A) 3.74
B) 4.74
C) 5.74
D) 6.74
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} Initial\,20\,\times \,0.1\,=\text{2}\,\text{mmol} \\ \text{At}\,\text{time}\,\text{t}\,\text{(2-1)}\,\text{=}\,\text{mmol} \end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,+\,\,\,\,\underset{\begin{smallmatrix} 10\,\times \,0.1=1\,\text{mmol} \\ (1-1)\,=\,0\,\text{mmol} \end{smallmatrix}}{\mathop{NaOH}}\,\] \[\xrightarrow{{}}\underset{\text{1}\,\text{mmol}}{\mathop{C{{H}_{3}}COONa}}\,\,+{{H}_{2}}O\] From Hendersons equation, \[pH=p{{K}_{a}}+\log \frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}\] \[=4.74+\log \frac{1}{1}\] \[=4.74\]
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