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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 $1 \times 10^{-6}$ )
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(Given $\mathrm{K}_{\mathrm{b}}$ of $\mathrm{BOH}$ is $1 \times 10^{-6}$ )
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$10 \%$
$\begin{aligned} & \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{B}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{BOH}]}=1 \times 10^{-6} \\ & \mathrm{pH}=10 \Rightarrow \mathrm{pOH}=14-10=4 \\ & \Rightarrow\left[\mathrm{OH}^{-}\right]=10^{-4} \mathrm{M} . \\ & \text { Now, } \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{Ca}^2}{1-\alpha}=\frac{\left[\mathrm{OH}^{-}\right] \alpha^2}{1-\alpha} \\ & \Rightarrow \frac{\mathrm{K}_{\mathrm{b}}}{\left[\mathrm{OH}^{-}\right]}=\frac{\alpha^2}{1-\alpha} \cong \alpha^2=\frac{10^{-6}}{10^{-4}}=10^{-2} \\ & \Rightarrow \alpha=10^{-1} \text { or } 0.1 \text { or } 10 \%\end{aligned}$
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