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 / 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) / 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:**

6.80 x 10¯^{37}= (5s)^{5}(3s)^{3}(s)6.80 x 10¯

^{37}= 84375s^{9}s = 2.7166 x 10¯

^{5}MThis is 2.7166 x 10¯

^{6}mol / 100mL(2.7166 x 10¯

^{6}mol / 100mL) (502.3069 g/mol) = 1.36 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 9.434.

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