How a titration curve is affected when a poorly soluble salt is formed?

Question

Suppose that I have $\pu{100 mL}$ solution of strong acid, $\ce{HA}$, at $\pu{0.10 mol L^{-1}}$. This solution is titrated with a strong base solution, $\ce{BOH}$, at $\pu{0.10 mol L^{-1}}$. Suppose yet that a poorly/sparingly soluble salt, $\ce{BA}$ $(K_\ce{sp}=\pu{2.30×10^{-3}})$, is formed during titration.

The reactions involved are:

1. ionization of strong acid: $$\ce{HA + H2O -> H3O+ + A-}$$

2. dissociation of strong base: $$\ce{BOH -> B+ + OH-}$$

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

4. Equilibrium of poorly soluble salt: $$\ce{BA <=> B+ + A-}\qquad K_{\ce{sp}}=\ce{[B+][A-]}$$

Charge balance of the solution is given by the equation: $$\ce{[H3O+] + [B+] = [OH-] + [A-]}$$

Once that $\ce{B+}$ and $\ce{A-}$ are spectator ions, are come from strong substances, and disregarding the ionic strength effect, I think that titration curve don't be affected. I'm right?

If not, how to write a mass balance expression to take account into the precipitation of $\ce{BA}$?

Answer 1

How a titration curve is affected when a poorly soluble salt is formed?

If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction.

how to write a mass balance expression to take account into the precipitation of BA?

In general terms, the mass balance (really the amount balance) can be expressed in concentrations if the system is homogeneous (all in solution). For example, it could look like this:

$$[E] + [F] + [G] = \mathrm{constant}$$

If there is a solid species $H$ that you have to consider, use an expression based on amount of substance:

$$n_E + n_F + n_G + n_H= \mathrm{another\ constant}$$

If you want to, you can express parts of this as concentration again:

$$([E] + [F] + [G]) \cdot V_\mathrm{solution} + n_H= \mathrm{another\ constant}$$