`AgBr_((s)) + 2S_(2)O_(3(aq))^(2-) hArr [Ag(S_(2)O_(3))_(2)]_((aq))^(3-) + Br_((aq))^(-)`
[`Ksp(AgBr) = 5 xx 10^(-13)`,
`K_(f) [Ag(S_(2)O_(3))_(2)]^(3-) = 5 xx 10^(13)`]
What is the molar solubility of `AgBr` in `0.1 M Na_(2)S_(2)O_(3)`?

Options

A. `0.5 M`
B. `0.45 M`
C. `0.045 M`
D. `0.65 M`

Answer

C

Explanation

`Ag_((aq))^(+) + 2S_(2)O_(3(aq))^(2-)hArr[Ag(S_(2)O_(3))_(2)]_((aq))^(3-)`
`AgBr_((s)) hArr Ag_((aq))^(+) + Br_(aq)^(-)`
`ArBr_((s)) + underset((0.1-2x))(2S_(2)O_(3(aq))^(2-))hArr[Ag(S_(2)O_(3))_(2)]_((aq))^(3-)+Br_((aq))^(-)`
`Ksp xx K_(f) = 25`
As value of `K_(f)` is very higher so we can assume almost `Ag^(+)` inverts into complex
`25 = (x^(2))/((0.1 - 2x)^(2)), 5 = (x)/(0.1-2x), x = 0.045 M`