Determine Molar Solubilty

Question

Given solubility product of PbBr2 is 4.45*10^-6, determine molar solubility in a 0.318 M Pb(NO3)2 solution.

Answer 1

First write out the dissociation reaction of PbBr2:

PbBr2 ⇔ Pb2+(aq) + 2 Br- (aq)

The K_sp of PbBr2 can be expressed as the following based on the definition of the solubility product.

K_sp = [Pb²⁺][Br-]²

Pb(NO3)2 fully dissociates because it is fully soluble. This means there will be 0.318 M of Pb2+ and 0.636 M of NO3. Since the Pb2+ is also combined with the Pb2+ from the dissociation of PbBr2, this is known as a common ion.

Taking this into account, using the "ICE" method, and referencing PbBr2 ⇔ Pb2+(aq) + 2 Br- (aq), we have

K_sp = (0.318 + x)(2x)² = 4.45*10^-6

Leaving the rest of the solution to you -- if you have any questions about this process or need help with getting the final answer, please schedule some time directly with me!

Answer 2

This problem deals with the common ion effect. Since you have Pb²⁺(aq) in both the Pb(NO₃)₂ and the PbBr₂, Pb²⁺ is the common ion.

Let us first look at the solubility of the PbBr₂. PbBr₂(s) <==> Pb²⁺(aq) + 2Br^-(aq). And note that the [Br^-] is twice that of the [Pb²⁺] because for each PbBr₂ you get 1 Pb²⁺(aq) and 2Br^-(aq).

Now we write the solubility expression for PbBr₂. Ksp = [Pb²⁺][Br^-]²

From here there are 2 different ways to go about solving the problem. One way is to let x be the molar solubility of the PbBr₂ and the 2nd way is to let x be the molar solubility of Br^-. Either way, you will get the same answer (see below).

Let x = molar solubility of PbBr₂. I believe this is the preferred approach

Ksp = [Pb²⁺][Br^-]² = 4.45x10^-6 = (0.318)(2x)²

1.399x10^-5 = 4x²

x = 1.87x10^-3 M = molar solubility of PbBr₂.

Let x = molar solubility of Br^- (less preferred approach)

Ksp = [Pb²⁺][Br^-]² = 4.45x10^-6 = (0.318)(x)²

1.399x10^-5 = x²

x = 3.74x10^-3 M = molar solubility of Br^-

Since the [Pb²⁺] is 1/2 the [Br^-], that makes [Pb²⁺] = 1.87x10^-3 M = molar solubility of PbBr₂