A) 2.5
B) 1.5
C) 1.0
D) 2.0
E) none of these
Correct Answer: D
Solution :
Firstly we convert molarity into normality. \[Normality\times eq.\text{ }wt.=molarity\times mol.\text{ }wt.\] \[N\times 49=0.005\times 98\] \[N={{10}^{-2}}\] \[N=[{{H}^{+}}]={{10}^{-2}}\] \[pH=\log \frac{1}{[{{H}^{+}}]}\] \[=\log {{10}^{2}}\] \[=2\log 10=2\]
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