An 
–NH 3 buffer is supposed to keep the pH of the solution constant within 0.3 pH unit during the reaction.
CH 3 COOCH 3 (aq) + 2H 2 O(aq) ⎯→ CH 3 COO – (aq) + H 3 O + (aq) + CH 3 OH(aq)
If this solution had initial concentrations : [NH 4 + ] 0 = 0.1 M, [NH 3 ] 0 = 0.06 M, [CH 3 COOCH 3 ] 0 = 0.02 M, determine the magnitude of pH change as a result of reaction.
Multiply the magnitude by 10 & add 1 if it is a satisfactory buffer, otherwise subtract 1. Report the answer rounding it off to the nearest whole number.
[K b (NH 3 ) = 1.8 × 10 –5 ]
Text Solution
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Sol. K b (NH 3 ) = 1.8 × 10 –5
CH 3 COOCH 3 (aq) + 2H 2 O (aq) 
CH 3 COO – (aq) + H 3 O + (aq) + CH 3 OH(aq)
[NH 4 + ] 0 = 0.1 M, [NH 3 ] = 0.06 M, [CH 3 COOCH 3 ] 0 = 0.02 M
pOH = pK b + log 
= 4.74 + log 
(pOH) initial = 4.74 + 0.22 = 4.96 ∴ (pH) initial = 9.04
NH 3 (aq) + H + (aq) 
NH 4 + (aq)
0.06 0.02 0.1 mole
0.04 – 0.12 mole
(pOH) final = 4.74 + log 
= 4.74 + log 
= 4.74 + log 3 = 4.74 + 0.48 = 5.22
∴ (pH) final = 8.78
Δ pH = 9.04 – 8.78 = 0.26
Yes this is satisfactory buffer.
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