Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of the solubility product constant, Ksp, is 5.0 x 10-13 at 298 K.

a) Calculate the value of [Ag+] in 50.0 mL of a saturated solution of AgBr at 298 K.

b) A 50.0 mL sample of distilled water is added to the solution described in part (b), which is in a beaker with some solid AgBr at the bottom. The solution is stirred and equilibrium is re-established. Some solid AgBr remains in the beaker. Is the value of [Ag+] greater than, less than, or equal to the value you calculated in part

c) Calculate the minimum volume of distilled water, in liters, necessary to completely dissolves a 5.0 g sample of AgBr at 298 K.

d) A student mixes 10.0 mL of 1.5 x 10-4 M AgNO3 with 2.0 mL of 5.0 x 10-4 M NaBr and stirs the resulting mixture. What will the student observe?

Is this a practice problem to check your answers? If so show your work and I'll check for you. If not, please explain what you don't understand about the problem and I'll help you through it.

I don't understand how to set it up and get started on the problem and this is a problem that our teacher gave us for practice for the AP Chem exam

Here is how you do a part. That should get you started.

.........AgBr ==> Ag^+ + Br^-
I........solid.....0......0
C........solid.....x......x
E........solid.....x......x

Substitute the E line into the Ksp expression and solve for x = (Ag^+) = (Br^-)
My presumption is that the answer for a is to be (Ag^+) in mols/L and the concn is the same whether the volume is 50 or 500.
Knowing this you should be able to answer b.
c is the just the reverse question in a.
D is a common ion problem and the concn of AgBr is less in that solution.
Post your work if you get stuck.

Okay thank you

so in the ICE box when you are solving for "x" am I supposed to use the Ksp for the solid

You are supposed to use Ksp but not for the solid. Ksp = (Ag^+)(Br^-). Remember that solids and pure liquid are not part of equilibrium constant equations. There must be at least the tiniest crystal but as long as there is SOME solid there then equilibrium can be established.

okay I think I got, I got 7.1 x 10^-7 for part a, and I got for part b that they should be the same, and for part c I got 3.7 x 10^4, and finally for part d I got since 1.1 x 10^-8 is greater than 5.0 x 10^-13 a precip will form.

Is that the correct answers?

I agree with all of your numbers. a is right; b is right. I obtained 3.8E4 L but that's probably because we didn't use the same molar mass for AgBr. When you ask us to check answers you should tell us what you used for molar mass and atomic masses if we are to get the same answer. For the last one I ended up with 1E-8 and I assume that's for the same reason as in c. Good work. I believe you have it too. (and another reason for a difference is I never take the number out of my calculator; therefore, I kept 7.07E-7 in my calculator from part a to work the others instead of using 7.1E-7.

okay thank you very much for helping me and everyone else that asks questions on here I think you did a really good job answering my questions and I will keep what you said about the molar mass and the atomic masses in mind for next time. Again I can't thank you enough.