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 dos 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.

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

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 ratio between AgBr and Ag^{+} and there is a 1:1 ratio between AgBr and Br¯. This means that, when 5.71 x 10¯^{7} mole per liter of AgBr dissolves, it produces 5.71 x 10¯^{7} moles per liter of Ag^{+} and 5.71 x 10¯^{7} moles 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}

Video: Calculating the K_{sp} from the Molar Solubility

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

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

CaF_{2}(s) <===> Ca^{2+}(aq) + 2 F¯ (aq)

The K_{sp} expression is:

K_{sp}= [Ca^{2+}] [F¯]^{2}

There is a 1:1 ratio between CaF_{2} and Ca^{2+}, BUT there is a 1:2 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} moles per liter of Ca^{2+}, BUT 4.28 x 10¯^{4} moles 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. I took the Ca^{2+} concentration and doubled it to get the F¯ concentration.

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

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

Hg_{2}Br_{2}(s) <===> Hg_{2}^{2+}(aq) + 2 Br¯ (aq)

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

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} moles per liter of Hg_{2}^{2+}, BUT 5.04 x 10¯^{8} moles 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}

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

The K_{sp} expression is:

K_{sp}= [Ce^{4+}] [IO_{3}¯]^{4}

We know the following:

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

Therefore:

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

_{sp}= 4.84 x 10¯^{17}

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

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}.

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

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

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

Answers only. No detailed solutions