Find the pH of a 0.1 M solution of acetic acid (CH3COOH). Given Ka=1.8×10−5.
Solution — Step by Step
CH3COOH⇌CH3COO−+H+.
Initial: 0.1,0,0. At equilibrium: 0.1−x,x,x.
Ka=x2/(0.1−x)≈x2/0.1 (since x≪0.1).
x2=Ka×0.1=1.8×10−6.
x=1.8×10−6=1.34×10−3 M.
x/0.1=1.34%, well under 5%, so the approximation is valid.
pH=−log(1.34×10−3)=3−log1.34≈2.87.
Final answer: pH ≈2.87.
Why This Works
The equilibrium constant Ka measures how strongly an acid dissociates. For weak acids, only a tiny fraction ionises, so we approximate C−x≈C. This linearises the math without sacrificing accuracy.
The shortcut formula pH=21(pKa−logC) comes from this analysis. For our case, pKa=4.74, log0.1=−1, giving pH=(4.74+1)/2=2.87. Same answer.
Alternative Method
Use the shortcut directly: pH=21(pKa−logC)=21(4.74+1)=2.87. One line, no setup.
The shortcut pH=21(pKa−logC) for weak monoprotic acids saves 30 seconds per question. For NEET, that’s gold.
Common Mistake
Treating acetic acid as a strong acid and computing pH=−log(0.1)=1. Strong acids give that, but acetic acid is weak — pH is much higher than 1.
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