Ionic EquilibriumHard

Question

The solubility of Pb(OH)2 in water is 6.0 × 10–6 M. The solubility of Pb(OH)2 in a buffer solution of pH = 8 is

Options

A.8.64 M
B.2.16 × 10−16 M
C.8.64 × 10−16 M
D.8.64 × 10−4 M

Solution

For saturated solution of Pb(OH)2 in water, Ksp = 4.S3 = 4 × (6 × 10–6)3 = 8.64 × 10–16

Now, for the given buffer, [OH–] = 10–6 M

∴ Solubility, S = $\left\lbrack Pb^{2 +} \right\rbrack = \frac{K_{sp}}{\left\lbrack OH^{-} \right\rbrack^{2}} = \frac{8.64 \times 10^{- 16}}{\left( 10^{- 6} \right)^{2}} = 8.64 \times 10^{- 4}M$

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