InterviewSolution
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InterviewSolution
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How much `AgBr` could dissolve in `1.0L` of `0.4M NH_(3)`? Assume that `Ag(NH_(3))_(2)^(o+)` is the only complex formed. Given: the dissociation constant for `Ag(NH_(3))_(2)^(o+) hArr Ag^(o+) xx 2NH_(3)`, `K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`. |
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Answer» Let solubility of `AgBr` be `xM`. Thus, `[Br^(Θ)] = xM`, but `[Ag^(o+)] != xM` since it will react with `NH_(3)` to form a complex and thus, its concentration will be decied by the disscoiation of complex. So, let `[Ag^(o+)] = yM`. `AgBr hArr Ag^(o+)(aq) +Br^(Θ) (aq)` `rArr K_(sp) = [Ag^(o+)] [Br^(-)] = yx = 5.0 xx 10^(-13)` Since the formation constant `(K_(f))` of the complex is very high, assume that whole of `Ag^(o+)` formed is consumed. `{:(Ag^(o+)+,2NH_(3)rarr,Ag(NH_(2))_(2).^(oplus),,),(x,0.4,,,),(,0.4-2x,x,,):}` `{:(Ag(NH_(3))_(2)^(o+)hArr,Ag^(o+)(aq),+2NH_(3),(K_(d)=5.0xx10^(-13)),,),(x,,0.4-2x,,),(x-y,,y0.4-2x+2y,,):}` Thus, `K_(d) = ([Ag^(oplus)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(oplus)])=(y(0.4-2x+2y)^(2))/(x-y) =6xx10^(-8)` Assuming `x-y ~~` since `K_(d)` is low and `x lt lt 0.4`, we get : `K_(d) = (y(0.4)^(2))/(x)` Solving for `x` : `x = 1.15 xx 10^(-3)M` (Verify the approximation yourself). |
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