Q34Short answer
The solubility product of Al (OH)3 is 2.7 × 10 . Calculate its solubility in gL–1 and also find out pH of this solution. (Atomic mass of Al = 27 u). 91 Equilibrium
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Let S be the solubility of Al(OH)3. Al (OH)3 Al 3+ (aq) + 3OH (aq) – Concentration of species at t = 0 1 0 0 Concentration of various species at equilibrium 1–S S 3S 3+ – 3 3 4 Ksp = [Al ] [OH ] = (S) (3S) = 27 S –11 K sp 27 × 10 –12 S = = = 1 × 10 27 27 × 10 –3 –1 S = 1× 10 mol L (i) Solubility of Al(OH)3 Molar mass of Al (OH)3 is 78 g. Therefore, –1 –3 –3 –1 Solubility of Al (OH)3 in g L = 1 × 10 × 78 g L–1 = 78 × 10 g L –2 –1 = 7.8 × 10 g L (ii) pH of the solution –3 –1 S = 1×10 mol L – –3 –3 [OH ] = 3S = 3×1×10 = 3 × 10 pOH = 3 – log 3 pH = 14 – pOH = 11 + log 3 = 11.4771 –8