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Calculate the pH of 0.1 M of H_(2)SO_(4) (concentration of hydrogen = 0.1xx2=0.2). |
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Answer» Solution :`PH=-log_(10)[H^(+)]-log_(10)[0.1xx2]=-log_(10)[0.2]-log_(10)[2xx10^(-1)]=1-log2` pH = 1 - 0.3010 = 0.699. |
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