A) \[0.05\]
B) \[0.5\]
C) \[1\]
D) \[2\]
Correct Answer: C
Solution :
Normality of \[{{H}_{2}}S{{O}_{4}}=0.05\times 2\] \[=0.1N\] \[\therefore \] \[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-1}}]\] \[\because \] \[=1\]You need to login to perform this action.
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