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a. Calculate [Ag^(o+)] in a solution of ...

a. Calculate `[Ag^(o+)]` in a solution of `[Ag(NH_(3))_(2)^(o+)]` prepared by adding `1.0 xx 10^(-3)mol AgNO_(3)` to `1.0L of 1.0M NH_(3)` solution `K_(f) Ag(NH_(3))_(2)^(o+) = 10^(8)`.
b. Calculate `[Ag^(o+)]` which is in equilibrium with `0.15M [Ag(NH_(3))_(2)]^(o+)` and `1.5 NH_(3)`.

Text Solution

Verified by Experts

The correct Answer is:
A, B
i.Since `K_(f)` is so large, most of `Ag^(o+)` will form complex ions.
`{:("Initial mol"),("Reacted"),("At equilibrium")]{:(Ag^(o+)+,2NH_(3)hArr,Ag(NH_(3))_(2)^(o+),,,),(10^(-3),0.1,0,,),(-x,-2x,0,,),(10^(-3)-x~~10^(-3),0.1-2x,10^(-3)-x~~10^(-3),,),(,0.1-2(10^(-3)),,,),(,~~0.098,,,):}`
`K_(f) =([Ag(NH_(3))_(2)^(o+)])/([Ag^(o+)][NH_(3)]^(2))`
`10^(8) = (10^(-3))/((x)(0.098)^(2)) :. x = [Ag^(o+)] = 1 xx 10^(-9)M`
ii. `10^(8) = (0.15)/((x)(1.5)^(2)) :. x = [Ag^(o+)] = 7 xx 10^(-10)M`.
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