Complete the following table for concentrations in equilibrium with PbBr2.
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.
a) [Br-] = 0.007 substituting:
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:
c) [Br- = 1/2[Pb2+]
This looks harder but really isn't!
[Pb2+] = 3 x 10-2 M or 0.03 M
d) I will leave to you to solve!
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.
Plugging in Ksp and solving we find:
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:
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+:
[F-] = (7 x 10-11/0.015 )1/2 = 6.8 x 10-5Now we must recall that each MgF2 releases 2 F- ions so the moles of MgF2 is half the number of moles of F-.
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!