Calculating the Ksp from Molar Solubility (2024)

Calculating the K_sp 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 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:

1) When AgBr dissolves, it dissociates like this:

AgBr(s) ⇌ Ag⁺(aq) + Br¯(aq)

2) The K_sp expression is:

K_sp = [Ag⁺] [Br¯]

3) 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.

4) 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 (CaF₂), given that its molar solubility is 2.14 x 10¯⁴ moles per liter.

Solution:

1) When CaF₂ dissolves, it dissociates like this:

CaF₂(s) ⇌ Ca²⁺(aq) + 2F¯(aq)

2) The K_sp expression is:

K_sp = [Ca²⁺] [F¯]²

3) 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.

4) 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 CaF₂ dissolving.

Example #3: Determine the K_sp 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 K_sp expression and the wrong answer to the problem.

The K_sp expression is:

K_sp = [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⁺, as well as 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(IO₃)₄, given that its molar solubility is 1.80 x 10¯⁴ mol/L

Solution:

The K_sp expression is:

K_sp = [Ce⁴⁺] [IO₃¯]⁴

We know the following:

There is a 1:1 molar ratio between the molar solubility and the cerium(IV) ion concentration.

These is a 1:4 molar ratio between the molar solubility and the iodate ion concentration.

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 Mg₃(PO₄)₂, given that its molar solubility is 3.57 x 10^-6 mol/L.

Solution:

The K_sp expression is:

K_sp = [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:

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.

	Molar Solubility	Ksp
Example #6:AgCN	7.73 x 10¯9 M	5.97 x 10¯17
Example #7:Zn3(AsO4)2	1.236 x 10¯6 M	3.12 x 10¯28
Example #8:Hg2I2	2.37 x 10¯10 M	5.33 x 10¯29

The values given for the K_sp answers are from a reference source. Due to rounding, the K_sp value you calculate may be slightly different, but it should be close.

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 K_sp for MgF₂.

Solution:

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

MgF₂(s) ⇌ Mg²⁺(aq) + 2F¯(aq)

2) The K_sp expression is this:

K_sp = [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 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 Ba₃(PO₄)₂ is 8.89 x 10¯⁹ M in pure water. Calculate the K_sp for Ba₃(PO₄)₂

Solution:

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

Ba₃(PO₄)₂(s) ⇌ 3Ba²⁺(aq) + 2PO₄³¯(aq)

2) Write the K_sp expression for barium phosphate:

K_sp = [Ba²⁺]³ [PO₄³¯]²

3) Determine [Ba²⁺] and [PO₄³¯]:

[Ba²⁺] = (3) (8.89 x 10¯⁹ M) = 2.667 x 10¯⁸ M
[PO₄³¯] = (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 Ba₃(PO₄)₂ is not involved at any point. This is because we were given a molarity for how much Ba₃(PO₄)₂ dissolved, as opposed to a gram amount. If a gram amount had been given, then the formula weight would have been involved.

Example #11: The molar solubility of Ag₂SO₄ is 1.44 x 10¯² mol/L. Calculate the K_sp of this compound.

Solution:

1) Write the chemical equation and the K_sp expression:

Ag₂SO₄(s) ⇌ 2Ag⁺(aq) + SO₄²⁺(aq)

K_sp = [Ag⁺]² [SO₄²⁺]

2) Based on the stoichiometry of the reaction, we determine the molar solubility.

For every one Ag₂SO₄ formula unit that dissolves and ionizes, one sulfate ion is present in solution. Because of this 1:1 molar ratio, we know the molar solubility of sulfate is 1.44 x 10¯² mol/L

For every one Ag₂SO₄ formula unit that dissolves and ionizes, two silver ions are present in solution. Because of this 1:2 molar ratio, we know the molar solubility of the silver ion is 2.88 x 10¯² mol/L.

3) Determine the K_sp:

K_sp = (2.88 x 10¯²)² (1.44 x 10¯²) = 1.2 x 10¯⁵

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