6.13 Solubility Equilibria of Sparingly Soluble Salts

We have already known that the solubility of ionic solids in water varies a great deal. Some of these (like calcium chloride) are so soluble that they are hygroscopic in nature and even absorb water vapour from the atmosphere. Others (such as lithium fluoride) have so little solubility that they are commonly termed as insoluble. The solubility depends on a number of factors important amongst which are the lattice enthalpy of the salt and the solvation enthalpy of the ions in a solution. For a salt to dissolve in a solvent the strong forces of attraction between its ions (lattice enthalpy) must be overcome by the ion-solvent interactions. The solvation enthalpy of ions is referred to in terms of solvation which is always negative i.e. energy is released in the process of solvation. The amount of solvation enthalpy depends on the nature of the solvent. In case of a non-polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent. As a general rule , for a salt to be able to dissolve in a particular solvent its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former. Each salt has its characteristic solubility which depends on temperature. We classify salts on the basis of their solubility in the following three categories.

We shall now consider the equilibrium between the sparingly soluble ionic salt and its saturated aqueous solution.

Solubility Product Constant

Let us now have a solid like barium sulphate in contact with its saturated aqueous solution. The equilibrium between the undisolved solid and the ions in a saturated solution can be represented by the equation:

The equilibrium constant is given by the equation:
\(
K=\left\{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]\right\} /\left[\mathrm{BaSO}_4\right]
\)
For a pure solid substance the concentration remains constant and we can write
\(
K_{\mathrm{sp}}=\mathrm{K}\left[\mathrm{BaSO}_4\right]=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right] \dots(7.43)
\)
We call \(K_{\mathrm{sp}}\) the solubility product constant or simply solubility product. The experimental value of \(K_{\mathrm{sp}}\) in above equation at \(298 \mathrm{~K}\) is \(1.1 \times 10^{-10}\). This means that for solid barium sulphate in equilibrium with its saturated solution, the product of the concentrations of barium and sulphate ions is equal to its solubility product constant. The concentrations of the two ions will be equal to the molar solubility of the barium sulphate. If molar solubility is \(\mathrm{S}\), then
\(
\begin{aligned}
& 1.1 \times 10^{-10}=(\mathrm{S})(\mathrm{S})=\mathrm{S}^2 \\
& \text { or } \quad \mathrm{S}=1.05 \times 10^{-5} .
\end{aligned}
\)
Thus, molar solubility of barium sulphate will be equal to \(1.05 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\).

A salt may give on dissociation two or more than two anions and cations carrying different charges. For example, consider a salt like zirconium phosphate of molecular formula \(\left(\mathrm{Zr}^{4+}\right)_3\left(\mathrm{PO}_4^{3-}\right)_4\). It dissociates into 3 zirconium cations of charge +4 and 4 phosphate anions of charge -3. If the molar solubility of zirconium phosphate is \(\mathrm{S}\), then it can be seen from the stoichiometry of the compound that
\(
\begin{aligned}
& {\left[\mathrm{Zr}^{4+}\right]=3 \mathrm{~S} \text { and }\left[\mathrm{PO}_4^{3-}\right]=4 \mathrm{~S}} \\
& \text { and } K_{\mathrm{sp}}=(3 \mathrm{~S})^3(4 \mathrm{~S})^4=6912(\mathrm{~S})^7 \\
& \text { or S }=\left\{K_{\mathrm{sp}} /\left(3^3 \times 4^4\right)\right\}^{1 / 7}=\left(K_{\mathrm{sp}} / 6912\right)^{1 / 7}
\end{aligned}
\)
A solid salt of the general formula \(\mathrm{M}_{\mathrm{x}}^{\mathrm{p}+} \mathrm{X}_{\mathrm{y}}^{\mathrm{q}-}\) with molar solubility \(\mathrm{S}\) in equilibrium with its saturated solution may be represented by the equation:
\(
\mathrm{M}_{\mathrm{x}} \mathrm{X}_{\mathrm{y}}(\mathrm{s}) \rightleftharpoons \mathrm{xM}^{\mathrm{p}+}(\mathrm{aq})+\mathrm{yX}^{\mathrm{q}-}(\mathrm{aq})
\)
\(
\left(\text { where } x \times p^{+}=y \times q^{-}\right)
\)
And its solubility product constant is given by:
\(
K_{\mathrm{sp}}=\left[\mathrm{M}^{\mathrm{p}+}\right]^{\mathrm{x}}\left[\mathrm{X}^{\mathrm{q}-}\right]^{\mathrm{y}}=(\mathrm{xS})^{\mathrm{x}}(\mathrm{yS})^{\mathrm{y}} \dots(7.44)
\)
\(
\begin{aligned}
& =\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{S}^{(\mathrm{x}+\mathrm{y})} \\
& \mathrm{S}^{(\mathrm{x}+\mathrm{y})}=K_{\mathrm{sp}} / \mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}
\end{aligned}
\)
\(
\mathrm{S}=\left(K_{\mathrm{sp}} / \mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}\right)^{1 / \mathrm{x}+\mathrm{y}} \dots(7.45)
\)
The term \(K_{\mathrm{sp}}\) in equation is given by \(Q_{\mathrm{sp}}\) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions \(K_{\mathrm{sp}}=Q_{\mathrm{sp}}\) but otherwise it gives the direction of the processes of precipitation or dissolution. The solubility product constants of a number of common salts at \(298 \mathrm{~K}\) are given in Table 7.9.

