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How much AgBr could dissolve in 1.0L of ...

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+) + 2NH_(3)`,
`K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`.

Text Solution

Verified by Experts

The correct Answer is:
A, C
`AgBr hArr Ag^(+) + Br^(Theta)`,
`Ag^(o+) + 2NH_(3) hArr Ag(overset(o+)NH_(3))_(2)`
Let `x =` Solubility, Then,
From charge neutrality
`x = [Br^(Theta)] = [Ag^(o+)] + [Ag(NH_(3))_(2)^(o+)]`
Because `[NH_(3)] gt gt [Br^(Theta)]`,
`K_(f) =([Ag(NH_(3))_(2)^(o+)])/([Ag^(o+)][NH_(3)]^(2))`
`1 xx 10^(8) = ([Ag(NH_(3))_(2)^(o+)])/([Ag^(o+)][0.4]^(2))`
[Since most of `Ag^(o+)` is in the form of complex]
`x = [Br^(Theta)] = [Ag(NH_(3)^(o+))_(2)]`
`:. [Ag^(o+)] = (x)/(10^(8) xx 1.6 xx 10^(-1)) = (x)/(1.6 xx 10^(7))`
`K_(sp) Ag Br = [Ag^(o+)] [Br^(Theta)] = (x)/(1.6 xx 10^(7)) xx (x)`
`= (x^(2))/(1.6 xx 10^(7)) = 5.0 xx 10^(-13)`
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