The acid ionization (hydrolysis) constant of $$Zn2+$$ is $$1.0$$ × $$10-9$$. Hence, pH of $$0.001$$ M solution of $$ZnCI2$$ is......

$$Zn2+H2O$$⇌$$Zn(OH)++H+Kh=1$$×$$10$$−$$9Kh=KwKbZn(OH)2)$$∴ $$Kb=KwKh=1$$×$$10$$−$$5$$ $$pKb=5$$∴ $$pH=7$$−$$pKb+log$$⁡$$C2=7$$−$$(5$$−$$3)2=6$$

Ionic equilibrium

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Question

The acid ionization (hydrolysis) constant of Zn²⁺ is 1.0 × 10^-9. Hence, pH of 0.001 M solution of ZnCI₂ is......

Moderate

Solution

$\begin{array}{l} {Zn}^{2} + H_{2} O ⇌ Zn (OH)^{+} + H^{+} \\ K_{h} = 1 \times 10^{- 9} \\ K_{h} = \frac{K_{w}}{K_{b} (Zn (OH)_{2})} \\ ∴ K_{b} = \frac{K_{w}}{K_{h}} = 1 \times 10^{- 5} \\ {pK}_{b} = 5 \\ ∴ pH = 7 - (\frac{{pK}_{b} + \log C}{2}) = 7 - \frac{(5 - 3)}{2} = 6 \end{array}$

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