In 1 ltr. saturated solution of #AgCl(K_"sp"=1.6xx10^(-10))#, #0.1# mol of #CuCl(K_"sp"=1.0xx10^(-6))# is added. What is the resultant concentration of #Ag^+# in the solution in molarity?

1 Answer
Oct 3, 2017

#sf([Ag^+]=1.6xx10^(-7)color(white)(x)"mol/l")#

Explanation:

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.