$0.16 g$ of $N_{2} H_{4}$ are dissolved in water and the total volume made up to $500 m L$ . Calculate the percentage of $N_{2} H_{4}$ that has reached with water at this dilution. The $K_{b}$ for $N_{2} H_{4}$ is $4.0 \times 10^{- 6} M$ .

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Solution

Concentration of $N_{2} H_{4}$
$[N_{2} H_{4}] = \frac{0.16}{32} \times \frac{1000}{500} = 0.01 M$
$N_{2} H_{4} + H_{2} O ⇌ N_{2} H_{5}^{+} + {O H}^{-}$
At equi. $C (1 - α)$ $C α$ $C α$
$K_{b} = {C α}^{2}$
$α^{2} = \frac{K_{b}}{C} = \frac{4.0 \times 10^{- 6}}{0.01} = 4 \times 10^{- 4}$
or $α = 2 \times 10^{- 2}$ i.e., $2$ %

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