A) 3.37
B) 4.37
C) 4.74
D) 0.474
Correct Answer: A
Solution :
\[{{H}^{+}}=c\,.\,\alpha =\sqrt{{{K}_{a}}\,.\,c}\] \[pH=-\log {{\left( \sqrt{{{K}_{a}}.\,c} \right)}^{1/2}}\]\[=\frac{1}{2}[-\log {{K}_{a}}-\log c]\] \[=\frac{1}{2}[4.74-\log {{10}^{-2}}]\]\[=\frac{1}{2}[4.74+2]=3.37\].
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