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Ksp : Solubility Product Chemistry Tutorial

Key Concepts

Solubility as an Equilibrium Process

When you were learning your solubility rules for ionic compounds dissolving in water at 25°C, you were probably taught that barium sulfate is insoluble.
But this doesn't mean that none of the barium sulfate dissolves in water at 25°C, it just means that very little of it does.
If you refer to a table of solubilities, you will find that about 10-3 g of barium sulfate can be dissolved in 100 mL of water at 25°C. OK, it's not much, but it is more than zero!

So let's imgaine an experiment in which I have 100 mL water at 25°C and I add solid barium sulfate (BaSO4) 0.0002 g at a time, stirring between each addition and analysing the solution to determine the concentration of barium ions (Ba2+(aq)) and sulfate ions (SO42-).

The results of the experiment might look like this:

total mass BaSO4
added to 100 mL water		 [Ba2+(aq)] mol L-1 [SO42-(aq)] mol L-1 Comments
0 0 0 Concentration of ions in solution is increasing as more BaSO4 is added to the solution.
All added BaSO4 is dissolved, no BaSO4 precipitates out of solution.
BaSO4(s) → Ba2+(aq) + SO42-(aq)
Solution is unsaturated (that is, more BaSO4 can be dissolved the solution).
0.0002 8.6 × 10-6 8.6 × 10-6
0.0004 1.7 × 10-5 1.7 × 10-5
0.0006 2.6 × 10-5 2.6 × 10-5
0.0008 3.4 × 10-5 3.4 × 10-5
0.001 3.9 × 10-5 3.9 × 10-5 Concentration of ions in solution is constant.
Equilibrium has been established:
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution is saturated (adding more BaSO4 causes "excess" BaSO4 to precipitate)
0.0012 3.9 × 10-5 3.9 × 10-5
0.0014 3.9 × 10-5 3.9 × 10-5
0.0016 3.9 × 10-5 3.9 × 10-5

The results above could be graphed, as shown on the right.
The flat area of the graph represents the equilibrium concentration of Ba2+(aq) when BaSO4(s) dissolves in water at 25°C,
[Ba2+(aq)] = 3.9 × 10-5 mol L-1
At this concentration of barium ions and sulfate ions the solution is saturated, adding more BaSO4 results in the precipitation of BaSO4(s).
If the concentration of barium ions (and the concentration of sulfate ions) is less than 3.9 × 10-5 mol L-1 the solution is unsaturated, that is all the BaSO4 dissolves to produce ions, and, more BaSO4 can be added to produce even more ions in solution.
[Ba2+(aq)]
(mol L-1)

total mass BaSO4 (g) added

So dissolving BaSO4 in water should be considered an equilibrium reaction:

BaSO4(s) Ba2+(aq) + SO42-(aq)

so we can write a mass-action expression for the system:

Q = [Ba2+(aq)][SO42-(aq)]
[BaSO4(s)]

But the concentration of a solid is constant, that is, if there is more solid present it will occupy a greater volume as determined by the density of the solid.
Therefore the concentration of solid barium sulfate, BaSO4(s), is a constant, and can be incorporated into the expression for the mass-action expression, Q
Q = [Ba2+(aq)][SO42-(aq)]

Note that Q is the product of the concentration of dissolved ions in the solution, the expression [Ba2+(aq)][SO42-(aq)] is known as the ion product (for barium sulfate in this example).
ion product = Q = [Ba2+(aq)][SO42-(aq)]

Let's use our concentration data above to calculate the value of the ion product at each stage of our experiment:

total mass BaSO4
added to 100 mL water		 [Ba2+(aq)] mol L-1 [SO42-(aq)] mol L-1 [Ba2+(aq)][SO42-(aq)]
(ion product)		 Comments
0 0 0 0 Value of ion product is increasing.
Equilibrium has not yet been established.
0.0002 8.6 × 10-6 8.6 × 10-6 7.4 × 10-11
0.0004 1.7 × 10-5 1.7 × 10-5 2.9 × 10-10
0.0006 2.6 × 10-5 2.6 × 10-5 6.8 × 10-10
0.0008 3.4 × 10-5 3.4 × 10-5 1.2 × 10-9
0.001 3.9 × 10-5 3.9 × 10-5 1.5 × 10-9 Value of ion product is constant.
Equilibrium has been established:
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution is saturated (adding more BaSO4 causes "excess" BaSO4 to precipitate)
0.0012 3.9 × 10-5 3.9 × 10-5 1.5 × 10-9
0.0014 3.9 × 10-5 3.9 × 10-5 1.5 × 10-9
0.0016 3.9 × 10-5 3.9 × 10-5 1.5 × 10-9

