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Calculate the pH of 0.01 MH_2SO_4 by assuming complete ionisation. |
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Answer» SOLUTION :`pH=-log[H^(+)]""underset(0.01M)(H_(2)SO_(4))tounderset(0.01xx2)(2H^(+))+SO_(4)^(-2)` `pH=-log[2xx10^(-2)]""[H^(+)]=0.02M` `=1.4990` |
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