In this case, please note that the `K_(sp)` values of two slats are very similar. So the concentration of `Ag^(o+)` ions (the common ions) cannot be calculated from a single salt alone and we have to consider the equilibrium of the two slats simultaneously.
Let the simulatenous solubilities of `AgSCN` and `AgBr` be `x` and `y`, respectively, in `molL^(-1)`.
`{:(AgSCN(s)hArrAg^(o+)underset(x)(aq)+SCunderset(x)N^(Θ)(aq),,,,),(AgBr(s)hArrAg^(o+)underset(y)((aq))+Brunderset(y)((aq)),,,,):}`
At equibrium:
`[Ag^(o+)] = x +y, [SCN^(Θ)] = x, [Br^(Θ)] = y`
`[Ag^(o+)] [Br^(Θ)] = K_(sp(AgBr))`
and `[Ag^(o+)] [SCN^(Θ)] = K_(ap(AgSCN))`
According to electircal change neutrality equation. Total positive change = Total negative change
Note that: `[Ag^(o+)] = [Br^(Θ)] +[SCN^(Θ)]`
[This is an Electrical change neturality equation]
`[Ag^(o+)] = (K_(sp)(AgBr))/([Ag^(o+)]) + (K_(sp)(AgSCN))/([Ag^(o+)])`
`rArr [Ag^(o+)] = sqrt(K_(sp(AgBr))+K_(sp(AgSCN)))`
`rArr x +y = 1.22 xx 10^(-6) ....(i)`
Also, `([Br^(Θ)])/([SCN^(Θ)]) = (y)/(x) = (K_(sp(AgBr)))/(K_(sp(AgSCN))) = 0.5 ...(ii)`
Using equaitons (i) and (ii), we get
`x = 8.0 xx 10^(-7)`
