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.

Solution

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

(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:

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?

## Answer:

0.1 *M*