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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
A
4
B
10
C
7
D
6
Detailed Solution
$B^-$ is the conjugate base of the weak acid HB; $pK_b = -\log(10^{-10}) = 10$
Henderson equation for the base: $pOH = pK_b + \log\frac{[HB]}{[B^-]}$
Since $[B^-] = [HB]$, the log term is $\log1 = 0$, so $pOH = pK_b = 10$
$pH = 14 - pOH = 14 - 10$
pH = 4
Check: $K_a$ of HB = $\frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}$, so $pH = pK_a = 4$.
Henderson equation for the base: $pOH = pK_b + \log\frac{[HB]}{[B^-]}$
Since $[B^-] = [HB]$, the log term is $\log1 = 0$, so $pOH = pK_b = 10$
$pH = 14 - pOH = 14 - 10$
pH = 4
Check: $K_a$ of HB = $\frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}$, so $pH = pK_a = 4$.