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)?

AgBr_((s)) + 2S_(2)O_(3(aq))^(2-) hArr [Ag(S_(2)O_(3))_(2)]_((aq))^(3-

1.	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)?
Answer» `0.5 M` `0.45 M` `0.045 M` `0.65 M` SOLUTION :`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`