Solubility Product Constants, K_sp

Solubility product constants are used to describe saturated solutions of ionic compounds of relatively low solubility. A saturated solution is in a state of dynamic equilibrium between the dissolved, dissociated, ionic compound and the undissolved solid.

M_xA_y(s) --> x M^y+(aq) + y A^x-(aq)

The general equilibrium constant for such processes can be written as:

K_c = [M^y+]^x[A^x-]^y

Since the equilibrium constant refers to the product of the concentration of the ions that are present in a saturated solution of an ionic compound, it is given the name solubility product constant, and given the symbol K_sp. Solubility product constants can be calculated, and used in a variety of applications.

Calculating K_sp's from solubility data
Calculating the solubility of an ionic compound in pure water from its K_sp
Calculating the solubility of an ionic compound in a solution that contains a common ion
Determination whether a precipitate will or will not form when two solutions are combined

Calculating K_sp's from Solubility Data

In order to calculate the K_sp for an ionic compound you need the equation for the dissolving process so the equilibrium expression can be written. You also need the concentrations of each ion expressed in terms of molarity, or moles per liter, or the means to obtain these values.

Example: Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it.

First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process.

PbCl₂(s) --> Pb²⁺(aq) + 2 Cl^-(aq)

K_sp = [Pb²⁺][Cl^-]²

Second, convert the amount of dissolved lead(II) chloride into moles per liter.

(0.2207 g PbCl₂)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl₂/278.1 g PbCl₂) = 0.0159 M PbCl₂

Third, create an "ICE" table.

	PbCl₂ (s)	Pb²⁺(aq)	Cl^-(aq)
Initial Concentration	All solid	0	0
Change in Concentration	- 0.0159 M (dissolves)	+ 0.0159 M	+ 0.0318 M
Equilibrium Concentration	Less solid	0.0159 M	0.0318 M

Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for K_sp.

K_sp = [0.0159][0.0318]² = 1.61 x 10^-5

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Calculating the Solubility of an Ionic Compound in Pure Water from its K_sp

Example: Estimate the solubility of Ag₂CrO₄ in pure water if the solubility product constant for silver chromate is 1.1 x 10^-12.

Write the equation and the equilibrium expression.

Ag₂CrO₄(s) --> 2 Ag⁺(aq) + CrO₄^2-(aq)

K_sp = [Ag⁺]²[CrO₄^2-]

Make an "ICE" chart.

Let "x" be the number of moles of silver chromate that dissolves in every liter of solution (its solubility).

Ag₂CrO₄(s) Ag⁺(aq) CrO₄^2-(aq)

Initial Concentration All solid 0 0

Change in Concentration - x dissolves + 2 x + x

Equilibrium Concentraion Less solid 2 x x

Substitute the equilibrium amounts and the K_sp into the equilibrium expression and solve for x.

1.1 x 10^-12 = [2x]²[x]
x = 6.50 x 10^-5 M

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Calculating the Solubility of an Ionic Compound in a Solution that Contains a Common Ion

The solubility of an ionic compound decreases in the presence of a common ion. A common ion is any ion in the solution that is common to the ionic compound being dissolved. For example, the chloride ion in a sodium chloride solution is common to the chloride in lead(II) chloride. The presence of a common ion must be taken into account when determining the solubility of an ionic compound. To do this, simply use the concentration of the common ion as the initial concentration.

Example: Estimate the solubility of barium sulfate in a 0.020 M sodium sulfate solution. The solubility product constant for barium sulfate is 1.1 x 10^-10.

Write the equation and the equilibrium expression for the dissolving of barium sulfate.

BaSO₄(s) --> Ba²⁺(aq) + SO₄^2-(aq)

Ksp = [Ba²⁺][SO₄^2-]

Make an "ICE" chart.

Let "x" represent the barium sulfate that dissolves in the sodium sulfate solution expressed in moles per liter.

BaSO₄(s) Ba²⁺(aq) SO₄^2-(aq)

Initial Concentration All solid 0 0.020 M (from Na₂SO₄)

Change in Concentration - x dissolves + x + x

Equilibrium Concentration Less solid x 0.020 M + x

Substitute into the equilibrium expression and solve for x. We will make the assumption that since x is going to be very small (the solubility is reduced in the presence of a common ion), the term "0.020 + x" is the same as "0.020." (You can leave x in the term and use the quadratic equation or the method of successive approximations to solve for x, but it will not improve the significance of your answer.)

1.1 x 10^-10 = [x][0.020 + x] = [x][0.020]
x = 5.5 x 10^-9 M

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Determining Whether a Precipitate will, or will not Form When Two Solutions are Combined

When two electrolytic solutions are combined, a precipitate may, or may not form. In order to determine whether or not a precipitate will form or not, one must examine two factors. First, determine the possible combinations of ions that could result when the two solutions are combined to see if any of them are deemed "insoluble" base on solubility tables (K_sp tables will also do). Second, determine if the concentrations of the ions are great enough so that the reaction quotient Q exceeds the K_sp value. One important factor to remember is there is a dilution of all species present and must be taken into account.

Example: 25.0 mL of 0.0020 M potassium chromate are mixed with 75.0 mL of 0.000125 M lead(II) nitrate. Will a precipitate of lead(II) chromate form. K_sp of lead(II) chromate is 1.8 x 10^-14.

First, determine the overall and the net-ionic equations for the reaction that occurs when the two soltutions are mixed.

K₂CrO₄(aq) + Pb(NO₃)₂(aq) --> 2 KNO₃(aq) + PbCrO₄(s)
Pb²⁺(aq) + CrO₄^2-(aq) --> PbCrO₄(s)
The latter reaction can be written in terms of K_sp as:
PbCrO₄(s) --> Pb²⁺(aq) + CrO₄^2-(aq)

Ksp = [Pb²⁺][CrO₄^2-]

Using the dilution equation, C₁V₁ = C₂V₂, determine the initial concentration of each species once mixed (before any reaction takes place).

(0.0020 M K₂CrO₄)(25.0 mL) = (C₂)(100.0 mL)
C₂ for K₂CrO₄ = 0.00050 M
Similar calculation for the lead(II) nitrate yields:
C₂ for Pb(NO₃)₂ = 0.0000938 M

Using the initial concentrations, calculate the reaction quotient Q, and compare to the value of the equilibrium constant, K_sp.

Q = (0.0000938 M Pb²⁺)(0.00050 M CrO₄^2-) = 4.69 x 10^-8
Q is greater than K_sp so a precipitate of lead(II) chromate will form.

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	Ag₂CrO₄(s)	Ag⁺(aq)	CrO₄^2-(aq)
Initial Concentration	All solid	0	0
Change in Concentration	- x dissolves	+ 2 x	+ x
Equilibrium Concentraion	Less solid	2 x	x

	BaSO₄(s)	Ba²⁺(aq)	SO₄^2-(aq)
Initial Concentration	All solid	0	0.020 M (from Na₂SO₄)
Change in Concentration	- x dissolves	+ x	+ x
Equilibrium Concentration	Less solid	x	0.020 M + x

Solubility Product Constants, Ksp

Solubility Product Constants, K_sp