We're going to work through this problem as three separate problems. The first one is to calculate the molar solubility of the silver chromate in pure water. So we will begin for each of these parts by writing the reaction of the dissolution of the solid. So there's a silver chromate. It will dissolve to produce 2 silver plus ions and a chromate ion. So we can set up an ICE table. We will place some of the solid, doesn't matter how much, some of the solid into the solution, into the water, and we will have no silver and no chromate present before the reaction happens. As it dissolves, we will subtract S, that is, the molar solubility. How much of the silver chromate would dissolve, and we would produce 2S here and s here. Giving me still some of the solid sitting in the bottom of the beaker. 2S and S. Write the KSP expression for the reaction you see above, so it's silver concentration squared, because it has a 2 coefficient, times a chromate concentration. And then we'll plug in the values that we know. 1.12 x 10 to the -12 is the KSP, and that's going to be equal for the silver concentration we have 2S, that's 2S, and that will be squared, plus the S for the chromate. This will give me 1.12 x 10 to the -12 = 4S cubed. So we squared the 2, we squared the S, now we can multiply that S squared with that other S and that's our values. Now, to obtain our value for S we have to divide both sides by 4. And then take the cube root of both sides, okay? So we're going to take the cube root. And that will leave us with S. And S = 6.54 x 10 to the minus 5, and since it's molar solubility that we're obtaining, we'll put an M for molarity.