Calculating the Ksp from the Molar Solubility

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The molar solubility of a substance is the number of moles that dissolve per liter of solution. For very soluble substances (like sodium nitrate, NaNO3), 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 Ksp 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 Ksp expression.
3) Insert the concentration of each ion and multiply out.

Example #1: Determine the Ksp 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 Ksp expression is:

Ksp = [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 Ksp expression, we obtain:

Ksp = (5.71 x 10¯7) (5.71 x 10¯7) = 3.26 x 10¯13

Example #2: Determine the Ksp of calcium fluoride (CaF2), given that its molar solubility is 2.14 x 10¯4 moles per liter.

Solution:

When CaF2 dissolves, it dissociates like this:

CaF2(s) ⇌ Ca2+(aq) + 2F¯(aq)

The Ksp expression is:

Ksp = [Ca2+] [F¯]2

There is a 1:1 molar ratio between CaF2 and Ca2+, BUT there is a 1:2 molar ratio between CaF2 and F¯. This means that, when 2.14 x 10¯4 mole per liter of CaF2 dissolves, it produces 2.14 x 10¯4 mole per liter of Ca2+ and it produces 4.28 x 10¯4 mole per liter of F¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (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 CaF2 dissolving.


Example #3: Determine the Ksp of mercury(I) bromide (Hg2Br2), given that its molar solubility is 2.52 x 10¯8 mole per liter.

Solution:

When Hg2Br2 dissolves, it dissociates like this:

Hg2Br2(s) ⇌ Hg22+(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 Ksp expression and the wrong answer to the problem.

The Ksp expression is:

Ksp = [Hg22+] [Br¯]2

There is a 1:1 ratio between Hg2Br2 and Hg22+, BUT there is a 1:2 ratio between Hg2Br2 and Br¯. This means that, when 2.52 x 10¯8 mole per liter of Hg2Br2 dissolves, it produces 2.52 x 10¯8 mole per liter of Hg22+, BUT 5.04 x 10¯8 mole per liter of Br¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (2.52 x 10¯8) (5.04 x 10¯8)2 = 6.40 x 10¯23

Example #4: Calculate the Ksp for Ce(IO3)4, given that its molar solubility is 1.80 x 10¯4 mol/L

Solution:

The Ksp expression is:

Ksp = [Ce4+] [IO3¯]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:

Ksp = (1.80 x 10¯4) (7.20 x 10¯4)4

Ksp = 4.84 x 10¯17


Example #5: Calculate the Ksp for Mg3(PO4)2, given that its molar solubility is 3.57 x 10-6 mol/L.

Solution:

The Ksp expression is:

Ksp = [Mg2+]3 [PO43¯]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:

Ksp = (1.071 x 10¯5)3 (7.14 x 10¯6)2

Ksp = 6.26 x 10¯26

For what it's worth, my "Handbook of Chemistry and Physics" gives the Ksp as 9.86 x 10¯25


For each compound, the molar solubility is given. Calculate its Ksp.

Example #6: AgCN, 7.73 x 10¯9 M

Example #7: Zn3(AsO4)2, 1.236 x 10¯6 M

Example #8: Hg2I2, 2.37 x 10¯10 M

Answers only. No detailed solutions


Example #9: A saturated solution of magnesium fluoride , MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 2.34 x 10-4 M. Calculate the Ksp for MgF2.

Solution:

1) Here's the chemical equation for the dissolving of MgF2:

MgF2(s) ⇌ Mg2+(aq) + 2F¯(aq)

2) The Ksp expression is this:

Ksp = [Mg2+] [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 Ksp, do this:

Ksp = (2.34 x 10-4) (4.68 x 10-4)2

To three sig figs, the Ksp is 5.12 x 10-11

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