### Solving Ksp Problems: Part Four - 108s5

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)3

108s5 = 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+ + 2PO43

Ksp = [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