Question

Suppose I take a 1.0 L saturated, room temp solution of lead(2)bromide and dissolve 1.3g of potassium bromide in the solution. Will lead(2) bromide precipitate out? If yes how much( in grams)? If no, why not?

I found the ksp for PbBr2 at room temp, and used it to find the Molarity of PbBr2 saturated, .0116M. I know I have to use the ICE box method but I'm just confused as to what the reactants/products are and how to set it up. Thanks'!

Answer 1

OK. We'll take your number of 0.116M for the solubility of KBr in a saturated solution. I don't know what value you used for Ksp but I used one from the web and ended up with 0.0118M so the answers are pretty close. You probably used a value for Ksp close to what I used of 6.6E-6.

(KBr) = 1.3/119 = approx 0.0109M and it ionizes 100% so Br^- will be 0.0109.

I tried the easy way on this as
Ksp = (Pb^2+)(Br^-)^2
Ksp = (x)(0.0109)^2
and x = approx 0.055 which obviously is not right because the solubility can't be greater when adding a common ion. Therefore, I think you have to go the long route which is as follows:


............PbBr2 ==> Pb^2+ + 2Br^-
I...........solid..0.0116...2*0.0116
add..........................0.0109
C...........solid......-x....-2x
E...........solid..0.0116-x..0.0341-2x
Substitute the E line into Ksp expression and solve for x = molarity PbBr2 that ppts.
You get a cubic equation which I solved with the help of an on-line calculator. Here is one you can use.
http://www.1728.org/cubic.htm
Post your work if you get stuck.