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Caculate the pH of 0.001 N `H_(2)SO_(4)` solution. |
Answer» Correct Answer - 3 `H_(2)SO_(4)` is a strong acid and it is completely ionised in the aqueous solution as : `H_(2)SO_(4)overset((aq))(to) H_(3)O^(+) +SO_(4)^(2-) (aq)` As `H_(2)SO_(4)` is a dibasic acid , the molarity of the solution is half of its normality or 0.0005 M `[H_(3)O^(+)] =2xx 0.0005 =0.001 M =10^(-3) M` `pH =- log [H_(3)O^(+)]=- log [(10^(-3))] =(-3) (-log 10) =3` |
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