A) 2
B) 1
C) 1.7
D) 0.3
Correct Answer: B
Solution :
Normality of \[{{H}_{2}}S{{O}_{4}}=\frac{Wt.}{Eq.wt.}\] \[=\frac{0.49}{49}\](Eq. wt. of \[{{H}_{2}}S{{O}_{4}}=49\]) \[=0.1={{10}^{-1}}\] This normality represents the hydrogen ion concentration. So, \[[{{H}^{+}}]={{10}^{-1}}\] Then, \[pH=\log \frac{1}{[{{H}^{+}}]}=\log \frac{1}{{{10}^{-1}}}\] \[=\log \,10\,\,=1\]You need to login to perform this action.
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