Ionic EquilibriumHard
Question
The acid ionization constant of Zn2+ is 2.0 × 10–10. What is the pH of 0.001 M solution of ZnCl2? (log 2 = 0.3)
Options
A.9.7
B.4.85
C.6.35
D.3.35
Solution
$Zn^{2 +} + H_{2}O \rightleftharpoons Zn(OH)^{+} + H^{+};$
Equilibrium (0.001 – x)M xM xM
$K_{a} = 2 \times 10^{- 10}$
$\Rightarrow x = \sqrt{2 \times 10^{- 13}} $$$\therefore P^{H} = - \log x = - {\log\left( 2 \times 10^{- 13} \right)}^{1/2} = 6.35$$
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