Question

The solubility product of \[\ce{Al(OH)3}\] is 2.7 × 10^–11. Calculate its solubility in gL^–1 and also find out pH of this solution. (Atomic mass of \[\ce{Al = 27 u}\]).

Answer 1

Let us assume \[\ce{S}\] be the solubility of \[\ce{Al(OH)3}\].

\[\ce{K_{sp} = [Al^{3+}] [OH-]3 = (S) (3S)^3 = 27S^4}\]

S⁴ = 27K_sp = 27 × 10²⁷ × 10^–11 = 1 × 10^–12

S = 1 × 10⁻³ mol L^–1.

(i) Solubility of \[\ce{Al(OH)3}\]: Molar mass of \[\ce{Al(OH)3}\] is 78g. Hence, solubility of \[\ce{Al(OH)3}\] in gL^–1 = 1 × 10³ × 78gL^–1 = 78 × 10^–3 gL^–1 = 7.8 × 10^–2gL^–1

(ii) \[\ce{pH}\] of the solution: S = 1 × 10⁻³ mol L⁻¹

\[\ce{[OH]}\] = 3S = 3 × 1 × 10⁻³ = 3 × 10⁻³

\[\ce{pOH}\] = 3 − log 3

\[\ce{pH}\] = 14 − \[\ce{pOH}\] = 11 + log 3 = 11.4771.