Basically, this type of problem will (1) calculate the molar solubility from the K_{sp}. Then, in an additional step, (2) calculate grams per liter from moles per liter. From there, it is an easy third step to g/100mL.

**Example #1:** Calculate the milligrams of silver carbonate in first 100 mL (then 250 mL) of a saturated solution of Ag_{2}CO_{3} (K_{sp} for silver carbonate = 8.4 x 10¯^{12})

**Solution:**

1) Solve for the molar solubility of Ag_{2}CO_{3}:

Ag_{2}CO_{3}⇌ 2Ag^{+}+ CO_{3}^{2}¯K

_{sp}= [Ag^{+}]^{2}[CO_{3}^{2}¯]8.4 x 10¯

^{12}= (2s)^{2}(s)s = 1.28 x 10¯

^{4}mol/LComment: The value of s is the molar concentration of the carbonate ion. Since there is a 1:1 molar ratio between it and silver carbonate, the value for s is also the molar solubility of silver carbonate.

2) Convert mol/L to gram/L:

1.28 x 10¯^{4}mol/L times 275.748 g/mol = 3.53 x 10¯^{2}g/L

3) Convert from g/L to g/100mL:

3.53 x 10¯^{2}g/L divided by 10 = 3.53 x 10¯^{3}g/100mLComment: this is done because there are 10 100mL amounts in 1 L.

4) Convert mol/L to g/250mL:

3.53 x 10¯^{2}g/L divided by 4 = 8.83 x 10¯^{3}g/250mLComment: this is done because there are 4 250mL amounts in 1 L.

**Example #2:** Calculate the milligrams of silver ion that are present in 250 mL of a saturated solution of silver carbonate.

**Solution:**

1) Restate the value for s from problem #1, the molar concentration of the carbonate ion:

s = 1.28 x 10¯^{4}mol/L

2) From the stoichiometry of the chemical equation, the molar concentration of the silver ion is twice that of the carbonate ion:

[Ag^{+}] = 2.56 x 10¯^{4}mol/L

3) Determine g/L of silver ion:

2.56 x 10¯^{4}mol/L times 107.8682 g/mol = 0.0276 g/L

4) Determine g/250mL and convert to mg/250mL

0.0276 g/L divided by 4 = 0.00690 g/250mL0.00690 g/250mL = 6.90 mg/250mL

**Example #3:** Calculate the mass of Ca_{5}(PO_{4})_{3}F (K_{sp} = 1.00 x 10^{-60}) which will dissolve in 100 mL of water.

**Solution:**

1) Determine moles of Ca_{5}(PO_{4})_{3}F in one liter:

Ca_{5}(PO_{4})_{3}F(s) ⇌ 5Ca^{2+}+ 3PO_{4}^{3-}+ F¯K

_{sp}= [Ca^{2+}]^{5}[PO_{4}^{3-}]^{3}[F¯]1.00 x 10

^{-60}= (5s)^{5}(3s)^{3}(s)1.00 x 10

^{-60}= 84375s^{9}s = 6.11 x 10

^{-8}M

2) Determine moles of Ca_{5}(PO_{4})_{3}F in 100 mL:

6.11 x 10^{-8}mole divided by 10 = 6.11 x 10^{-9}mol / 100mL

3) Determine how many grams this is:

6.11 x 10^{-9}mol times 504.298 g/mol = 3.08 x 10^{-6}g / 100mL

**Example #4:** Calculate the mass of Ca_{5}(PO_{4})_{3}OH (K_{sp} = 6.80 x 10^{-37}) which will dissolve in 100 ml of water. (Please ignore any complications which might result as a consequence of the basicity or acidity of the ions, ion pairing, complex ion formation, or auto-ionization of water.)

**Solution:**

1.00 x 10^{-37}= (5s)^{5}(3s)^{3}(s)1.00 x 10

^{-37}= 84375s^{9}s = 2.1955 x 10

^{-5}MThis is 2.1955 x 10

^{-6}mol / 100mL2.1955 x 10

^{-6}mol / 100mL times 502.3069 g/mol = 1.10 x 10^{-2}g / 100mL

By the way, we can use the molarity to determine the pH of a saturated solution of hydroxyapatite. It is 9.342.

You may be interested in seeing the above two questions answered on Yahoo Answers. The question there has an additional part concerning the chemical reasoning behind fluoridation of water.

**Example #5:** The K_{sp} for magnesium arsenate, Mg_{3}(AsO_{4})_{2}, is 2.10 x 10^{-20} at 25 °C. What is the solubility of magnesium arsenate in g/L?

**Solution:**

1) Determine the molar solubility:

K_{sp}= [Mg^{2+}]^{3}[PO_{4}^{3-}]^{2}2.10 x 10

^{-20}= (3s)^{3}(2s)^{2}2.10 x 10

^{-20}= 108s^{5}s = 0.0000454736 M <--- left some extra digits, will round off at end

2) Convert from mol/L to g/L:

0.0000454736 mol/L times 350.751 g/mol = 0.0159 g/L (to three sig figs)