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Calculate the pH of 0.125M of H_(2)SO_(4). |
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Answer» SOLUTION :`H_(2)SO_(4)to2H^(+)+SO_(4)^(2-)` `[H^(+)]=2[H_(2)SO_(4)]=2xx0.125=0.25` `pH=-log_(10)[H^(+)]=-log_(10)[0.25]=-log_(10)[2.5xx10^(-1)]` `pH=-log2.5-log10^(-1)=-0.3979+1=0.6021`. |
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