Example 7.26: Calculate the solubility of \(A_2 X_3\) in pure water, assuming that neither kind of ion reacts with water. The solubility product of \(\mathrm{A}_2 \mathrm{X}_3, K_{\mathrm{sp}}=1.1 \times 10^{-23}\).

Answer:

\(
\begin{aligned}
& \mathrm{A}_2 \mathrm{X}_3 \rightarrow 2 \mathrm{~A}^{3+}+3 \mathrm{X}^{2-} \\
& K_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3=1.1 \times 10^{-23} \\
& \text { If } \mathrm{S}=\text { solubility of } \mathrm{A}_2 \mathrm{X}_3 \text {, then } \\
& {\left[\mathrm{A}^{3+}\right]=2 \mathrm{~S} ;\left[\mathrm{X}^{2-}\right]=3 \mathrm{~S}}
\end{aligned}
\)
therefore,
\(
\begin{aligned}
K_{\mathrm{sp}}=(2 \mathrm{~S})^2(3 \mathrm{~S})^3 & =108 \mathrm{~S}^5 =1.1 \times 10^{-23}
\end{aligned}
\)
thus, \(S^5=1 \times 10^{-25}\)
\(\mathrm{S}=1.0 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\)

Example 7.27: The values of \(K_{\mathrm{sp}}\) of two sparingly soluble salts \(\mathrm{Ni}(\mathrm{OH})_2\) and \(\mathrm{AgCN}\) are \(2.0 \times 10^{-15}\) and \(6 \times 0^{-17}\) respectively. Which salt is more soluble? Explain.

Answer:

\(
\mathrm{AgCN} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{CN}^{-}
\)
\(
\begin{aligned}
& K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right]=6 \times 10^{-17} \\
& \mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}^{2+}+2 \mathrm{OH}^{-} \\
& K_{\mathrm{sp}}=\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=2 \times 10^{-15}
\end{aligned}
\)
Let \(\left[\mathrm{Ag}^{+}\right]=\mathrm{S}_1\), then \(\left[\mathrm{CN}^{-}\right]=\mathrm{S}_1\)
Let \(\left[\mathrm{Ni}^{2+}\right]=\mathrm{S}_2\), then \(\left[\mathrm{OH}^{-}\right]=2 \mathrm{~S}_2\)
\(
\mathrm{S}_1^2=6 \times 10^{-17}, \mathrm{~S}_1=7.8 \times 10^{-9}
\)
\(
\left(\mathrm{S}_2\right)\left(2 \mathrm{~S}_2\right)^2=2 \times 10^{-15}, \mathrm{~S}_2=0.58 \times 10^{-4}
\)
\(\mathrm{Ni}(\mathrm{OH})_2\) is more soluble than AgCN.

Common Ion Effect on Solubility of Ionic Salts

It is expected from Le Chatelier’s principle that if we increase the concentration of any one of the ions, it should combine with the ion of its opposite charge and some of the salt will be precipitated till once again \(K_{\mathrm{sp}}=Q_{\mathrm{sp}}\). Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till once again \(K_{\mathrm{sp}}=\mathcal{Q}_{\mathrm{sp}}\). This is applicable even to soluble salts like sodium chloride except that due to higher concentrations of the ions, we use their activities instead of their molarities in the expression for \(Q_{\mathrm{sp}}\). Thus if we take a saturated solution of sodium chloride and pass \(\mathrm{HCl}\) gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride ion available from the dissociation of \(\mathrm{HCl}\). Sodium chloride thus obtained is of very high purity and we can get rid of impurities like sodium and magnesium sulphates. The common ion effect is also used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation. Thus we can precipitate silver ion as silver chloride, ferric ion as its hydroxide (or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations.

