The molar solubility of a substance is the number of moles that dissolve per liter of solution. For very soluble substances (like sodium nitrate, NaNO_{3}), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases.

For insoluble substances like silver bromide (AgBr), the molar solubility can be quite small. In the case of AgBr, the value is 5.71 x 10¯^{7} moles per liter.

Given this value, how does one go about calculating the K_{sp} of the substance? Here is a skeleton outline of the process:

1) Write the chemical equation for the substance dissolving and dissociating.

2) Write the K_{sp}expression.

3) Insert the concentration of each ion and multiply out.

**Example #1:** Determine the K_{sp} of silver bromide, given that its molar solubility is 5.71 x 10¯^{7} moles per liter.

**Solution:**

When AgBr dissolves, it dissociates like this:

AgBr(s) ⇌ Ag^{+}(aq) + Br¯(aq)

The K_{sp} expression is:

K_{sp}= [Ag^{+}] [Br¯]

There is a 1:1 molar ratio between the AgBr that dissolves and Ag^{+} that is in solution. In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This means that, when 5.71 x 10¯^{7} mole per liter of AgBr dissolves, it produces 5.71 x 10¯^{7} mole per liter of Ag^{+} and 5.71 x 10¯^{7} mole per liter of Br¯ in solution.

Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (5.71 x 10¯^{7}) (5.71 x 10¯^{7}) = 3.26 x 10¯^{13}

**Example #2:** Determine the K_{sp} of calcium fluoride (CaF_{2}), given that its molar solubility is 2.14 x 10¯^{4} moles per liter.

**Solution:**

When CaF_{2} dissolves, it dissociates like this:

CaF_{2}(s) ⇌ Ca^{2+}(aq) + 2F¯(aq)

The K_{sp} expression is:

K_{sp}= [Ca^{2+}] [F¯]^{2}

There is a 1:1 molar ratio between CaF_{2} and Ca^{2+}, BUT there is a 1:2 molar ratio between CaF_{2} and F¯. This means that, when 2.14 x 10¯^{4} mole per liter of CaF_{2} dissolves, it produces 2.14 x 10¯^{4} mole per liter of Ca^{2+} and it produces 4.28 x 10¯^{4} mole per liter of F¯ in solution.

Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (2.14 x 10¯^{4}) (4.28 x 10¯^{4})^{2}= 3.92 x 10¯^{11}

Please note, I DID NOT double the F¯ concentration. The F¯ concentration is TWICE the value of the amount of CaF_{2} dissolving.

**Example #3:** Determine the K_{sp} of mercury(I) bromide (Hg_{2}Br_{2}), given that its molar solubility is 2.52 x 10¯^{8} mole per liter.

**Solution:**

When Hg_{2}Br_{2} dissolves, it dissociates like this:

Hg_{2}Br_{2}(s) ⇌ Hg_{2}^{2+}(aq) + 2Br¯(aq)

Important note: it is NOT 2Hg^{+}. IT IS NOT!!! If you decide that you prefer 2Hg^{+}, then I cannot stop you. However, it will give the wrong K_{sp} expression and the wrong answer to the problem.

The K_{sp} expression is:

K_{sp}= [Hg_{2}^{2+}] [Br¯]^{2}

There is a 1:1 ratio between Hg_{2}Br_{2} and Hg_{2}^{2+}, BUT there is a 1:2 ratio between Hg_{2}Br_{2} and Br¯. This means that, when 2.52 x 10¯^{8} mole per liter of Hg_{2}Br_{2} dissolves, it produces 2.52 x 10¯^{8} mole per liter of Hg_{2}^{2+}, BUT 5.04 x 10¯^{8} mole per liter of Br¯ in solution.

Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (2.52 x 10¯^{8}) (5.04 x 10¯^{8})^{2}= 6.40 x 10¯^{23}

**Example #4:** Calculate the K_{sp} for Ce(IO_{3})_{4}, given that its molar solubility is 1.80 x 10¯^{4} mol/L

**Solution:**

The K_{sp} expression is:

K_{sp}= [Ce^{4+}] [IO_{3}¯]^{4}

We know the following:

There is a 1:1 molar ratio between the molar solubility and the cerium(IV) ion's concentration.These is a 1:4 molar ratio between the molar solubility and the iodate ion.

