The $\mathrm{pH}$ of $0.01 \mathrm{M} \mathrm{BOH}$ solution is 10 . What is its degree of dissociation? (Given $\mathrm{K}_{\mathrm{b}}$ of $\mathrm{BOH}$ is $1 imes 10^{-6}$ )

Question

The $\mathrm{pH}$ of $0.01 \mathrm{M} \mathrm{BOH}$ solution is 10 . What is its degree of dissociation? (Given $\mathrm{K}_{\mathrm{b}}$ of $\mathrm{BOH}$ is $1 	imes 10^{-6}$ )

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{BOH} ightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \ & \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{B}^{+}ight]\left[\mathrm{OH}^{-}ight]}{[\mathrm{BOH}]}=1 	imes 10^{-6} \ & \mathrm{pH}=10 \Rightarrow \mathrm{pOH}=14-10=4 \ & \Rightarrow\left[\mathrm{OH}^{-}ight]=10^{-4} \mathrm{M} . \ & 	ext { Now, } \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{Ca}^2}{1-\alpha}=\frac{\left[\mathrm{OH}^{-}ight] \alpha^2}{1-\alpha} \ & \Rightarrow \frac{\mathrm{K}_{\mathrm{b}}}{\left[\mathrm{OH}^{-}ight]}=\frac{\alpha^2}{1-\alpha} \cong \alpha^2=\frac{10^{-6}}{10^{-4}}=10^{-2} \ & \Rightarrow \alpha=10^{-1} 	ext { or } 0.1 	ext { or } 10 \%\end{aligned}$

The $\mathrm{pH}$ of $0.01 \mathrm{M} \mathrm{BOH}$ solution is 10 . What is its degree of dissociation?…

Solution