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Question

Calculate the solubility of zinc (II) hydroxide in a solution maintained at (i) pH= 5.3 (ii) pH= 8.2. Ksp = 2.0 x 10-17

Answer 1

Zn(OH)₂(s) <==> Zn²⁺(aq) + 2OH^-(aq)

Ksp = [Zn²⁺][OH-]²

We are given the Ksp, and from the pH, we can calculate the pOH and then the [OH-].

pH = 5.3

pOH = 14 - 5.3 = 8.7

[OH-] = 1.99x10^-9 M

2.0x10^-17 = [Zn²⁺][1.99x10^-9]²

[Zn²⁺] = 2.0x10^-17 / 1.99x10^-9

[Zn²⁺] = 1.0 x10^-8 M = solubility of Zn(OH)2 at pH 5.3

For pH 8.2, follow the same procedure