The following solutions are mixed: `500mL of 0.01 M AgNO_(3)` and `500mL` solution that was both `0.01M` in `NaCI` and `0.01M` in `NaBr`. Given `K_(sp) AGCI = 10^(-10), K_(sp) AgBr = 5 xx 10^(-13)`.
Calculate the `[Br^(Theta)]` in the equilibrium solution.

Options

A. `2.0 xx10^(-8)M`
B. `2.0 xx 10^(-10)M`
C. `2.5 xx 10^(-5)M`
D. `2.5 xx 10^(-8)M`

Answer

C

Explanation

i. On dilution (equal volume) and if there were no precipitation. `[NO_(3)^(Theta)] = [Ag^(o+)] = [CI^(Theta)] = [Br^(Theta)] = (0.01)/(2) = 0.005 M = 5.0 xx 10^(-3) M`.
`AgBr` is the more soluble salt (less `Ksp` means more soluble)and would take precendence in the precipitating reaction.
ii. Assume `AgCI` does not precipitate. In this case `Ag^(o+)` and `Br^(Theta)` would be removed by precipitation and the concentration of these two ions in solution would remain eqial to each other.
iii. `[Ag^(o+)] [Br^(Theta)] = sqrt(K_(sp) AgBr) = (5.0 xx 10^(-13))^((1)/(2)) = 7.1 xx 10^(-7)M`
iv. `Q_(sp)` or I.P. of `AgCI = [Ag^(o+)] [CI^(Theta)]`
`= (7.1 xx 10^(-7)) (5.0 xx 10^(-3)M) = 3.5 xx 10^(-9)`
`Q_(sp)` of `AgCI gt K_(sp)` of AgCI. ( `:'` Some AgCI most also precipitate)
Hence the assumption in (ii) is wrong.
v. Since both holides precipitate, it is a case of simultaneous solubilities.
vi. By electronutrality
`[Na^(o+)] + [Ag^(o+)] = [CI^(Theta)] + [Br^(Theta)] + [NO_(3)^(Theta)]`
`0.01 + [Ag^(o+)] = [CI^(Theta)] +[Br^(Theta)] = 0.005`
or `[CI^(Theta)] + [Br^(Theta)] - [Ag^(o+)] = 0.005 ...(1)`
vii. `[Ag^(o+)] [CI^(Theta)] = 10^(-10) ...(2)`
`[Ag^(o+)] [Br^(Theta)] = 5 xx 10^(-13).. (3)`
viii. Divide (2) by (3), `([CI^(Theta)])/([Br^(Theta)]) = 200`
This shows `Br^(Theta)` plays a significant role in the total anion concentration of the solution.
xi. Moreover, `[Ag^(o+)]` must be negligible in (1) because of ionsolubility of two silver salts.
x. Therefore assume in (1) that `[CI^(Theta)] = 0.005 = 5 xx 10^(-3)`.
`[Br^(Theta)] = 2.5 xx 10^(-5)`
Check of assumption : Both `[Ag^(o+)]` and `[Br^(Theta)]` are negligible compared with `[CI^(Theta)]` of `0.005M`.