Q95 / 180 · NEET 2026 (Cancelled) · Chemistry
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 ?
- A$2$
- Correct answerB$4$
- C$6$
- D$10$
Solution
Given $K_b$ for $X^{-} = 10^{-10}$.
For a conjugate acid-base pair at $298$ K, $K_a \times K_b = K_w = 10^{-14}$.
$K_a = \dfrac{10^{-14}}{10^{-10}} = 10^{-4}$
The $pK_a$ of the weak acid $HX$ is:
$pK_a = -\log(K_a) = -\log(10^{-4}) = 4$
Using the Henderson-Hasselbalch equation for an acidic buffer:
$\text{pH} = pK_a + \log \left( \dfrac{[X^{-}]}{[HX]} \right)$
Since the concentrations of $X^{-}$ and $HX$ are equal, $[X^{-}] = [HX]$.
$\text{pH} = 4 + \log(1) = 4 + 0 = 4$
Answer: $4$
For a conjugate acid-base pair at $298$ K, $K_a \times K_b = K_w = 10^{-14}$.
$K_a = \dfrac{10^{-14}}{10^{-10}} = 10^{-4}$
The $pK_a$ of the weak acid $HX$ is:
$pK_a = -\log(K_a) = -\log(10^{-4}) = 4$
Using the Henderson-Hasselbalch equation for an acidic buffer:
$\text{pH} = pK_a + \log \left( \dfrac{[X^{-}]}{[HX]} \right)$
Since the concentrations of $X^{-}$ and $HX$ are equal, $[X^{-}] = [HX]$.
$\text{pH} = 4 + \log(1) = 4 + 0 = 4$
Answer: $4$
Hence, the correct answer is option: 'B'.