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Assuming `H_(2)SO_(4)` to be completely ionised the `pH` of a `0.05M` aqueous of sulphuric acid is approximatelyA. `0.01`B. `0.005`C. `2`D. `1` |
Answer» Correct Answer - D `0.05 M_(H_(2)SO_(4)) = 2xx0.05 N_(H_(2)SO_(4)) = 0.1N` `:. [H_(3)O^(o+)] = [0.1] = 10^(-1)` `-log [H_(3)O^(o+)] =- log [10^(-1)] = 1` `pH = 1` |
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