Solubility product of a sparingly soluble salt in presence of other dissolved salts

Question

I read here that solubility product of a salt (say $X^+ Y^- $) remains constant even in presence of another salt with a common ion (say $Z^+ Y^-$).

Though I can intuitively understand that as the concentration of $Y^-$ has increased due to the added salt, and hence by le-Chatelier's principle concentration of $X^-$ has to decrease, is it possible to get a strong proof for the fact that the value of Ksp remains constant? (Earlier, I had thought that Ksp would decrease with addition of another salt, but later realized that the concentration of the common ion due to both the contributing salts is considered in the calculation of Ksp)

Also, would the solubility product of a salt remain constant in the presence of another dissolved salt with no ion common with it? (i.e, would the the Ksp value of a salt XY be the same in water and in a solution of another salt AB ?)

Answer 1

Presence of salts with the common ion decreases the salt solubility via the activity product.

Presence of salts without the common ion increases the salt solubility via the activity coefficients.

---

Solubility product as a thermodynamic constant is defined for ion thermodynamic activities, like:

$$K_\mathrm{sp} = a(\ce{X+})a(\ce{Y-})$$

which are defined as

$$\mu_i = \mu_i^{\circ} + RT\ln(a_i)$$

where $\mu$ is a chemical potential of a substance, being a partial derivative of the system Gibbs function per its molar amount ( at constant temperature pressure and content of other compounds):

$$\mu_i = \left(\frac{\partial G }{\partial n_i}\right)_{p,T,n_j,j \ne i}$$

If a the solid salt XY is in the equilibrium with its hydrated ions:

$$\ce{XY(s) <=> X+(aq) + Y-(aq)}$$

then

$$\mu(\ce{XY(s)}) = \mu(\ce{X+}) + \mu(\ce{Y-})= \\ \mu^{\circ}(\ce{X+}) + \mu^{\circ}(\ce{Y-}) + RT(\ln{(a(\ce{X+})}) + \ln{(a(\ce{Y-})}))= \\ \mu^{\circ}(\ce{X+}) + \mu^{\circ}(\ce{Y-}) + RT\ln{(a(\ce{X+})a(\ce{Y-}))}$$

$$\exp{\left(\frac{\mu{(\ce{XY(s)}) - \mu^{\circ}(\ce{X+}) - \mu^{\circ}(\ce{Y-}) }}{ RT}\right)} = K_\mathrm{sp} =a(\ce{X+})a(\ce{Y-})$$

Standard chemical potentials for ions are defined in such a way that ion activities are numerically equal to concentrations for the latter converging to infinitely diluted solution.

The ratio of the ion activity and its molar concentration is called activity coefficient: $a = \gamma \cdot \frac{c}{c^{\circ}}$

For enough diluted solutions, activity coefficients can be approximated by Debye–Hückel equation (for ionic strength below 0.1 M)

$$\ln(\gamma_i) = -\frac{z_i^2 q^2 \kappa}{8 \pi \varepsilon_r \varepsilon_0 k_\text{B} T} = -\frac{z_i^2 q^3 N^{1/2}_\text{A}}{4 \pi (\varepsilon_r \varepsilon_0 k_\text{B} T)^{3/2}} \sqrt{10^3\frac{I}{2}} = -A z_i^2 \sqrt{I}$$

with $A \approx 0.51$, using the ionic strength $I$:

$$I = \begin{matrix}\frac{1}{2}\end{matrix}\sum_{i=1}^{n} c_i z_i^{2}$$

Answer 2

This is another example of everything you learned in HS, then in College, and then in Grad School; is wrong! Unfortunately or fortunately even after one does research there is still more to learn or unlearn. The site you give gives a thorough if simplified view. Equilibrium constants are constant for a specific reaction at a given temperature if the chemical potentials of the reactants are expressed as activities not necessarily the same as concentrations. At low concentrations of everything in solution concentrations approximate activities closely. This is similar to the use of PV=nRT in gas calculations.

PbCl2[crystals] dissolve in pure water to give hydrated Pb++ ions and hydrated Cl- ions. With crystals of appropriate size and proper mixing an equilibrium is established. Careful analysis shows that the measured amount of Pb and Cl in solution are constant regardless of the method of preparation when expressed as concentrations. The relationship is expressed as the equilibrium constant shortened to the solubility product provided the solid is a pure crystal with chemical activity equal to 1.. Ksp = [Pb++][Cl-]^2. The concentration of chloride is squared because the molecularity of chloride at equilibrium is 2.

If an ionic compound not related to lead chloride is present, the solvation of the lead and chloride ions will change. This change is specified as a change in the chemical activity of the ions and tables of the activity coefficients are found in handbooks and books on quantitative analysis [and are usually ignored]. If the concentration of lead or chloride ions is changed by a change in total ionic strength the amount of PbCl2 that dissolves changes usually to greater dissolution. This is because the activity of the ions in solution has lowered, and more are needed to satisfy the Ksp. There are two possible approaches to handle this: 1. generate a new Ksp for every possible solution [quite a job] or 2. Try to measure activity coefficients as a function of ionic strength for each ion. The second works [and as usual is ignored until necessary].

The common ion effect means adding an ion contained in the solid. This can be done in two ways: 1. electrolytically by using an appropriate cell. The best example of this that I can think of is the operation of the lead acid battery used in cars. Measures had to be taken to prevent PbSO4 from precipitating everywhere. and 2. The addition of a salt containing one of the ions. This has two effects: it increases the concentration of one of the ions and slightly increases ionic strength the first effect is greater so adding a common ion will cause the other ion to be reduced in solution to satisfy Ksp. This means that concentrations approximate activities, best at low concentrations. There can be complications with some salts if the ion added can react further with the salt to give complex ions. This happens with ions such as chloride, cyanide, oxalate. Silver chloride or cyanide are insoluble in water but are appreciably soluble in excess added chloride or cyanide and cyanide solutions will dissolve gold under mildly oxidizing conditions.