Ionic EquilibriumHard
Question
The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be :
(pKa for CH3COOH = 4.74)
(pKa for CH3COOH = 4.74)
Options
A.4.74
B.8.87
C.9.10
D.8.57
Solution
CH3COOH + OH- CH3COO- + H2O
t = 0 20 20
t = eq - - 20
So, [CH3COO-] =
= 0.1 M
pH = 7 +
pKa + 
log C = 7 + 2.37 + 
log 10-1 = 7 + 2.37 - 0.5 = 8.87
t = 0 20 20
t = eq - - 20
So, [CH3COO-] =
= 0.1 MpH = 7 +
pKa + log C = 7 + 2.37 + log 10-1 = 7 + 2.37 - 0.5 = 8.87Create a free account to view solution
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