For weak acid HA with Ka = 10⁻⁵, [H+] in 0.1 M solution is approximately:
A10⁻⁴
B10⁻⁵
C10⁻³
D10⁻⁶
Answer & Solution
Correct answer: C. 10⁻³
For weak acid, [H+] ≈ √(Ka × C) = √(10⁻⁵ × 0.1) = √(10⁻⁶) = 10⁻³ M. So pH ≈ 3.
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