Question

A monobasic acid HA has pH of 3 in 0.1 M solution at 298 K. What is the pKa of the acid at 298 K?

• A. 3
• B. 4
• C. 5
• D. 6
• E. 2

Hint:

For weak acids, if dissociation is small, take initial concentration $\approx$ equilibrium concentration.

Answer 1

The Correct Option is C

Concept:
• \( \text{pH} = -\log[H^+] \)
• \( K_a = \frac{[H^+][A^-]}{[HA]} \)
• \( \text{p}K_a = -\log K_a \)

Step 1: Find concentration of \(H^+\).
\[ \text{pH} = 3 \Rightarrow [H^+] = 10^{-3} \, \text{M} \]

Step 2: Set up equilibrium.
Initial concentration of HA = \(0.1\) M Dissociation: \[ HA \rightleftharpoons H^+ + A^- \] At equilibrium: \[ [H^+] = [A^-] = 10^{-3}, \quad [HA] \approx 0.1 \]

Step 3: Calculate \(K_a\).
\[ K_a = \frac{(10^{-3})(10^{-3})}{0.1} = 10^{-5} \]

Step 4: Find pKa.
\[ \text{p}K_a = 5 \]