Question

The basic ionisation constant for hydrazine, $N_2{H}_4$ is $9.6\times {10}^{-7}$. What would be the percentage hydrolysis of $0.1M{N}_2{H}_{5}Cl$ ?
$(Given:\sqrt{10.4}\approx 3.2and\sqrt{1.04}=1.013)$

• A. $0.01\%$
• B. $3.2\%$
• C. $1.013\%$
• D. $0.032\%$

Answer 1

The correct option is D $0.032\%$

$N_2{H}_5^{+}+H_{2}O\rightleftharpoons N_2{H}_4+H_3{O}^{+}{K}_h=\frac{\left[N_2{H}_4\right]\left[H_3{O}^{+}\right]}{\left[N_2{H}_5^{+}\right]}=\frac{k_w}{k_b}=\frac{{10}^{-14}}{9.6\times {10}^{-7}}=1.04\times {10}^{-8}\therefore h=\sqrt{\frac{K_h}{c}}\Rightarrow h=\sqrt{\frac{1.04\times {10}^{-8}}{0.1}}\Rightarrow h=3.2\times {10}^{-4}\text{Percentage of hydrolysis}=3.2\times {10}^{-4}\times 100=0.032\%$