The solubility of salts of weak acids like phosphates increases at lower \(\mathrm{pH}\). This is because at lower \(\mathrm{pH}\) the concentration of the anion decreases due to its protonation. This in turn increase the solubility of the salt so that \(K_{\mathrm{sp}}=Q_{\mathrm{sp}}\). We have to satisfy two equilibria simultaneously i.e.,
\(
K_{\mathrm{sp}}=\left[\mathrm{M}^{+}\right]\left[\mathrm{X}^{-}\right],
\)
\(
\begin{array}{r}
\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \\
K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}(\mathrm{aq})\right]\left[\mathrm{X}^{-}(\mathrm{aq})\right]}{[\mathrm{HX}(\mathrm{aq})]}
\end{array}
\)
\(\left[\mathrm{X}^{-}\right] /[\mathrm{HX}]=K_{\mathrm{a}} /\left[\mathrm{H}^{+}\right]\)
Taking inverse of both side and adding 1 we get
\(
\begin{aligned}
& \frac{[\mathrm{HX}]}{\left[\mathrm{X}^{-}\right]}+1=\frac{\left[\mathrm{H}^{+}\right]}{K_{\mathrm{a}}}+1 \\
& \frac{[\mathrm{HX}]+\left[\mathrm{H}^{-}\right]}{\left[\mathrm{X}^{-}\right]}=\frac{\left[\mathrm{H}^{+}\right]+K_{\mathrm{a}}}{K_{\mathrm{a}}}
\end{aligned}
\)
Now, again taking inverse, we get
\([\mathrm{X}] /\{[\mathrm{X}]+[\mathrm{HX}]\}=\mathrm{f}=K_{\mathrm{a}} /\left(K_{\mathrm{a}}+\left[\mathrm{H}^{+}\right]\right)\)and it can be seen that ‘ \(\mathrm{f}\) ‘ decreases as \(\mathrm{pH}\) decreases. If \(\mathrm{S}\) is the solubility of the salt at a given \(\mathrm{pH}\) then
\(
\begin{aligned}
& K_{\mathrm{sp}}=[\mathrm{S}][\mathrm{f} \mathrm{S}]=\mathrm{S}^2\left\{K_{\mathrm{a}} /\left(K_{\mathrm{a}}+\left[\mathrm{H}^{+}\right]\right)\right\} \text {and } \\
& \mathrm{S}=\left\{K_{\mathrm{sp}}\left(\left[\mathrm{H}^{+}\right]+K_{\mathrm{a}}\right) / K_{\mathrm{a}}\right\}^{1 / 2} \dots(7.46)
\end{aligned}
\)
Thus solubility \(\mathrm{S}\) increases with increase in \(\left[\mathrm{H}^{+}\right]\)or decrease in \(\mathrm{pH}\).

Example 7.28: Calculate the molar solubility of \(\mathrm{Ni}(\mathrm{OH})_2\) in \(0.10 \mathrm{M} \mathrm{NaOH}\). The ionic product of \(\mathrm{Ni}(\mathrm{OH})_2\) is \(2.0 \times 10^{-15}\).

Answer: Let the solubility of \(\mathrm{Ni}(\mathrm{OH})_2\) be equal to \(\mathrm{S}\).
Dissolution of \(\mathrm{S} \mathrm{mol} / \mathrm{L}\) of \(\mathrm{Ni}(\mathrm{OH})_2\) provides \(\mathrm{S} \mathrm{mol} / \mathrm{L}\) of \(\mathrm{Ni}^{2+}\) and \(2 \mathrm{~S} \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{OH}^{-}\), but the total concentration of \(\mathrm{OH}^{-}=(0.10+\) \(2 \mathrm{~S}) \mathrm{mol} / \mathrm{L}\) because the solution already contains \(0.10 \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{OH}^{-}\)from \(\mathrm{NaOH}\).
\(
K_{\mathrm{sp}}=2.0 \times 10^{-15}=\left[\mathrm{Ni}^{2+}\right][\mathrm{OH}]^2=(\mathrm{S})(0.10+2 \mathrm{~S})^2
\)
As \(K_{\mathrm{sp}}\) is small, \(2 \mathrm{~s} \ll 0.10\), thus, \((0.10+2 \mathrm{~S}) \approx 0.10\)
Hence,
\(
\begin{aligned}
& 2.0 \times 10^{-15}=\mathrm{S}(0.10)^2 \\
& \mathrm{~S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]
\end{aligned}
\)