# 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

• It is possible to calculate reasonable curves, but I'm too lazy to sort all this out. I'll ask one question however. Did you account for the volume change? 100 ml of 0.1 molar HCl netralizes 100 ml of 0.1 molar NaOH, but the result is 200 ml of 0.05 NaCl. – MaxW Jul 13 at 16:48
• Yes I believe I have hence the $V_{a}+V_{b}$ term. It can also be seen in some of the other graph I produced using this model, one such example is in a strong acid-strong base titration, the conductivity curve plateaus when large volumes of base are added, which is a result of volume increase – sab hoque Jul 13 at 23:50
• Ok, I'll nibble on this some more what are the acid and base pK values? Looks like the acid has a pKa of about 3.8. The top of the titration curve is pretty flat so I'm guessing a pKb of 2 or less. So to me the titration curve looks like a weak acid being titrated with a relatively strong base. – MaxW Jul 14 at 0:18
• All the values are at the left of the image of the graph, but you can change them ass appropriate and it's not relevant to what I'm asking. I would like for the values of ka and kb to be variable rather than set values and i don't see how setting them to a certain value helps. – sab hoque Jul 14 at 5:50
• If $Q_\ce{sp}=\ce{[HB+][A-]}$ through titration don't exceed $K_\ce{sp}$ value, don't will be need incorporate the salt formation equilibrium and the curve don't be affected. – GRSousaJr Jul 28 at 16:54

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

1. ionization of weak acid: $$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}} \tag{1} \label{eq:KWeakAcid}$$

2. ionization of weak base: $$\ce{B + H2O <=> BH+ + OH-}\qquad K_\ce{b}=\frac{\ce{[HB+][OH-]}}{\ce{[B]}} \tag{2} \label{eq:KWeakBase}$$

3. self-ionization of water: $$\ce{2 H2O <=> H3O + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]} \tag{3} \label{eq:KWater}$$

4. 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:

1. 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}

1. 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:

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

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 $$Q_\ce{sp} we have $$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

• What software did you use to plot the graphs? It looks really clean. – Blaise Aug 1 at 16:03
• @Blaise I used gnuplot. – GRSousaJr Aug 1 at 16:08
• @sabhoque When $Q_\ce{sp} > K_\ce{sp}$ the chemical intuition say that there will be precipitation, and the amount of precipitate will be calculated by (15) equation. I'm still working on understanding of this equation and its consequeces on titration and conductivity curves. How soon as possible I'll complete the answer with some conclusions. – GRSousaJr Aug 2 at 14:32
• With regard to hypothesis 1, I believe you are correct in saying the pH value can be calculated from (10) equation and the titration curve don't be affected because the precipitation reaction does not change the concentration of $\ce{H+}$ (as I am aware of). It is mainly the conductivity curve I am struggling to get a grasp of. I will test out your derivations and everything after my upcoming exams – sab hoque Aug 3 at 7:19
• @sabhoque I'm refer to hypothesis #3. I made some edits on answer to clarify. Additionally, I included some graphs on hypothesis #1. – GRSousaJr Aug 3 at 17:33