In a saturated solution of sf(AgCl) the solid is in equilibria with its constituent ions:
sf(AgCl_((s))rightleftharpoonsAg_((aq))^(+)+Cl_((aq))^(-))
For which sf(K_(sp)=[Ag_((aq))^+][Cl_((aq))^-]=1.6xx10^(-10))
We can work out the concentration of sf(Ag_((aq))^(+)) before any sf(CuCl) is added:
Since sf([Ag_((aq))^(+)]=[Cl_((aq))^-]) we can say:
sf([Ag_((aq))^(+)]^2=1.6xx10^(-10))
:.sf([Ag_((aq))^+]=sqrt(1.6xx10^(-10))=1.26xx10^(-5)color(white)(x)"mol/l")
Now we add 0.1 mol of sf(CuCl). In a saturated solution the solid is in equilibria with its ions:
sf(CuCl_((s))rightleftharpoonsCu_((aq))^(+)+Cl_((aq))^(-))
sf(K_(sp)=[Cu_((aq))^+][Cl_((aq))^-]=1.0xx10^(-6))
Le Chatelier's Principle would predict that introducing extra sf(Cl^-) would cause the silver chloride equilibrium to shift to the left thus reducing the concentration of sf(Ag_((aq))^+).
We can check if all the added sf(CuCl) will dissolve:
If the solubility is s then:
sf([Cu_((aq))^(+)][Cl_((aq))^-]=1.0xx10^(-6))
Since sf([Cu_((aq))^+]=[Cl_((aq))^-]) then sf(s^2=1.0xx10^(-6))
:.sf(s=sqrt(1.0xx10^(-6))=10^(-3)color(white)(x)"mol/l")
The sf(M_r) of sf(CuCl) is taken to be 99.
:.sf(s=99xx10^(-3)=0.099color(white)(x)"g/l")
We add 0.1 mol which is 9.9 g to 1 litre of solution. You can see that this greatly exceeds the solubility by a factor of 100. We can conclude that this forms a saturated solution.
In such a solution we can say that sf([Cl_((aq))^-]=10^(-3)color(white)(x)"mol/l"). There will also be some sf(Cl^-) ions from the silver chloride equilibrium but these will be much less.
Here I am going to make the assumption that virtually all the sf(Cl_((aq))^-) come from the sf(CuCl). This makes the calculations a lot simpler. It is an approximation used with "common ion" examples like this.
We know that sf([Ag_((aq))^+][Cl_((aq))^-]=1.6xx10^(-10))
:.sf([Ag_((aq))^+]=(1.6xx10^(-10))/([Cl_((aq))^-])=(1.6xx10^(-10))/(10^(-3))=1.6xx10^(-7)color(white)(x)"mol/l")
As predicted, the concentration of sf(Ag_((aq))^+) has been reduced.