Therefore:

K_{sp}= (1.80 x 10¯^{4}) (7.20 x 10¯^{4})^{4}K

_{sp}= 4.84 x 10¯^{17}

**Example #5:** Calculate the K_{sp} for Mg_{3}(PO_{4})_{2}, given that its molar solubility is 3.57 x 10^{-6} mol/L.

**Solution:**

The K_{sp} expression is:

K_{sp}= [Mg^{2+}]^{3}[PO_{4}^{3}¯]^{2}

We know the following:

These is a 3:1 ratio between the concentration of the magnesium ion and the molar solubility of the magnesium phosphate.There is a 2:1 ratio between the concentation of the phosphate ion and the molar solubility of the magnesium phosphate.

Therefore:

K_{sp}= (1.071 x 10¯^{5})^{3}(7.14 x 10¯^{6})^{2}K

_{sp}= 6.26 x 10¯^{26}

For what it's worth, my "Handbook of Chemistry and Physics" gives the K_{sp} as 9.86 x 10¯^{25}

For each compound, the molar solubility is given. Calculate its K_{sp}.

**Example #6:** AgCN, 7.73 x 10¯^{9} M

**Example #7:** Zn_{3}(AsO_{4})_{2}, 1.236 x 10¯^{6} M

**Example #8:** Hg_{2}I_{2}, 2.37 x 10¯^{10} M

Answers only. No detailed solutions

**Example #9:** A saturated solution of magnesium fluoride , MgF_{2}, was prepared by dissolving solid MgF_{2} in water. The concentration of Mg^{2+} ion in the solution was found to be 2.34 x 10^{-4} M. Calculate the K_{sp} for MgF_{2}.

**Solution:**

1) Here's the chemical equation for the dissolving of MgF_{2}:

MgF_{2}(s) ⇌ Mg^{2+}(aq) + 2F¯(aq)

2) The K_{sp} expression is this:

K_{sp}= [Mg^{2+}] [F¯]^{2}

3) Based on the stoichiometry of the chemical equation, the [F¯] is this:

4.68 x 10^{-4}M

4) To calculate the K_{sp}, do this:

K_{sp}= (2.34 x 10^{-4}) (4.68 x 10^{-4})^{2}To three sig figs, the K

_{sp}is 5.12 x 10^{-11}

**Example #10:** The molar solubility of Ba_{3}(PO_{4})_{2} is 8.89 x 10¯^{9} M in pure water. Calculate the K_{sp} for Ba_{3}(PO_{4})_{2}

**Solution:**

1) Write the chemical equation for the dissolving of barium phosphate in water:

Ba_{3}(PO_{4})_{2}(s) ⇌ 3Ba^{2+}(aq) + 2PO_{4}^{3}¯(aq)

2) Write the K_{sp} expression for barium phosphate:

K_{sp}= [Ba^{2+}]^{3}[PO_{4}^{3}¯]^{2}

3) Determine [Ba^{2+}] and [PO_{4}^{3}¯]:

[Ba^{2+}] = (3) (8.89 x 10¯^{9}M) = 2.667 x 10¯^{8}M

[PO_{4}^{3}¯] = (2) (8.89 x 10¯^{9}M) = 1.778 x 10¯^{8}M

4) Put values into and then solve the K_{sp} expression:

K_{sp}= (2.667 x 10¯^{8})^{3}(1.778 x 10¯^{8})^{2}= 5.996979 x 10¯^{39}to three sig figs, 6.00 x 10¯

^{39}

5) Note that the formula weight of Ba_{3}(PO_{4})_{2} is not involved at any point. This is because we were given a molarity for how much Ba_{3}(PO_{4})_{2} dissolved, as opposed to a gram amount. If a gram amount had been given, then the formula weight would have been involved.