A) 0.5
B) 1.0
C) 0.05
D) 2.00
Correct Answer: B
Solution :
\[0.05\text{ }M\text{ }{{H}_{2}}S{{O}_{4}}=2\times 0.05\] \[=0.1\,M\,[{{H}^{+}}]\] \[\therefore \] \[[{{H}^{+}}]=1\times {{10}^{-1}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-1}}]=\log 10\] \[pH=1\]You need to login to perform this action.
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