A certain buffer solution contains equal conc | vedclass

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Question

(A) For the base $X^{-}$,the equilibrium is: $X^{-} + H_2O ightleftharpoons OH^{-} + HX$. Given $K_b = 10^{-10}$. We know that $K_a 	imes K_b = K_w

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$4$

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(A) For the base $X^{-}$,the equilibrium is: $X^{-} + H_2O ightleftharpoons OH^{-} + HX$. Given $K_b = 10^{-10}$. We know that $K_a 	imes K_b = K_w = 10^{-14}$. Therefore,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$. For a buffer solution,the Henderson-Hasselbalch equation is: $pH = pK_a + \log\frac{[salt]}{[acid]}$. Since the concentrations of $X^{-}$ (salt) and $HX$ (acid) are equal,$[X^{-}] = [HX]$. Thus,$pH = pK_a + \log(1) = pK_a$. $pH = -\log(K_a) = -\log(10^{-4}) = 4$.

$A$ certain buffer solution contains equal concentrations of $3.9 \times 10^{-5} \ M$ $X^{-}$ and $HX$. The $K_b$ for $X^{-}$ is $10^{-10}$. The $pH$ of the buffer is:

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