A) 3.37
B) 1.37
C) 4.74
D) 8.01
Correct Answer: B
Solution :
\[pH\] of \[0.01\,M\,C{{H}_{3}}COOH\] \[pH=\frac{1}{2}\,[p{{K}_{a}}-\log \,C]\] \[=\frac{1}{2}\,(4.74-\log \,\,0.10)\] \[=\frac{1}{2}\,(4.74+2)=33.7\] When 0.01 mol \[C{{H}_{3}}COONa\] is added to it, it is now a buffer and \[[[C{{H}_{3}}COONa]=0.1]\,\,M\]. 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 \[pH=4.74-3.37=1.37\]
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