A) Zero
B) One
C) 2
D) 0.3010
E) \[-0.3010\]
Correct Answer: E
Solution :
\[M\,{{H}_{2}}S{{O}_{4}}=2N{{H}_{2}}S{{O}_{4}}\] \[[{{H}^{+}}]=N.\alpha \] \[=2\times 1=2\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [2]\] \[pH=-0.3010\]
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