Solving Ksp Problems:
Part One - s2Solving Ksp Problems:
Part Two - 4s3Solving Ksp Problems:
Part Three - 27s4Solving Ksp Problems:
Part Five - 256s5Back to Equilibrium Menu
The generic problem 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.
Example #1: Bismuth sulfide, Ksp = 1.82 x 10¯99
Solution:
Bi2S3(s) ⇌ 2Bi3+(aq) + 3S2¯(aq)Ksp = [Bi3+]2 [S2¯]3
1.82 x 10¯99 = (2s)2 (3s)3
108s5 = 1.82 x 10¯99
s = 7.00 x 10¯21 M
Note how 's' is the moles of the Bi2S3 that dissolved.
There may also be a bit of a calculator problem. When you divide the Ksp by 108, the answer is 1.685 x 10¯101. Some older calculators (as well as some low-end newer calculators) are not equipped to handle three digit exponents. If you have one of those, you can (1) use a spreadsheet on your computer, (2) access a calculator on-line, (3) borrow from a friend, or (4) go shopping.
Example #2: Copper(II) phosphate, Ksp = 1.93 x 10¯37
Solution:
Cu3(PO4)2(s) ⇌ 3Cu2+(aq) + 2PO43¯(aq)Ksp = [Cu2+]3 [PO43¯]2
1.93 x 10¯37 = (3s)3 (2s)2
108s5 = 1.93 x 10¯37
s = 1.78 x 10¯8 M
Example #3: Iron(III) sulfide, Ksp = 1 x 10¯88
Solution:
1) Here's the dissociation equation:
Fe2S3(s) ⇌ 2Fe3+(aq) + 3S2¯(aq)
2) The Ksp expression:
Ksp = [Fe3+]2 [S2¯]3
3) Solve the Ksp expression
1 x 10¯88 = (2s)2 (3s)3108s5 = 1 x 10¯88
s = 9.847 x 10¯19 M
4) When you round the answer off to one sig fig, you get:
s = 1 x 10¯18 M
Example #4: Magnesium phosphate, Ksp = 9.86 x 10¯25
Solution:
Mg3(PO4)2(s) ⇌ 3Mg2+(aq) + 2PO43¯(aq)Ksp = [Mg2+]3 [PO43¯]2
9.86 x 10¯25 = (3s)3 (2s)2
9.86 x 10¯25 = 108s5
s = 6.20 x 10¯6 M
Example #5: Magnesium arsenate, Ksp = 2.1 x 10¯20
Solution:
Mg3(AsO4)2(s) ⇌ 3Mg2+(aq) + 2AsO43¯(aq)Ksp = [Mg2+]3 [AsO43¯]2
2.1 x 10¯20 = (3s)3 (2s)2
2.1 x 10¯20 = 108s5
s = 0.000045 M
Example #6: Cadmium(II) phosphate, Ksp = 2.5 x 10¯33
Solution:
Cd3(PO4)2(s) ⇌ 3Cd2+(aq) + 2PO43¯(aq)Ksp = [Cd2+]3 [PO43¯]2
2.5 x 10¯33 = (3s)3 (2s)2
Example #7: Ni3(PO4)2, Ksp = 4.74 x 10¯32
Solution:
Ni3(PO4)2 ⇌ 3Ni2+ + 2PO43Ksp = [Ni2+]3 [PO43¯]2
4.74 x 10¯32 = (3s)3 (2s)2
Example #8: Pb3(PO4)2, Ksp = 1.0 x 10¯54
Solution:
Pb3(PO4)2 ⇌ 3Pb2+ + 2PO43¯Ksp = [Pb2+]3 [PO43¯]2
1.0 x 10¯54 = (3s)3 (2s)2
Solving Ksp Problems:
Part One - s2Solving Ksp Problems:
Part Two - 4s3Solving Ksp Problems:
Part Three - 27s4Solving Ksp Problems:
Part Five - 256s5Back to Equilibrium Menu