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Calculate the solubility of AgCN in a buffer solution of pH 3*00. K_(sp) for AgCN is 2.2xx10^(-16)and K_(a) forHCN is 6.2xx10^(-12). |
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Answer» Solution :`AgCN rarr AG^(+) CN^(-)` pH = 3.0means `[H^(+)] = 10^(-3) M ` `CN^(-)` ions now combine with the `H^(+)` ion to form HCN but `[H^(+)]` remains almost CONSTANT because we have buffer solution . Now `HCN hArr H^(+) + CN^(-)` `K_(a) = ([H^(+)][CN^(-)])/([HCN]) or ([HCN])/([CN^(-)]) = ([H^(+)])/(K_(a)) = (10^(-3))/(6.2xx10^(-10)) = 1.6 xx 10^(6) ` ..(i) Suppose the solubility in the buffer solution is 'x' mol `L^(-1)`. Then `x=[Ag^(+)]=[CN^(-)]+ [HCN] ~= [HCN]` (as from eqn. (i) `[CN^(-)] = [ HCN]//(1.6xx10^(6))` is negligible in comparison to [HCN]) `:. [CN^(-)] = ([HCN])/(1.6xx10^(6))` `K_(sp) = [ Ag^(+)][CN^(-)]=x xx (x)/(1.6xx10^(6))=2.2xx10^(-16)` (Given) or `x^(2) = 3.52 xx 10^(-10)"" x=1.9xx10^(-5) "mol" L^(-1)` |
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