Ionic EquilibriumHard

Question

The solubility product of Zn(OH)2 is 10–14 at 25oC. What would be the concentration Zn+2 ion in 0.1 M-NH4OH solution which is 50% ionized?

Options

A.2 × 10–13
B.4 × 10–12
C.4 × 10–8
D.2 × 10–11

Solution

$\left\lbrack OH^{-} \right\rbrack = 0.1 \times \frac{50}{100} = 0.05\text{ M}$

$\therefore\left\lbrack Zn^{2 +} \right\rbrack = \frac{K_{sp}}{\left\lbrack OH^{-} \right\rbrack} = \frac{10^{- 14}}{(0.05)^{2}} = 4 \times 10^{- 12}\text{ M}$

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