What is the pH of a 0.15 M solution of hydrazine, ${\mathrm{N}}_2{\mathrm{H}}_4$?

Hydrazine	${\mathrm{K}}_{\mathrm{b}}$
${\mathrm{N}}_2{\mathrm{H}}_4$	$1.0\times {10}^{-6}$

 

As a result, the pH of the 0.15 M hydrazine solution is 10.59.${\mathrm{N}}_2{\mathrm{H}}_4+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{N}}_2{\mathrm{H}}_5^{+}+{\mathrm{OH}}^{-}$

$\begin{array}{c}{\mathrm{k}}_{\mathrm{b}}=\frac{0.18\mathrm{\alpha }\times 0.15\mathrm{\alpha }}{0.5(1-\mathrm{\alpha })}=\frac{0.15{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}\\ {10}^{-6}=0.15{\mathrm{\alpha }}^{2}3\mathrm{\alpha }=\sqrt{\frac{{10}^{-6}}{0.15}}\end{array}$

$\left[{\mathrm{OH}}^{-}\right]=0.15x=\sqrt{0.15\times {10}^{-6}}=0.39\times {10}^{-3}$

$\begin{array}{l}\left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-14}}{0.39\times {10}^{-3}}\\ \left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-14}}{0.39\times {10}^{-3}}=2.58\times {10}^{-11}\end{array}$

$\mathrm{pH}=14-\frac{1}{2}\left[{\mathrm{pk}}_{\mathrm{b}}-\mathrm{logc}\right]$

$\begin{array}{l}\mathrm{pH}=-{\mathrm{logH}}^{+}=-\log \left(2.58\times {10}^{-11}\right)\\ \mathrm{pH}=-\log (2.58)-\log {10}^{-11}\end{array}$

$\begin{array}{l}\mathrm{pH}=11-0.411\\ \mathrm{pH}=10.59\end{array}$