Some uses of Ksp with problems in Chapter 16

(C) - Copyright, 2001 F.W. Boyle, Jr., Ph.D.

Problem 16.6

Complete the following table for concentrations in equilibrium with PbBr2.

[Pb2+] [Br-]
a) 0.007
b) 0.4
c) 1/2[Pb2+]
d) 4[Br-]

First consider what Ksp implies. It implies the [ ] of ions dissolved in solution in equilibrium with some solid which contains the ions.

In this case was have PbBr2. We look up the Ksp in a table and find Ksp = 6.6 x 10-6

Next we must determine the dissolution (dissolving) reaction for the compound.

PbBr2 Pb2+ + 2Br-

Next we must write the Ksp expression.

Ksp = [Pb2+][Br-]2

This expression is numerically equal to the value 6.6 x 10- 6.


Ksp = [Pb2+][Br-]2 = 6.6 x 10-6

You should notice that there are two (2) unknowns in this expression, [Pb2+] and [Br-]. You should also notice that the table provides one concentration for each of its four parts. So simply substitute in the expression and solve.

a) [Br-] = 0.007 substituting:

[Pb2+](0.007)2 = 6.6 x 10-6

And solving: [Pb2+] = 0.1 M

You probably got 0.135 M but forgot there was only one significant figure in 0.007.

This is the [Pb2+] that can be in solution with Br at 0.007 M with PbBr2 solid.

b) [Pb2+] = 0.4 M substituting:

(0.4)[Br-]2 = 6.6 x 10-6

And solving: [Br-] = 4 x 10-3 M

c) [Br- = 1/2[Pb2+]

This looks harder but really isn't!


[Pb2+](1/2[Pb2+])2 = 6.6 x 10-6

Next square the second portion of the left side and:

[Pb2+](1/4)[Pb2+]2 = 6.6 x 10-6

Then combine terms as you would in any mathematical expression:

(1/4)[Pb2+]3 = 6.6 x 10-6

Multiply by 4 and take the cube root:

[Pb2+] = 3 x 10-2 M or 0.03 M

d) I will leave to you to solve!

Problem 16.18:

Calculate the solubility (in grams per liter) of MgF2 in:

a) pure water
b) 0.015 M Mg(NO3)2
c) 0.0050 M SbF3

Part a) is a simple solubility. Write out the dissolution reaction.

MgF2 Mg2+ + 2F-

Next write the Ksp expression and go look up the Ksp.

Ksp = [Mg2+][F-] = 7 x 10-11

Now when MgF2 dissolves it release 1 Mg and 2 F ions. If we call the concentration of Mg, s then the concentration of F is 2 s.


[s][2s] = 7 x 10-11

Solving we find:

s = (Ksp/4)1/3

This is the same as the XY2 form.

Plugging in Ksp and solving we find:

s = [Mg2+] = 2.6 x 10-4 M

This is 0.02 g of MgF2.

Part b) is a study of the common ion effect. Common ion means an ion is dissolving into solution from two different sources. For part b) the common ion is Mg. The NO3- has absolutely nothing to do with solving this problem.

The Ksp is still 7 x 10/11 and the expression is still:

Ksp = [Mg2+][F-] = 7 x 10-11

However in this case we have to figure out the dissolution reaction for Mg(NO3)2.

Mg(NO3)2 Mg2+ + 2NO3- So 1 Mg(NO3)2 releases 1 Mg2+ and 2 F-. Thus 0.015 M releases 0.015 M Mg2+.

The MgF2 also releases some Mg and some F. How much? We don't know so we call it x. Thus Mg is 0.015 + x and F is equal to x.

Back to our Ksp expression where we substitute the amount of Mg2+:

(0.015+x)(x) = 7 x 10-11

But we can guess from part a that 0.015 is large compared to x and we ignore the x added to 0.015. Now the equation is simple and we solve for x. And solve for x which equals [F-]

    [F-] = (7 x 10-11/0.015 )1/2

         = 6.8 x 10-5

Now we must recall that each MgF2 releases 2 F- ions so the moles of MgF2 is half the number of moles of F-.


MgF2 = 6.8 x 10-5/2 = 3.4 x 10-5 moles

The problem wants to know the g so we must convert moles to grams.

molar mass of MgF2 is 62 so multiply 62 x 3.4 x 10-5 which results in 2 x 10-3 g.

c) Try it yourself!