This topic applies equilibrium ideas to two practical situations: how a shared ion suppresses ionisation (the common ion effect) and how sparingly soluble salts dissolve (the solubility product).
Common ion effect
When a salt that shares an ion with a weak electrolyte is added, the ionisation of the weak electrolyte is suppressed (Le Chatelier). For example, adding $CH_3COONa$ to acetic acid increases $[CH_3COO^-]$, pushing the equilibrium $CH_3COOH\rightleftharpoons H^+ + CH_3COO^-$ to the left, lowering $[H^+]$ and raising the pH.
Buffer solutions
A buffer resists changes in pH on adding small amounts of acid or base. An acidic buffer is a weak acid + its salt (e.g. $CH_3COOH + CH_3COONa$); a basic buffer is a weak base + its salt ($NH_4OH + NH_4Cl$). The pH of an acidic buffer is given by the Henderson–Hasselbalch equation:
$$pH=pK_a+\log\frac{[salt]}{[acid]}$$
When $[salt]=[acid]$, $pH=pK_a$ and the buffer capacity is maximum. Blood is buffered near $pH\,7.4$ by the $H_2CO_3/HCO_3^-$ system.
Solubility product $K_{sp}$
For a sparingly soluble salt the dissolved part is in equilibrium with the solid: $A_xB_y(s)\rightleftharpoons xA^{y+}+yB^{x-}$. The solubility product is
$$K_{sp}=[A^{y+}]^x[B^{x-}]^y$$
If the molar solubility is $s$, then for the types:
- $AB$ (e.g. $AgCl$): $K_{sp}=s^2$, so $s=\sqrt{K_{sp}}$.
- $AB_2$ or $A_2B$ (e.g. $CaF_2$): $K_{sp}=4s^3$, so $s=\sqrt[3]{K_{sp}/4}$.
- $AB_3$ (e.g. $Fe(OH)_3$): $K_{sp}=27s^4$.
Predicting precipitation
Compute the ionic product $Q_{sp}$ from the actual ion concentrations and compare with $K_{sp}$:
- $Q_{sp}
- $Q_{sp}=K_{sp}$: saturated — equilibrium.
- $Q_{sp}>K_{sp}$: supersaturated — precipitation occurs until $Q_{sp}=K_{sp}$.
The common ion effect also lowers solubility: adding $NaCl$ to a saturated $AgCl$ solution raises $[Cl^-]$, so $[Ag^+]$ must fall to keep $K_{sp}$ constant, and $AgCl$ precipitates.
The solubility of $AgCl$ is $1.3\times10^{-5}\ \text{mol L}^{-1}$. Calculate its $K_{sp}$.
Solution- $AgCl\rightleftharpoons Ag^+ + Cl^-$, so $[Ag^+]=[Cl^-]=s$.
- $K_{sp}=s^2=(1.3\times10^{-5})^2$.
- Compute: $K_{sp}=1.69\times10^{-10}$.
Answer: $K_{sp}\approx1.7\times10^{-10}$.
The $K_{sp}$ of $CaF_2$ is $3.2\times10^{-11}$. Calculate its molar solubility $s$.
Solution- $CaF_2\rightleftharpoons Ca^{2+}+2F^-$, so $[Ca^{2+}]=s$, $[F^-]=2s$.
- $K_{sp}=s(2s)^2=4s^3$, so $s=\sqrt[3]{K_{sp}/4}=\sqrt[3]{3.2\times10^{-11}/4}$.
- $=\sqrt[3]{8\times10^{-12}}=2\times10^{-4}\ \text{mol L}^{-1}$.
Answer: $s=2\times10^{-4}\ \text{mol L}^{-1}$.
Calculate the pH of a buffer that is $0.2\ \text{M}$ in acetic acid and $0.2\ \text{M}$ in sodium acetate. ($pK_a=4.74$.)
Solution- Use $pH=pK_a+\log\dfrac{[salt]}{[acid]}$.
- Here $[salt]=[acid]=0.2$, so $\log\dfrac{0.2}{0.2}=\log 1=0$.
- $pH=4.74+0=4.74$.
Answer: $pH=4.74$ (equal to $pK_a$).
A buffer contains $0.1\ \text{M}$ acetic acid and $0.5\ \text{M}$ sodium acetate ($pK_a=4.74$). Find its pH.
Solution- $pH=pK_a+\log\dfrac{[salt]}{[acid]}=4.74+\log\dfrac{0.5}{0.1}$.
- $\log 5=0.70$.
- $pH=4.74+0.70=5.44$.
Answer: $pH=5.44$.
Equal volumes of $0.002\ \text{M}$ $BaCl_2$ and $0.002\ \text{M}$ $Na_2SO_4$ are mixed. Will $BaSO_4$ precipitate? ($K_{sp}=1.1\times10^{-10}$.)
Solution- On mixing equal volumes, each concentration halves: $[Ba^{2+}]=[SO_4^{2-}]=0.001\ \text{M}=10^{-3}$.
- Ionic product $Q_{sp}=[Ba^{2+}][SO_4^{2-}]=10^{-3}\times10^{-3}=10^{-6}$.
- Compare: $Q_{sp}=10^{-6}>K_{sp}=1.1\times10^{-10}$.
Answer: $Q_{sp}>K_{sp}$, so $BaSO_4$ precipitates.
Explain, using $K_{sp}$, why the solubility of $AgCl$ decreases when $NaCl$ is added (common ion effect).
Solution- $AgCl(s)\rightleftharpoons Ag^+ + Cl^-$ with $K_{sp}=[Ag^+][Cl^-]$ fixed at constant temperature.
- Adding $NaCl$ raises $[Cl^-]$ (common ion).
- To keep the product equal to $K_{sp}$, $[Ag^+]$ must fall, so $AgCl$ precipitates and its solubility decreases.
Answer: Added $Cl^-$ shifts equilibrium left (Le Chatelier), lowering $[Ag^+]$ and the solubility of $AgCl$.