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Equal volumes of `0.02M AgNO_(3)` and `0.01M HCN` are mixed. Calculate `[Ag^(o+)]` in solution after attaining equilibrium. `K_(a) HCN = 6.2 xx 10^(-10)` and `K_(sp)` of `AgCN = 2.2 xx 10^(-16)`. |
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Answer» Solution: `{:(,Ag^(o+)+,HCNrarr,AgCN+,H^(o+)),("Initial",(0.02xxV)/(2V),(0.02xxV)/(2V),0,0),("Final",0,0,0.01,0.01):}` `AgCN hArr Ag^(o+) +CN^(Theta)` `:. K_(sp) = 2.2 xx 10^(-16) = [Ag^(o+)] [CN^(Theta)]` `HCN hArr H^(o+) + CN^(Theta) K_(a) = 6.2 xx 10^(-10) = ([H^(o+)][CN^(Theta)])/([HCN])` Now, soluble `CN^(Theta)` formed will hydrolyses as `CN^(Theta) + H_(2)O hArr HCN + overset(Theta)OH` `:. [Ag^(o+)]` of soluble `AgCN = CN^(Theta)` left after hydrolysis `+HCN` formed after hydrolysis `:. [(2.2 xx 10^(-10))/([CN^(Theta)])] = [CN^(Theta)] + ([H^(o+)][CN^(Theta)])/(6.2 xx 10^(-10))` `= [CN^(Theta)] + (10^(-2)[CN^(Theta)])/(6.2 xx 10^(-10)) [because CN^(Theta)lt lt lt (10^(-2)[CN^(Theta)])/(6.2xx10^(-10))]` or `[CN^(Theta)]^(2) = (2.2xx10^(-16)xx6.2xx10^(-10))/(10^(-2))` `:. [CN^(Theta)] = 3.7 xx 10^(-12)` `:. [Ag^(o+)] = (K_(sp))/([CN^(Theta)]) = (2.2 xx 10^(-16))/(3.7xx10^(-12)) = 5.96 xx 10^(-5)M` |
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