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Question
The pH of 0.3 M acetic acid (CH3COOH) with Ka=1.8×10−5 is :
Take log10(2.32)=0.36
The pH of 0.3 M acetic acid (CH3COOH) with Ka=1.8×10−5 is :
Take log10(2.32)=0.36
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Solution
The correct option is B 2.64
Ka=1.8×10−5
the dissociation of acetic acid is given by
CH3COOH⇌H++CH3COO−Initial: 0.3 0 0At equilibrium: 0.3−x x xHere x is the dissociation of CH3COOH at equilibrium
⇒ Ka=1.8×10−5=[CH3COO−][H+][CH3COOH]=x×x0.3−xSince CH3COOH is a weak acid x<<0.3
∴ 0.3−x≈0.3Ka=1.8×10−5=x20.3 ⇒x=2.32×10−3x=[H+]=2.32×10−3pH=−log10[H+]pH=−log10(2.3×10−3)pH=2.638
Ka=1.8×10−5
the dissociation of acetic acid is given by
CH3COOH⇌H++CH3COO−Initial: 0.3 0 0At equilibrium: 0.3−x x xHere x is the dissociation of CH3COOH at equilibrium
⇒ Ka=1.8×10−5=[CH3COO−][H+][CH3COOH]=x×x0.3−xSince CH3COOH is a weak acid x<<0.3
∴ 0.3−x≈0.3Ka=1.8×10−5=x20.3 ⇒x=2.32×10−3x=[H+]=2.32×10−3pH=−log10[H+]pH=−log10(2.3×10−3)pH=2.638
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