Coupled Equilibria – Increased Solubility in Acidic Solutions

Sparingly soluble compounds that produce a basic anion can react in acidic solution – via coupled equilibria – and become more soluble. For example, calcium fluoride ($CaF_2$) is a sparingly soluble salt: $$CaF_2 \, (s)\; \rightleftharpoons \; Ca^{2+}\,(aq)\, + \,2\, F^{-}\, (aq) \qquad\qquad K_{sp} = 5.3\times 10^{-9}$$

The fluoride anion ($F^{-}$) is a weak base – the conjugate base of hydrofluoric acid ($HF$): $$HF\, (aq) \; \rightleftharpoons \;H^+ \, (aq) \, + \, F^- \, (aq) \qquad\qquad K_a = 6.3 \times 10^{-4}$$

In an acidic solution, the $H^+$ (from the acid) and the $F^-$ produced by the calcium fluoride can react in a very favourable reaction (the reverse of the $K_a$ above): $$ H^+ \, (aq) \, + \, F^- \, (aq)\; \rightleftharpoons \; HF\, (aq) \qquad\qquad K = 1.6 \times 10^{+3}$$

This reaction strongly favours the products (notice it’s large K value) – and by consuming $F^-$ in solution, following Le Châtelier’s Principle, we can expect it to “pull” the solubility equilibrium of CaF to the right / products side — dissolving more CaF and increasing overall solubility.

We can show this mathematically by combining the $K_{sp}$ reaction for CaF with the neutralization of $F^-$ to show the overall reaction: $$\require{cancel} CaF_2 \, (s)\; \rightleftharpoons \; Ca^{2+}\,(aq)\, + \,\cancel{2\, F^{-}}\, (aq) \qquad\qquad K_{sp} = 5.3\times 10^{-9} \\ 2\,H^+ \, (aq) \, + \, \cancel{2\,F^-} \, (aq)\; \rightleftharpoons \; 2\, HF\, (aq) \qquad\qquad K = (1.6 \times 10^{3})^2 \\ \; \\ CaF_2 \, (s)\,+\, 2\,H^+ \, (aq) \; \rightleftharpoons \; Ca^{2+}\,(aq)\, + \, 2\, HF\, (aq) \qquad\qquad K = \mathbf{1.3 \times 10^{-2}}$$

(the HF-producing reaction was doubled to make the stoichiometry work – refer to the page on math with equilibrium constants).

The overall reaction of the CaF in acid has a much larger equilibrium constant than the “neutral” CaF dissociation reaction – so more of the solid CaF will dissolve in an acidic solution.

Increased Solubility in Acidic SolutionsCompute and compare the molar solublities for aluminum hydroxide, Al(OH)3, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.

(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:

$$Al(OH)_3(s)⇌Al^{3+}(aq)+3OH^-(aq)\qquad K_{sp}=2×10^{-32}$$ $$\text{molar solubility in water }=[Al^{3+}]=(2×10^{-32}/27)^{1/4}=5×10^{-9}\;M$$

(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:

$$\text{pH}=\text{p}K_a+log[CH_3COO^-]/[CH_3COOH]$$ $$\text{pH}=4.74+log(0.100/0.100)=4.74$$

At this pH, the concentration of hydroxide ion is

$$\text{pOH}=14.00-4.74=9.26$$ $$[OH^-]=10^{-9.26}=5.5×10^{-10}$$

The solubility of Al(OH)3 in this buffer is then calculated from its solubility product expressions:

$$K_{sp}=[Al^{3+}][OH^-]^3$$ $$\text{molar solubility in buffer }=[Al^{3+}]=K_{sp}/[OH^-]^3=(2×10^{-32})/(5.5×10^{-10})^3=1.2×10^{-4}\;M$$

Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).

Check Your Learning

What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?


0.1 M