Calculate the pH of 0.01 M acetic acid solution. Ka=1.8×10−5.
Solution — Step by Step
CH3COOH⇌CH3COO−+H+.
Initial: C=0.01 M, products =0.
Change: −Cα, +Cα, +Cα.
At equilibrium: C(1−α), Cα, Cα.
Ka=C(1−α)Cα⋅Cα=1−αCα2
For weak acids, α≪1, so 1−α≈1:
Ka≈Cα2⟹α=Ka/C.
α=1.8×10−5/0.01=1.8×10−3≈0.0424.
[H+]=Cα=0.01×0.0424=4.24×10−4 M.
pH=−log(4.24×10−4)=4−log4.24=4−0.627=3.37.
Final answer:pH≈3.37.
Why This Works
Acetic acid is weak, so it dissociates only partially. The factor α (degree of dissociation) is the key — for weak acids at moderate concentrations, α<0.05, so the 1−α≈1 approximation is valid.
Cross-check: our α=0.042<0.05 — the approximation holds. If α>0.1, you’d need to solve the full quadratic.
Alternative Method
Use the shortcut: pH=21(pKa−logC)=21(4.74−log0.01)=21(4.74+2)=3.37. Same answer in one line — useful for MCQs.
Common Mistake
Students often use the strong-acid formula pH=−logC for weak acids — that gives pH=2 for 0.01 M, which is wildly off. Strong acid formula assumes complete dissociation; for weak acid, you must use Ka. Always check the acid’s strength first.
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