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 dos 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.

Problem #1: Determine the Ksp 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 Ksp expression is:

Ksp = [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 Ksp expression, we obtain:

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

Video: Calculating the Ksp from the Molar Solubility


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

When CaF2 dissolves, it dissociates like this:

CaF2 (s) <===> Ca2+ (aq) + 2 F¯ (aq)

The Ksp expression is:

Ksp = [Ca2+] [F¯]2

There is a 1:1 ratio between CaF2 and Ca2+, BUT there is a 1:2 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 moles per liter of Ca2+, BUT 4.28 x 10¯4 moles 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. I took the Ca2+ concentration and doubled it to get the F¯ concentration.


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

When Hg2Br2 dissolves, it dissociates like this:

Hg2Br2 (s) <===> Hg22+ (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 Ksp expression.

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 moles per liter of Hg22+, BUT 5.04 x 10¯8 moles 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

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

The Ksp expression is:

Ksp = [Ce4+] [IO3¯]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:

Ksp = (1.80 x 10¯4) (7.20 x 10¯4)4

Ksp = 4.84 x 10¯17


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

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.

Problem #6: AgCN, 7.73 x 10¯9 M

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

Problem #8: Hg2I2, 2.37 x 10¯10 M

Answers only. No detailed solutions


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