A) 2
B) 3
C) 2.301
D) 3.6990
Correct Answer: A
Solution :
Key Idea \[{{H}_{2}}S{{O}_{4}}\] is a dibasic acid. Hence, 0.005 M \[{{H}_{2}}S{{O}_{4}}=2\times 0.005\,N=0.01\,N\] \[[{{H}^{+}}]=2\times 0.005\,N=0.01\,N\] \[pH=-\log \,\,[{{H}^{+}}]\] \[pH=-\log \,\,[0.01]\] \[pH=-\log \,[1\times {{10}^{-2}}]\] = 2
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