The concentration of `Ag^(o+)` ions in a saturated solution of `Ag_(2)C_(2)O_(4)` is `2.0 xx 10^(-4)M`. Calculate the solubility of `Ag_(2)C_(2)O_(4)` in a solution which is `0.01M` in `H_(2)C_(2)O_(4)`.

Answer

A

Explanation

`{:(Ag_(2)C_(2)O_(4)hArr,2Ag^(o+)+,C_(2)O_(4)^(2-),,),(,2S,S,,):}`
`2S = 2.0 xx 10^(-4)`
`:. S = (2.0 xx 10^(-4))/(2) = 1.0 xx 10^(-4)M`
`K_(sp) = [Ag^(o+)]^(2) [C_(2)O_(4)^(2-)]`
`= (2.0 xx 10^(-4))^(2) (1.0 xx 10^(-4)) = 4 xx 10^(-12)`
Solubility in `0.01 M H_(2)C_(2)O_(4)`:
`{:(H_(2)C_(2)O_(4)hArr,2H^(o+)+,C_(2)O_(4)^(2-),,),(,2xx0.01M,0.01M,,):}`
Due to the common is effect `(C_(2)O_(4)^(2-))`, the solubility of `Ag_(2)C_(2)O_(4)` is supressed.
Using the formuls:
`S_(new) = ((K_(sp))/(4C))^(1//2) = ((cancel4xx10^(-12))/(cancel4xx0.01))=10^(-5)M`
Alternatively:
let `S'` is the solubility of `Ag_(2)C_(2)O_(4)`.
`{:(Ag_(2)C_(2)O_(4)hArr,2Ag^(o+)+,C_(2)O_(4)^(2-),,),(,2S'+,(S'+0.01),,),(,,~~0.01M,,):}`
`K_(sp) = [2S']^(2) (0.01)`
`S = ((K_(sp))/(4xx0.01))^(1//2) = ((4 xx 10^(-12))/(4xx0.01))^(1//2) = 10^(-5)M`.