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The pH of a solution of AgCI(s) with solubility product `1.6 xx 10^(-10)` in 0.1 M Nacl solution would be :A. `1.26 xx 10^(-5) M`B. `1.6 xx 10^(-9) M`C. `1.6 xx 10^(-11) M`D. Zero |
Answer» Correct Answer - B Let S be the solubility of AgCI in moles per litre. `underset(S)(AgCI (aq)) hArr underset(S)(Ag^(+))(aq) + underset((S+0.1))(CI^(-)(aq))` (0.1 NaCI solution also provides 0.1 M `CI^(-)` ions) `Ks_(p) =[Ag^(+)] [CI^(-)]` `1.6 xx 10^(-10) =S [S +0.1] = S(0.1)` `("Because " S lt lt lt 0.1)` `S=(1.6xx 10^(-10))/(0.1) =1.6 xx 10^(9) M` |
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