InterviewSolution
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InterviewSolution
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Determine the number of mole of AgI which may be dissolved in `1.0` litre of `1M CN^(-)` solution. `K_(SP)` for AgI and `K_(C)` for `Ag(CN)_(2)^(-)` are `1.2xx10^(-17)M^(2)` and `7.1xx10^(19)M^(-2)` respectively. |
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Answer» Given, `AgI(s) hArr Ag^(o+) (aq) +I^(Theta) (aq)`, `K_(sp) = [Ag^(o+)] [I^(Theta)] = 1.2 xx 10^(-17) ..(1)` `Ag^(o+) (aq) +2CN^(Theta) (aq) hArr [Ag(CN)_(2)]^(Theta) (aq),` `K_(f) = ([Ag(CN)_(2)^(Theta)])/([Ag^(o+)][CN^(Theta)]^(2)) = 7.1 xx 10^(19)....(2)` Let `x` mole of `AgI` be dissolved in `CN^(Theta)` solution then, Now `{:(,AgI(s)+,2CN^(Theta)hArr,[Ag(CN)_(2)^(Theta)]+,I^(Theta)),("Mole before",,1,0,0),("reaction",,,,),("Mole after",,(1-2x),x,x),("reaction,,,,):}` By equaiton (1) and (2), `K_(ep) = K_(sp) xx K_(f)` `K_(ep) = ([Ag(CN)_(2)^(Theta)][I^(Theta)])/([CN^(Theta)]^(2)) =1.2 xx 10^(-17 xx 7.1 xx 10^(19)` `K_(ep) = 8.52 xx 10^(2) ....(3)` `:. K_(ep) = 8.52 xx 10^(2) = (x)/((1-2x)^(2)) = (x^(2))/((1-2x)^(2))` or `(x)/(1-2x) = 29.2` Thus, `x = 29.2 - 58.4x` or `x = 0.49` mol |
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