A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5 ) is titrated with 0.40 M NaOH . Calculate the pH after the addition of 15.0 mL of NaOH . Express your answer numerically.
CH3COOH millimoles = 52 x 0.35 = 18.2
NaOH millimoles = 15 x 0.4 = 6
CH3COOH + NaOH ---------------> CH3COONa + H2O
18.2 6 0 0 ---------------> before reaction
-6 -6 +6 +6 -----------------> changed
12.2 0 6 6 -----------------> after reaction
now mixture contain acid and salt it forms buffer
pKa = -log Ka = -log(1.8 x10^-5)= 4.74
pH = pKa + log[salt/acid]
pH = 4.74 + log[6 /12.2]
pH = 4.43
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