Question

The $pH$ of $0.004 M$ hydrazine solution is $9.7 .$ Its ionisation constant $\left(K_{b}\right)$ is

• A. $7.79 \times 10^{-8}$
• B. $4.49 \times 10^{-9}$
• C. $1.67 \times 10^{-10}$
• D. $6.25 \times 10^{-7}$

Answer 1

Answer: (D)

For weak bases: $\left[ OH ^{-}\right]=\sqrt{K_{b} \times C}$

$pH =9.7$ Thus $pOH =14-9.7=4.3$

$-\log \left[ OH ^{-}\right]=4.3$

$\left[ OH ^{-}\right]=5 \times 10^{-5}$

$5 \times 10^{-5}=\sqrt{K_{b} \times 0.004}$

$ \Rightarrow K_{b} \times 0.004=25 \times 10^{-10}$

$K_{b}=\frac{25}{4 \times 10^{-3}} \times 10^{-10}=6.25 \times 10^{-7}$