Ionic EquilibriumMedium
Question
The degree of dissociation of pure water at 25°C is found to be 1.8 × 10–9. The dissociation constant, Kd of water, at 25°C is
Options
A.10−14
B.1.8 × 10–16
C.5.56 × 10–13
D.1.8 × 10–14
Solution
$K_{d} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack}{\left\lbrack H_{2}O \right\rbrack}$
$= \frac{\alpha^{2}.C}{1 - \alpha} \approx \alpha^{2}.C = \left( 1.8 \times 10^{- 9} \right)^{2} \times \frac{1000}{18} = 1.8 \times 10^{- 16}$
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