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Question

Calculate the pH at the equivalence point of the titration between 0.1 M CH3COOH (25 mL) with 0.05 M NaOH.
(Given pKa for CH3COOH=4.74; log(3)=0.48).

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Solution

Milli equivalents of CH3COONa formed = 0.1×25
CH3COOH+NaOH CH3COONa+H2O

At the equivalnce point,
Milli equivalents of NaOH= Milli equivalents of CH3COOH
0.05×V=0.1×25
V= 50 mL
The total volume of the solution = 75 mL
Concentration of CH3COONa = 0.1×2575=130
pH=7+12pKa+12logC
(for a salt of W.A. + S.B.)
=7+12×4.74+12log 130
=8.63


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