an irreversible inhibitor of AChE, with a Kd of 300 μM, where it forms a covalent species with the enzyme.[Abramson, 1989]
Q.2: What does the adduct of Onchidal with AChE look like?
This molecule was isolated from the mucus obtained by squeezing the mollusc O.binneyi. [Ireland et al. 1978] It is a sesquiterpene acetate containing an α/β-unsaturated aldehyde and an acetate ester as shown above. It has also been speculated that onchidal’s active form is believed to be ancistrodial.

Abramson (1989) proposed multiple modes of inhibition:
(i) Schiff base formation with a lysine;
(ii) attack on the beta-unsaturated carbon in conjugation with the aldehydes by a Cys, Lys, His, or Tyr; and
(iii) the formation of 1,4-dialdehyde, which reacts with a Lys to form a pyrrole covalent adduct.
A look at the AChE co-crystal structure with an inhibitor (paraoxon in this case to locate the active site) as deposited with RCSB as 5HF5.pdb (DOI: 10.2210/pdb5HF5/pdb) , indicates that the closest reactive ε-amino group of a Lys residue is ~18  Å away. This distance is presumed too much for a covalent attachment of onchidal to AChE active site and thereby affecting AChE’s catalytic function. Likewise, there is no free sulhydryl available in the active site either, so a different residue may be involved as the point of attachment. The following figure below shows AChE with paraoxon shown with speres; Lys in green; and Cys in orange. The distance is in Å on the following figure I prepared:

Stoddard (2014) proposed three docking poses for onchidal. Here is the figure S2 from their paper:

Then, Stoddard (2014) further noted:

Only one His, the active site His-440, is located in the gorge. Several Tyr are located in the gorge; Tyr 442, 334, 130, 121, and 70. Based on the crystal structure and Abramson’s findings, the most probable mechanism is attack of the beta-unsaturated carbon in conjugation with the aldehyde by a Tyr; however a candidate Tyr for this mechanism was not identified from our docking analysis.

I’m wondering whether non-specific cellular esterases may convert Onchidal to an alcohol? If true, my speculation may change the docking analyses and ensuing inference. In any case, I’m not sure we have the full picture of Onchidal molecular mode of action, just yet.
Q.3: Why is the inhibition of AChE by Onchidal irreversible?
Onchidal contains an aldehyde and vinyls. Such reactive functionalities have literature precedents describing irreversible adduct formation with specific protein residues.

References
Abramson S.N., Radic Z., Manker D., Faulkner D.J., Taylor P. Onchidal (1989) “A naturally occurring irreversible inhibitor of acetylcholinesterase with a novel mechanism of action.” Mol. Pharmacol. 36:349–354.
Gavagnin M., Mollo E., Castelluccio F., Montanaro D., Ortea J., Cimino G. (1997) “A novel dietary sesquiterpene from the marine sacoglossan Tridachia crispate. Nat. Prod. Lett. 10:151–156. doi: https://10.1080/10575639708043731
Gavagnin M., Mollo E., Cimino G., Ortea J. (1996) “A new γ-dihydropyrone-propionate from the Caribbean sascoglossan Tridachia crispate” Tetrahedron Lett. 37:4259–4262. doi: 10.1016/0040-4039(96)00811-8
Gavagnin M., Mollo E., Montanaro D., Ortea J., Cimino G. (2000) “Chemical studies of Caribbean sacoglossans: Dietary relationships with green algae and ecological implications.” J. Chem. Ecol. 26:1563–1578. doi: 10.1023/A:1005526526884
Ireland C., Faulkner D.J. (1978) “The defensive secretion of the opisthobranch mollusc Onchidella binneyi. Bioorg. Chem. 7:125–131. doi: 10.1016/0045-2068(78)90043-3
Stoddard, S.V., Hamann, M.T., and Wadkins, R.M. (2014) “Insights and Ideas Garnered from Marine Metabolites for Development of Dual-Function Acetylcholinesterase and Amyloid-β Aggregation Inhibitors.”Mar Drugs 12:2114-2131. doi: 10.3390/md12042114


H NMR of 4-chlorochalcone In an aldol condensation reaction, I synthesized a molecule out of two unknown reactants. These reactants I found to be 4-chlorobenzaldehyde and acetophenone which made 4-chlorochalcone. I was given an H NMR spectrum to identify and label the hydrogens for the molecule but am having trouble reading it!

There is a quintuplet with an integration of 6 but no where in the molecule am I seeing 6 hydrogens that share a chemical environment (I counted 7 environments). Is this possible or could I have come up with the wrong product structure?

Any help deciphering this would be great!

As with all NMR analysis problems like this, it is good to have a strategy to interpret data. Perhaps rather than looking at your spectrum and trying to rationalise your structure, look at your structure and predict what you might expect to see in your spectrum. You'll find it much easier to rationalise the appearance of your peaks, especially in the case of overlapping peaks as you have here.
So, in your molecule, what splitting would you expect?

You have 11 proton environments, so your integrals look fine. Your chloroform doublet is  a bit of a worry though. Perhaps your sample needs to be mixed a little better? You've got 3 distinct spin systems that will give characteristic splittings:

trans double bond. These two peaks will appear as distinct doublets, with couplings of around 12-17Hz. I can see one very clearly. The other, which we know to be upfield of the one at 7.77 because of the tenting is probably under the mess of stuff between 7.5-7.6. Being next to the carbonyl will have strong influence on the chemical shifts of these, and through a bit of electron-arrow pushing we can show that the proton closest to the carbonyl is shielded, and the other deshielded. Hence, the signal at 7.77 is likely to be your proton β to the carbonyl group (and closest to the para-substituted benzene ring)
para-substituted benzene ring. Again a distinct splitting pattern; these will look like a pair of doublets (actually a second order pattern) with coupling about 7.5Hz. I can see one probably at 7.4. Its partner will also be under the mess between 7.5 and 7.6
a mono-substituted benzene ring, which will have three sets of signal, with ratio 2:2:1. A carbonyl group is strongly electron withdrawing, deshielding the ortho protons, which is probably the signal up at about 8. The others.... under the mess between 7.5-7.6.

This is clearly a homework/lab problem, so this should be enough to keep you going until the next question.