Hi, I have a few chemistry problems that I'm having trouble solving. I put my attempted solutions after the problems in *'s. There was one problem I had no idea how to solve.

What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve to make a saturated solution in water?

*For this one the formula is Ag2CrO4, which is [Ag]^2[CrO4]. I plugged x in for CrO4, so x^3 is equal to Ksp. I then got that x = 2.0e-4. However, that x is for CrO4, not for silver chromate. Is the answer for silver chromate just the Ksp value?*

What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve in 500. mL of 0.01 M aqueous silver nitrate?

A saturated solution of CaSO4(aq) is made in a beaker until there is excess calcium sulfate resting at the bottom. Then solid potassium sulfate is added. Which of the following is true? (The Ksp for potassium sulfate is larger than the Ksp for calcium sulfate)
a. More calcium sulfate will start to precipitate out of solution
b. More calcium sulfate will start to dissolve in solution

*I know that when a smaller Ksp is added to a bigger Ksp it precipitates out of solution. Is that the same in the reverse?*

How does the solubility of AgCl in pure 1.0 M NaCl(aq) compare with its solubility in pure water?

*It is less, right? because there are already chloride ions present.*

Worked above.

For the first problem, to find the concentration of silver chromate that will make a saturated solution in water, you need to determine the concentration of [Ag+] and [CrO4^2-] ions in the solution. The Ksp expression for silver chromate (Ag2CrO4) is:

Ksp = [Ag+]^2[CrO4^2-]

Since Ag2CrO4 dissociates into 2Ag+ ions and 1CrO4^2- ion, we can assume that the concentration of [Ag+] is 2x, where x is the concentration of [CrO4^2-]. Thus, the Ksp expression becomes:

Ksp = (2x)^2 * x = 4x^3

Now, you can substitute the given Ksp value (9.0 x 10^-12) into the equation and solve for x:

4x^3 = 9.0 x 10^-12

Take the cube root of both sides:

x = (9.0 x 10^-12)^(1/3)

This will give you the concentration of [CrO4^2-]. Since the concentration of [Ag+] is 2x, you can multiply the value of x by 2 to get the concentration of [Ag+].

For the second problem, you have 500 mL of a 0.01 M solution of silver nitrate (AgNO3). The silver nitrate will dissociate into Ag+ and NO3- ions. Since Ag2CrO4 is sparingly soluble and reacts with Ag+ ions to form a precipitate, you need to determine how much Ag+ is present in the solution.

First, calculate the amount of Ag+ in moles present in 500 mL of the 0.01 M AgNO3 solution:

moles of Ag+ = (0.01 M) * (0.5 L) = 0.005 mol

Since each Ag2CrO4 molecule requires 2 Ag+ ions, you need to divide the moles of Ag+ by 2 to determine the moles of Ag2CrO4 that can be formed:

moles of Ag2CrO4 = 0.005 mol / 2 = 0.0025 mol

Finally, convert the moles of Ag2CrO4 to concentration by dividing by the volume of the solution (0.5 L):

concentration of Ag2CrO4 = 0.0025 mol / 0.5 L = 0.005 M

Therefore, the concentration of silver chromate that will dissolve in the solution is 0.005 M.

For the third problem, you are correct. When a smaller Ksp is added to a solution containing a larger Ksp, the ions from the compound with the smaller Ksp will associate with the ions from the compound with the larger Ksp, leading to the precipitation of the compound with the smaller Ksp. In this case, since the Ksp for potassium sulfate is larger than the Ksp for calcium sulfate, more calcium sulfate will start to precipitate out of the solution.

Lastly, you are also correct in your understanding of solubility. The solubility of AgCl in 1.0 M NaCl(aq) is less than its solubility in pure water. This is because the presence of chloride ions from NaCl in the solution reduces the activity of Cl- ions and therefore decreases the solubility of AgCl.