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In a buffer solution containing equal concentration of $B^{-}$ and HB, the K_b for $B^{-}$ is $10^{- 10}$ . The pH of buffer solution is:

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answer is D.

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Detailed Solution

$p_{OH} = p K_{b} + \log [B^{-}] / [HB]$ Since $[B^{-}]$ $= [HB]$ (given),the $p_{OH} = {pK}_{b}$ or $P_{OH}$

$\begin{aligned} = - \log 10^{- 10} = - (- 10) \log 10 \\ 10 = p_{H} = 14 - p_{OH} = 14 - 10 = 4 \end{aligned}$

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