Question

The solubility of solid silver chromate, `Ag_(2)Cro_(4)`, is determined in three solvents `K_(sp)` of `Ag_(2)CrO_(4) = 9 xx 10^(-12)`
I. pure water II. `0.1M gNO_(3)`
III. `0.1M Na_(2)CrO_(4)`
Predict the relative solubility of `Ag_(2)CrO_(4)` in the three solvents.
A. `I =II =III`
B. `I lt II lt III`
C. `II = III lt I`
D. `II lt III lt I`

Answer 1

Correct Answer - D
i. `{:(Ag_(2)CrO_(4)hArr,2Ag^(o+)+,CrO_(4)^(2-)("in"H_(2)O)),(,2S,S):}`
`4S^(3) = K_(sp)`,
`S_(H_(2)O) = ((K_(sp))/(4))^(1//3) = ((9 xx 10^(-12))/(4))^(1//3)`
`=(2.25)^(1//3) xx 10^(-4)M`
ii. On adding `0.1M AgNO_(3)`
`{:((AgNO_(3)rarr,Ag^(o+)+,NO_(3)),(,0.1M,0.1M)):}`
Due to common ion `(Ag^(o+))`, the solubility of `Ag_(2)CrO_(4)` os supressed.
Total `[Ag^(o+)] (2S +0.1) ~~ 0.1`
Let the solubility is `S_(1)`.
`:. S_(1) =(K_(sp))/([Ag^(o+)]_("externally added")^(2))`
`= (9xx10^(-12))/(0.1xx0.1) = 9x 10^(-10)M`
iii. On adding `0.1M Na_(2)CrO_(4)`
`{:((Na_(2)CrO_(4)rarr,2Na^(o+)+,CrO_(4)^(2-)),(,2xx0.1M,0.1M)):}`
Again due to common ion `(CrO_(4)^(2-))`, the solubility of `Ag_(2)CrO_(4)` is supressed.
Let the solubility is `S_(2)`.
Total `[CrO_(4)^(2-)](S+0.1) ~~ 0.1`
`{:(Ag_(2)CrO_(4)hArr,2Ag^(o+)+,CrO_(4)^(2-),,),(,2S_(2),(S_(2)+0.1)~~0.1,,):}`
`K_(sp) = [2S_(2)]^(2) [0.1]`
`:. S_(2) = ((K_(sp))/(4xx0.1))^(1//2) = ((9xx10^(-12))/(4xx0.1))^(1//2)`
`= 22.5 xx 10^(-6) = 2.25 xx 10^(-7)`
Hence, the solubility order is `II lt III lt I`.
Note: In such problem, instead of solving the exact value, try to get the approximate data to chech the comparative solubilities).