At 298 K, a certain buffer solution contains equal concentrations of $X^{-}$ and $HX$. $K_b$ for $X^-$ is $10^{-10}$. What is the pH of this buffer solution ?
Calculating Buffer pH with Equal Concentrations
The problem asks for the pH of a buffer solution containing equal concentrations of a weak acid ($HX$) and its conjugate base ($X^-$). We are given the base dissociation constant ($K_b$) for the conjugate base $X^-$.
Buffer Chemistry Basics
A buffer solution maintains a relatively stable pH. When the concentrations of the weak acid ($HX$) and its conjugate base ($X^-$) are equal, the pH of the buffer solution is equal to the $pK_a$ of the weak acid.
The Henderson-Hasselbalch equation describes the pH of a buffer:
$pH = pK_a + \log \frac{[X^{-}]}{[HX]}$In this specific case, $[X^{-}] = [HX]$, so the ratio $\frac{[X^{-}]}{[HX]} = 1$. The equation simplifies to:
$pH = pK_a + \log(1)$ $pH = pK_a + 0$ $pH = pK_a$Determining pKa from Kb
We need to find the $pK_a$ of the weak acid $HX$. We are given $K_b$ for the conjugate base $X^-$, which is $10^{-10}$. We know the relationship between the acid dissociation constant ($K_a$), the base dissociation constant ($K_b$), and the ion product of water ($K_w$):
$K_a \times K_b = K_w$At 298 K, $K_w = 1.0 \times 10^{-14}$. We can find $K_a$:
$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{10^{-10}}$ $K_a = 1.0 \times 10^{-4}$Now, we calculate $pK_a$:
$pK_a = -\log(K_a)$ $pK_a = -\log(1.0 \times 10^{-4})$ $pK_a = -(-4)$ $pK_a = 4$Final pH Calculation
Since $pH = pK_a$ when the concentrations of the acid and its conjugate base are equal:
$pH = 4$Therefore, the pH of the buffer solution is 4.
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