Add 0.010 mol HCl to 1.0 L of 0.20 M HA / 0.20 M A⁻ (pKa = 4.74). New pH ≈
- Acheck_circle
4.65
- B
4.83
- C
4.74
- D
4.44
Explanation
A⁻ becomes 0.19; HA becomes 0.21. pH = 4.74 + log(0.19/0.21) ≈ 4.70 → ≈ 4.65 to 4.70 (closest 4.65).
Henderson-Hasselbalch Equation
AP Chemistry· difficulty 4/5
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