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प्रश्न
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 \[\ce{Al = 27 u}\]).
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उत्तर
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 = 27Ksp = 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.
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