The $pH$ of $0.004 \, M$ hydrazine solution is $9.7$. Calculate its ionization constant $K_{b}$ and $pK_{b}$.

$NH_{2}NH_{2} + H_{2}O \rightleftharpoons NH_{2}NH_{3}^{+} + OH^{-}$
Given the $pH = 9.7$,we calculate the $pOH$ as:
$pOH = 14 - pH = 14 - 9.7 = 4.3$
Now,calculate the concentration of hydroxyl ions $[OH^{-}]$:
$[OH^{-}] = 10^{-pOH} = 10^{-4.3} = 5.01 \times 10^{-5} \, M$
Since the dissociation of hydrazine is $NH_{2}NH_{2} + H_{2}O \rightleftharpoons NH_{2}NH_{3}^{+} + OH^{-}$,the concentration of $[NH_{2}NH_{3}^{+}] = [OH^{-}] = 5.01 \times 10^{-5} \, M$.
The concentration of undissociated hydrazine is approximately $0.004 \, M$.
$K_{b} = \frac{[NH_{2}NH_{3}^{+}][OH^{-}]}{[NH_{2}NH_{2}]} = \frac{(5.01 \times 10^{-5})^{2}}{0.004} = 6.275 \times 10^{-7}$
$pK_{b} = -\log(K_{b}) = -\log(6.275 \times 10^{-7}) = 6.20$