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In a buffer solution containing equal concentration of $\mathrm{B}^{-}$and $\mathrm{HB}$, the $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{B}^{-}$is $10^{-10}$. The $\mathrm{pH}$ of buffer solution is
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Key Idea (i) For basic buffer
$$
\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]}
$$
(ii) $\mathrm{pH}+\mathrm{pOH}=14$
Given, $\quad \mathrm{K}_{\mathrm{b}}=1 \times 10^{-10}$, [salt] = [base]
$$
\begin{aligned}
\mathrm{pOH} & =-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { base }]} \\
\therefore \quad \mathrm{pOH} & =-\log \left(1 \times 10^{-10}\right)+\log 1 \\
& =10 \\
\mathrm{pH}+\mathrm{pOH} & =14 \\
\mathrm{pH} & =14-10=4
\end{aligned}
$$
$$
\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]}
$$
(ii) $\mathrm{pH}+\mathrm{pOH}=14$
Given, $\quad \mathrm{K}_{\mathrm{b}}=1 \times 10^{-10}$, [salt] = [base]
$$
\begin{aligned}
\mathrm{pOH} & =-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { base }]} \\
\therefore \quad \mathrm{pOH} & =-\log \left(1 \times 10^{-10}\right)+\log 1 \\
& =10 \\
\mathrm{pH}+\mathrm{pOH} & =14 \\
\mathrm{pH} & =14-10=4
\end{aligned}
$$
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