Ionic EquilibriumMedium
Question
The degree of dissociation of water at 25°C is 1.8 × 10–7% and the density is 1.0 g cm–3. The ionic product for water is
Options
A.1.0 × 10–14
B.1.8 × 10–16
C.1.0 × 10–16
D.1.0 × 10–8
Solution
$H_{2}O \rightleftharpoons H^{+} + OH^{-}$
$CM = \frac{1000}{18}M00$
Equilibrium C(1 – α)M C α M C α M
$K_{w} = \left\lbrack H^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack = C\alpha.C\alpha = \alpha^{2}.C^{2} $$$= \left( 1.8 \times 10^{- 9} \right)^{2} \times \left( \frac{1000}{18} \right)^{2} = 1.0 \times 10^{- 14}$$
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