CHEM PLEASE HELP!!!?

I'm not sure what you don't understand, Sarah, so I'll start at the beginning and hope you follow me. For a moment, forget about the AgNO3 solution. Let's just deal with the Ksp for Ag2SO4. If you write the equation for the ionization of the stuff you should get:

Ag2SO4 <=======> 2Ag+ + SO4=

Writing the Ksp for that is Ksp = [Ag]2 [SO4]

You're given the Ksp so if you solve for the concentration of the ions in solution, you will be able to calculate the solubility in moles/liter. So set up the problem:

1.2 x 10^-5 = [2x]2 [x]........now, if we were just going to solve this in an aqueous solution, we'd have 4x^3 = 1.2 x 10^-5 and have to solve for x. The thing is that we sort of already know the Ag+ ion concentration. If the solution is 2 molar in AgNO3, then the AgNO3 will produce 2 moles of silver ions per liter of solution. So the silver ion concentration from AgNO3 is 2 molar. Now, this is where the solution to the problem gets easy. The 2 molar Ag+ ion concentration from the AgNO3 is so high compared to the additional Ag+ ions from the extremely small amount of Ag2SO4 that dissolves, that we can forget about calculating the Ag+ ion concentration in the Ksp equation. The problem then becomes:

1.2 x 10^-5 = [2]2 [x]........(where x equals the SO4= ion concentration).so we have 4x = 1.2 x 10^-5 and x = 3 x 10^-6

Okay, if 3 x 10^-6 is the SO4= ion concentration, and you get 1 mole of SO4= ions for every 1 mole of Ag2SO4 that dissolves, we can say that the SO4= ion concentration is equal to the amount of Ag2SO4 that dissolves. Do you follow me, Sarah? I hope so, because we've just solved the problem. If the SO4=ion concentration (in moles per liter) is 1 x 10^-3, then the concentration of the Ag2SO4 in moles per liter is 3 x 10^-6 moles per liter.

Ksp is the product of the ions in concentration when the solution is saturated.

Ag2SO4 --> 2 Ag+ + SO4^-2

Ksp = [Ag+]²[SO4^-2]

In this case, however, we already have some silver ions in solution, which means that less of the compound will dissolve than if we did it in initially pure water.

We do not have to worry about the nitrate (NO3-) ions. It just means that silver nitrate is a high solubility compound. After is dissolves, the nitrates just remain in solution as spectator ions and do not affect the silver sulfate solution.

Let the molarity of Ag2SO4 dissolved be x.

At point of saturation,

[Ag+] = 2x

[SO4^-2] = 2.0 + x

Since Ksp is very small compared to 2.0, we can assume that [SO4^-2] is approx 2.0 and will not change noticeably.

1.2 x 10^-5 = (2x)²(2.0)

1.2 x 10^-5 = 8x²

1.5 x 10^-6 = x²

0.00122 = x

(Approximation was valid.)

Solubility is 0.00122 M