Here in part B we are going to see the effects of adding sodium chromate to the solution of silver chromate. Now very commonly what students will try to do when they come across a problem like this is to react these two as reactants and try to come up with a product, but that is not the way you work these problems. What you will want to do is write the reaction as you did in part A. The silver chromate dissolving in solution to give you two silver ions plus chromate. And the change or the difference in part B versus part A comes in this I line. You will put some of the solid into the beaker where there's already some solution present of sodium, or some ions present of sodium and chromate. So there's not any silver ions present but there is 2.6 times 10 to the minus 3 molar of the chromate present. And so I'll put that number here saying it's this number. Now word of caution. The reason it's the same number is because there is one of the chromate in sodium chromate. So the concentration of chromate is the same as this. But if there were two chromates in the formula, which there isn't, but if there were, then I would need to double that value that they're giving me here before I put it into the table, because that would be the concentration of the ions in solution. So we will use minuses here because that's how much of the sodium chromate will dissolve. That's the molar solubility. I will add 2s here. I will add s over here. That will still give me some solid in solution, I mean in the container, I have 2s for that, and I have 2.6 x 10 to the minus 3 plus s for this. Now, I will always want to write the Ksp expression it's the silver ion concentration squared. I'm looking at the reaction, I'm doing the powers a lot of times students try to skip this step to save time, but then they invariably forget to work with those coefficients. Then I will look at the silver concentration, and that was 2s, and it's squared, and then I will look at the chromate concentration, which I see here, 2.6 times 10 to the minus 3 plus s, so all of this is multiplied. Now we're going to assume that S is very small compared to the other values, and ignore this term for simplicity sake. We'll make sure we go back and check that it was a valid assumption. So all of this is going to be equal to the number of ksp which is given in the problem. 1.2 times 10 to the minus 12. And so I will multiply the 2 times 2 that will be 2 squared and then I'm going to multiply that to the 2.6 x 10 to the -3. And that is going to give me 1.04 times 10 to the minus 2 times s squared. So then I will divide both sides by the 1.04 times 10 to the minus 2, And that's going to give me s squared equal to 1.1 times 10 to the minus 10. So then I can take the square root of both sides, and that will give me s, and the value for s is 1.0 times 10 to the minus 5. So then we check our assumption, and if we take this value and subtract it from the 2.6 times 10 to the minus 3, we will still get 2.6 times 10 to the minus 3 or add it to it, add it, it won't change the values, so that's a good assumption for us. And so, this is the molar solubility of silver chromate In a solution that contains sodium chromate. Now let's see if it makes sense. The solubility of the silver chromate in pure water, that was determined in part A, was 6.54 times 10 to the minus 5. Okay, and this is 1 times 10 to the minus 5, so the molar solubility has decreased by the addition of this small amount of the sodium chromate, so it is supporting what we know about a common ion situation. Common ions always decrease the solubility of an insoluble salt.