### Calculating gram per 100 mL solubility, given the Ksp

Basically, this type of problem will (1) calculate the molar solubility from the Ksp. 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 Ag2CO3 (Ksp for silver carbonate = 8.4 x 10¯12)

Solution:

1) Solve for the molar solubility of Ag2CO3:

Ag2CO3 ⇌ 2Ag+ + CO32¯

Ksp = [Ag+]2 [CO32¯]

8.4 x 10¯12 = (2s)2 (s)

s = 1.28 x 10¯4 mol/L

Comment: 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/100mL

Comment: 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/250mL

Comment: 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/250mL

0.00690 g/250mL = 6.90 mg/250mL

Example #3: Calculate the mass of Ca5(PO4)3F (Ksp = 1.00 x 10¯60) which will dissolve in 100 mL of water.

Solution:

1) Determine moles of Ca5(PO4)3F in one liter:

Ca5(PO4)3F(s) ⇌ 5Ca2+ + 3PO43- + F¯

Ksp = [Ca2+]5[PO43-]3[F¯]

1.00 x 10¯60 = (5s)5(3s)3(s)

1.00 x 10¯60 = 84375s9

s = 6.11 x 10¯8 M

2) Determine moles of Ca5(PO4)3F 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 Ca5(PO4)3OH (Ksp = 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 = 84375s9

s = 2.7166 x 10¯5 M

This 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 Ksp for magnesium arsenate, Mg3(AsO4)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:

Ksp = [Mg2+]3 [PO43-]2

2.10 x 10¯20 = (3s)3 (2s)2

2.10 x 10¯20 = 108s5

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)