Calculate pH of the following mixtures. Given : K a of CH 3 COOH = 2 × 10 –5 and K b of NH 4 OH = 2 × 10 –5 .
Text Solution
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7 12.4, 4.7,
9.3, (e) 5.3, (f) 7
Sol. H + + OH – 
H 2 O
t = 0 5 mmol 5 mmol
so pH of resulting solution = 7 .
CH 3 COOH + OH – 
CH 3 COO – + H 2 O
t = 0 2.5 mmol 5 mmol
– 2.5 2.5 –
[OH – ] = 
M = 2.5 × 10 –2 M
pOH = 3 – log(2.5) = 1.6
∴ pH = 12.4
CH 3 COOH + OH – 
CH 3 COO – + H 2 O
t = 0 5 mmol 2.5 mmol
2.5 – 2.5
pH = pK a + log 
= pK a = 4.7
NH 4 OH + H + 
NH 4 + + H 2 O
t = 0 5 mmol 2.5 mmol
2.5 – 2.5
pOH = pK b + log 
= 4.7
∴ pH = 9.3
(e) NH 4 OH + H + 
NH 4 + + H 2 O
t = 0 5 mmol 5 mmol 0
0 0 5
pH = 
[14 – 4.7 – log 0.05]
pH = 5.3
(f) NH 4 OH + CH 3 COOH 
CH 3 COONH 4 + H 2 O
t = 0 2.5 mmol 2.5 mmol
– – 2.5
pH = 7 + 
pK a – 
pK b = 7
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