A) 3.37
B) 4.37
C) 4.74
D) 0.474
Correct Answer: A
${H^{+}}=c,.,\alpha =\sqrt{K_{a},.,c,.,10^{pH}}$
$ pH=-\log {\sqrt{K_{a}.,c}}$
$ =\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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