10.
A monobasic acid has pKa = 5.2. Calculate the pH at which the ratio between the acid
and the anion (A-) will be 1:50.
a. 3.5
b. 5.7
c. 6.9
d. 7.4
A monobasic acid with pKa 5.2 requires pH 6.9 for a 1:50 acid-to-anion (HA:A⁻) ratio, as calculated using the Henderson-Hasselbalch equation. This MCQ tests buffer equilibrium concepts crucial for CSIR NET Life Sciences biochemistry sections. The correct choice is option c.
Solution Using Henderson-Hasselbalch
The Henderson-Hasselbalch equation relates pH, pKa, and acid-base ratios:
pH = pKa + log10([A⁻]/[HA])
Given pKa = 5.2 and [HA]:[A⁻] = 1:50, so [A⁻]/[HA] = 50.
- log10(50) ≈ 1.70
- pH = 5.2 + 1.70 = 6.9
This matches option c exactly, confirming the anion dominates at this pH above pKa.
Option Analysis
| Option | pH Value | Calculated [A⁻]/[HA] Ratio | Explanation |
|---|---|---|---|
| a | 3.5 | 0.02 (1:50 reverse) | Below pKa; favors protonated acid (HA >> A⁻) |
| b | 5.7 | 3.16 | Slightly above pKa; modest anion excess, not 50-fold |
| c | 6.9 | 50 | Exact match for 1:50 HA:A⁻ ratio |
| d | 7.4 | 158 | Much higher; anion overwhelmingly dominates |
