Arrange the following solutions in decreasing order of `[Ag^(o+)]` ion:
I. `1M [Ag(CN)_(2)]^(Theta)`
II. Saturated `AgC1`
III. `1M [Ag(NH_(3))_(2)]^(o+) in 0.1M NH_(3)`
IV. Saturated `AgI`
`(K_(sp) of AgC1 = 10^(-10), K_(sp) of AgI = 8.3 xx 10^(-17) K_(f)` (formation constant) `[Ag(CN_(2))]^(Theta) = 10^(21), K_(f) [Ag(NH_(3))_(2)]^(o+) = 10^(8)`

Options

A. `I gt II gt III gt IV`
B. `II gt III gt I gt IV`
C. `IV gt II gt II gt I`
D. I gt IV gt III gt II`

Answer

B

Explanation

`II gt III gt IV gt I`
From the values of `K_(sp)` and `K_(f)`, it is evident that `AgC1` is more soluble than `AgI` nad `[Ag(CN)_(2)]^(Theta)` is more than `[Ag(NH)_(3)]^(o+)`. Calculate of `[Ag^(o+)]` is each case revels `II gt III gt I gt IV`.
I. `Ag (CN)_(2)^(Theta) rarr Ag^(o+) + 2Coverset(Theta)N`,
`K = (1)/(K_(f)) = ([Ag^(o+)][CN^(Theta)]^(2))/([Ag(CN)_(2)^(Theta)]) = 10^(-21)`
Let `[Ag^(o+)] = x, [CN^(Theta)] =2x`,
`[Ag(CN)_(2)^(Theta)] = 1.0 - x = 1.0`.
` :. K = ((x)(2x)^(2))/(1.0) = 10^(-21) = 4x^(3)`
`rArr x = 6.3 xx 10^(-8)m`.
II. `AgC1 hArr Ag^(o+) + C1^(Theta)`,
`K_(sp) = [Ag^(o+)] [C1^(Theta)] = 10^(-10)`
`:. x^(2) = 10^(-10), rArr x = 10^(-5)M`.
III. `{:(,[Ag(NH_(3))_(2)]^(o+)rarr,Ag^(o+)+,2NH_(3),),("Initial",1.0,0,0.1,),("Final",1.0-x,x,2x+0.1~~0.1M,):}`
`K = (1)/(K_(f)) = ([Ag^(o+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(o+)]) = 10^(-8)`
Let `[Ag^(o+)] = x, [NH_(3)] = (2x + 0.1) ~~ 0.1`
`[Ag(NH_(3))_(2)^(o+)] = 1.0 -x =1.0`
`K = (x xx 0.1 xx 0.1)/(1.0) = 10^(-8) rArr x = 10^(-6)M`
IV. `AgI hArr Ag^(o+) + I^(Theta)`,
`x^(2) = 8.3 xx 10^(-17) rArr x rArr 2.8 xx 10^(-9)`.
II `(AgC1 = 10^(-5)) gt III [Ag(NH_(3))_(2)^(o+)) = 10^(-6)] gt I [(Ag(CN)_(2)^(Theta)) = 6.3 xx 10^(-9)] gt IV (AgI = 2.8 xx 10^(-9))`
`:. IIgt III gt I gt IV`