Question

$0.16g$ of $N_2{H}_4$ are dissolved in water and the total volume made up to $500mL$. 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 $4.0\times {10}^{-6}M$.

Answer 1

Concentration of $N_2{H}_4$
$\left[N_2{H}_4\right]=\frac{0.16}{32}\times \frac{1000}{500}=0.01M$
$N_2{H}_4+H_{2}O\rightleftharpoons N_2{H}_5^{+}+{OH}^{-}$
At equi. $C(1-\alpha )$ $C\alpha$ $C\alpha$
$K_b={C\alpha }^2$
${\alpha }^2=\frac{K_b}{C}=\frac{4.0\times {10}^{-6}}{0.01}=4\times {10}^{-4}$
or $\alpha =2\times {10}^{-2}$i.e., $2$%