As in the other parts the problem stays the same in each example. It is:
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
I'm going to assume you've gone through the other files first, so the presentations here will be a bit abbreviated.
The two example substances are:
Bi2S3 Ksp = 1.82 x 10¯99 Cu3(PO4)2 Ksp = 1.93 x 10¯37
Bismuth sulfide
Bi2S3 <===> 2 Bi3+ + 3 S2¯Ksp = [Bi3+]2 [S2¯]3
1.82 x 10¯99 = (2x)2 (3x)3
108x5 = 1.82 x 10¯99
x = 7.00 x 10¯21 M
Note how 'x' still is the moles of the substance that dissolved.
There is also a bit of a problem with this problem. When you divide the Ksp by 108, the answer is 1.685 x 10¯101. Some calculators are not equipped to handle three digit exponents. However, bring up a spreadsheet on your computer and you will be able to enter values and a formula to calculate the value.
Copper(II) phosphate
Cu3(PO4)2 <===> 3 Cu2+ + 2 PO43¯Ksp = [Cu2+]3 [PO43¯]2
1.93 x 10¯37 = (3x)3 (2x)2
108x5 = 1.93 x 10¯37
x = 1.78 x 10¯8 M
Here are two more substances:
Fe2S3 Ksp = 1 x 10¯88 Mg3(PO4)2 Ksp = 9.86 x 10¯25
Solve them and go to the answers.
Go to Solving Ksp Problems - Part One
Go to Solving Ksp Problems - Part Two
Go to Solving Ksp Problems - Part Three