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Grade 12th passPhysical Chemistry

To a 100 mL solution of 0.1 M CH3COONa and 0.1 M CH3COOH, 0.4 gm of solid NaOH was added . Assuming volume remains constant, the change in pH value will be [Given that pKa (CH3COOH) = 4.74 ]
​A. 0.125
​B. 0.225
C. 0.01
​D. 0.0872
Ans ; – (D) 0.0872

Profile image of Yasir Rather
7 Years agoGrade 12th pass
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1 Answer

Profile image of Sahil Yadav
6 Years ago
A mixture containing 100 ml of 0.1 M acetic acid and 50 ml of 0.1 M NaOH is acidic buffer soluion. Its pH is given by the expression pH=pKa+logacidsalt.
Substitute values in the above expression.
5=pKa+log0.1×0.050.1×0.05 or pKa=5.
When 100 ml of 0.05 M NaOH  is added, the acid is completely neutralized and the solution contains salt sodium acetate.
The expression for the hydrogen ion concentration of the salt of weak acid and strong base is pH=21(pKw+pKa+logc).
Substitute values in the above expression.
pH=21(14+5+log(0.250.1))=8.8.
Hence, the change in pH is ΔpH=8.85=3.8.