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A certain buffer solution contains equal concentration of $X^{-}$and $H X$. The ${ }^{K_b}$ for $X^{-}$is $10^{-10}$. The $p H$ of the buffer is
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$\mathrm{X}^{-}+\mathrm{H}_2 \mathrm{O}=\mathrm{OH}^{-}+\mathrm{HX}$
$\begin{aligned} & K_b=\frac{\left[O H^{-}\right][H X]}{\left[X^{-}\right]} \\ & H X=H^{+}+X^{-} \\ & K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[H X]} \\ & \therefore K_a \times K_b=\left[H^{+}\right]\left[O H^{-}\right]=K_w=10^{-14}\end{aligned}$
Hence $K_\alpha=10^{-4}$
Now as $\left[X^{-}\right]=[H X], p H=p K_a=4$.
$\begin{aligned} & K_b=\frac{\left[O H^{-}\right][H X]}{\left[X^{-}\right]} \\ & H X=H^{+}+X^{-} \\ & K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[H X]} \\ & \therefore K_a \times K_b=\left[H^{+}\right]\left[O H^{-}\right]=K_w=10^{-14}\end{aligned}$
Hence $K_\alpha=10^{-4}$
Now as $\left[X^{-}\right]=[H X], p H=p K_a=4$.
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