A) 6.27
B) 2.67
C) 1.67
D) 4.74
Correct Answer: D
Solution :
As we know, number of millimoles = molarity \[\times \] volume in ml \[\therefore \] For 50 ml \[\operatorname{C}{{H}_{3}}COOH\] millimoles \[= \,1 M \times 50 ml\] \[= \,\,50 m moles\] For 50 ml NaOH millimoles \[=\, 0.5 M \times 50 ml\] \[= \,\,25 m moles\] Thus from the following equation, \[\underset{\begin{smallmatrix} 50\,\,m\,\,mol \\ 5\,\,m\,\,mol \end{smallmatrix}}{\mathop{{{\operatorname{CH}}_{3}}COOH}}\,+\underset{\begin{smallmatrix} 25\,\,m\,\,mol \\ \,\,\,\,\,\,\,\,\,\,\,0 \end{smallmatrix}}{\mathop{NaOH}}\,\,\,\to \,\,\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,0 \\ 25\,\,m\,\,mol \end{smallmatrix}}{\mathop{C{{H}_{3}}COONa}}\,\,+\,{{H}_{2}}O\] \[pH=p{{K}_{a}}+\log \,\frac{[salt]}{[acid]}\] \[=\,\,log\left( 1.8\times 1{{0}^{-}}^{5} \right)+\log \frac{25}{25}\] \[pH\,\,=\,\,4.74\]
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