$0.32$ g of $N_{2} H_{4}$ is dissolved in water to make up to a $1$ L solution. The percentage of $N_{2} H_{4}$ that has reacted with water in this solution is:
$(K_{b} (N_{2} H_{4}) = 4.0 \times 10^{- 6})$

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Solution

$N_{2} H_{4} + H_{2} O ⇌ N_{2} H_{5}^{+} + O H^{-}$
Before dissociation 1 $-$ 0 0
After dissociation $(1 - α)$ $-$ $α$ $α$
$K_{b} = \frac{C α^{2}}{(1 - α)}; (∵ 1 - α) \approx 1$
$K_{b} = C α^{2}$
$[N_{2} H_{4}] = C = \frac{0.32 \times 1000}{32 \times 1000} = 0.01$
$α^{2} = \frac{K_{b}}{C} = \frac{4 \times 10^{- 6}}{0.01} = 4 \times 10^{- 4}$
$α = 2 \times 10^{- 2}; i . e 2$ %.

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