Question

The $pH$ of $0.004M$ hydrazine solution is $\mathrm{9.7.}$ 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: $\left[O{H}^{-}\right]=\sqrt{K_b\times C}$

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

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

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

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

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

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