Question

The base dissociation constant for hydrazine $N_2{H}_4$ is $9.6\times {10}^{-7}$. What would be the percent hydrolysis of $0.1M{N}_2{H}_{5}Cl$, a salt containing the acid ion conjugate to hydrazine base at ${25}^{\circ }C$?

• A. $2.667\times {10}^{-3}$
• B. $1.525\times {10}^{-4}$
• C. $4.557\times {10}^{-6}$
• D. $3.225\times {10}^{-2}$

Answer 1

The correct option is D $3.225\times {10}^{-2}$

$N_2{H}_5^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N_2{H}_4\left(aq\right)+H_3{O}^{+}\left(aq\right)$
$K_h=\frac{K_w}{K_b}=\frac{{10}^{-14}}{9.6\times {10}^{-7}}=1.04\times {10}^{-8}$
now, $\frac{C}{K_h}=\frac{0.1}{1.04\times {10}^{-8}}\approx {10}^7$
Since $\frac{C}{K_h}>100\therefore 1-h\approx 1$

$h^2=\frac{K_h}{C}h=\sqrt{\frac{K_h}{C}}h=\sqrt{\frac{1.04\times {10}^{-8}}{0.1}}=3.225\times {10}^{-4}$
Percentage hydrolysis $\%h=h\times 100=3.225\times {10}^{-2}$