A) 3.37
B) 1.37
C) 4.74
D) 8.01
Correct Answer: B
Solution :
pH of \[\text{0}\text{.01}\,\text{M}\,\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\] \[pH=\frac{1}{2}[p{{K}_{a}}-log\,C]\] \[=\frac{1}{2}(4.74-log\,0.01)\] \[=\frac{1}{2}(4.74\,+\,2)\] \[=3.37\] When \[\text{0}\text{.01}\,\text{mol}\,\text{C}{{\text{H}}_{\text{3}}}\text{COONa}\]is added to it, it is now a buffer and \[[[C{{H}_{3}}COONa]=0.01]\,M.\] Now from \[pH+p{{K}_{a}}+\log \frac{[C{{H}_{3}}COO]}{[C{{H}_{3}}COOH]}\] \[=4.74+\log \frac{0.01}{0.01}=4.74\] \[\therefore \] Change in \[pH=4.74-3.37=1.37\]You need to login to perform this action.
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