Question

At \(298K\), a certain buffer solution contains equal concentrations of \(X^-\) and \(HX\). If \(K_b\) for \(X^-\) is \(10^{-10}\), what is the pH of this buffer solution?

• A. \(4\)
• B. \(6\)
• C. \(2\)
• D. \(10\)

Hint:

For a buffer having equal acid and salt concentrations, \(pH=pK_a\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the $pH$ of a buffer solution composed of a weak acid ($HX$) and its conjugate base ($X^-$). We are provided with the base dissociation constant ($K_b$) of the conjugate base and told that the concentrations of the acid and base are equal. To find the $pH$, we first need to find the acid dissociation constant ($K_a$) and then use the Henderson-Hasselbalch equation.
Step 2: Key Formula or Approach:

Conjugate Pair Relation: At $298 K$, $K_a \times K_b = K_w = 10^{-14}$.
Henderson-Hasselbalch Equation: $pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$ (for acidic buffers).
p-notation: $pK = -\log(K)$.
Step 3: Detailed Explanation:

Calculate $K_a$: Given $K_b = 10^{-10}$. Using the relation $K_a = \frac{K_w}{K_b}$, we get: \[ K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4} \]
Calculate $pK_a$: $pK_a = -\log(K_a) = -\log(10^{-4}) = 4$.
Analyze Concentration Ratio: The problem states that the concentrations of $X^-$ (salt/base) and $HX$ (acid) are equal. Therefore, $\frac{[X^-]}{[HX]} = 1$.
Apply Henderson-Hasselbalch: \[ pH = 4 + \log(1) \]
Since $\log(1) = 0$, we find that: \[ pH = 4 + 0 = 4 \]
In such scenarios where the concentration of the acid and its conjugate base are identical, the $pH$ simply equals the $pK_a$ of the acid.
Step 4: Final Answer:
The $pH$ of the buffer solution is $4$.