This is a pretty straightforward solubility problem.
The first thing I do is I look at the compound for which I am given Ksp. In this case, that compound is Silver Chloride. From there, I write out the equation for silver chloride's dissociation along with the equilibrium expression.
AgCl(s) <---> Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
Notice that we leave AgCl(s) out of the equilibrium expression since it is a solid.
From here, this question is basically asking us how much NaCl we can dissolve into this solution. This is a common ion effect question since NaCl and AgCl both dissociate into Cl- ions. The next step should be to make an ICE chart:
AgCl(s) <---> Ag+(aq) + Cl-(aq)
I......................6M...........0
C.....................+x.........+x
E.....................6+x..........x
We plug in the values that we got at equilibrium back into our expression for Ksp:
Ksp = (6+x)(x)
At this point, we can ignore the x inside of the (6+x) expression (if we didn't, we'd need to use a quadratic equation which is out of the scope of the MCAT)
Leaving us with:
Ksp = 6x
==> x = Ksp/6
x is our Molar Solubility. This is basically telling you how much x you can dissolve at equilibrium. Since we are interested in Cl- concentration and at equilibrium, Cl- concentration = x according to our ICE chart, x represents the molarity of Cl- that you can dissolve to saturate this solution.
We are interested in finding the amount of grams NaCl that can be used to saturate this solution. Molarity = # of moles/ L solution = grams/Molar Mass / L solution
==> after some algebra, you should be left with:
grams NaCl = [Ksp (L solution) (Molar Mass NaCl) ] / 6
Hope that helps! Please correct me if I made any mistakes.