Equilibrium law states that the condition for chemical equilibrium is when the value of the mass-action expression, Q, is equal to the value for the equilibrium constant, Kc.
When an ionic solid dissolves to form ions in solution, Q is the ion product, and the equilibrium constant is called a solubility product and is given the symbol Ksp:

At equilibrium: ion product = solubility product (Ksp)
At equilibrium the solution is saturated, that is, it is not possible to increase the concentration of ions in solution by adding more solid, the addition of more solid simply results in it not dissolving.

For our barium sulfate experiment:

BaSO4(s) Ba2+(aq) + SO42-(aq)
At equilibrium: ion product = [Ba2+(aq)][SO42-(aq)] = 1.5 × 10-9 = solubility product (Ksp)

When the value of the ion product is less than the value for the solubility product (Ksp), the solution is unsaturated, we can continue to add more solute and it will all dissolve, that is, no precipitate forms:

When the value of the ion product is greater than the value for the solubility product (Ksp), the solution is supersaturated with respect to one or both of the ions, the composition of the solution is unstable and the equilibrium position shifts to the left to produce a precipitate, reducing the concentration of ions in solution, until equilibrium is established.

Values for various solubility products, Ksp, are tabulated on the right.
Note the tabulated value of Ksp for barium sulfate at 25°C is 1.5 × 10-9.

Note that Ksp is an equilibrium constant so it is temperature dependent, and tables of values are produced for a specific temperature (usually 25°C).

We can use the values of Ksp to:

  • calculate the concentration of ionic species in a saturated solution
    for MA(s) M+(aq) + A-(aq)
    Ksp = [M+(aq)][A-(aq)]
    [M+(aq)] = [A-(aq)] = √Ksp

  • calculate the solubility of an ionic compound
    for MA(s) M+(aq) + A-(aq)
    the solubility of MA (mol L-1) = [M+(aq)] = [A-(aq)] = √Ksp

  • decide if a precipitate will form when two aqueous solutions are mixed together:

    ion product < Ksp : no precipitate (unsaturated solution)
    ion product = Ksp : no precipitate (equilibrium, saturated solution)
    ion product > Ksp : precipitate forms (supersaturated solution)

Solubility Products
at 25°C
Ionic
Compound		 Ksp
Ag2S 5.5 × 10-51
CuS 8.0 × 10-37
Al(OH)3 5.0 × 10-33
ZnS 1.2 × 10-23
Cu(OH)2 1.6 × 10-19
Fe(OH)2 2.0 × 10-15
Ag2CrO4 1.9 × 10-12
Mg(OH)2 8.9 × 10-12
AgCl 1.7 × 10-10
CaF2 1.7 × 10-10
BaSO4 1.5 × 10-9
Ca(OH)2 1.3 × 10-6
PbCl2 1.6 × 10-5
BaF2 1.7 × 10-6
CaSO4 2.4 × 10-5

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Example : Calculating the solubility of an ionic compound (MA).

(based on the StoPGoPS approach to problem solving)

Question: Calculate how much silver bromide in moles will dissolve in 1 L of water at 25°C given Ksp = 5.0 x 10-13 at 25oC.

Response:

  1. What have you been asked to do?

    2. calculate moles of silver bromide dissolved in 1 L of water at 25°C:

    n(AgBr(aq)) = ? mol (saturated solution)

  2. What information have you been given in the question?

    3. Ksp = 5.0 x 10-13 at 25oC.

  3. What is the relationship between what you know and what you need to find out?

    4. AgBr(s) Ag+(aq) + Br-(aq)

    Ksp = [Ag+(aq)][Br-(aq)]

    From balanced chemical equation: [Ag+(aq)] = [Br-(aq)]

    let x = [Ag+(aq)] = [Br-(aq)]

    Ksp = x2

    x = √Ksp

    For a saturated solution: x = [Ag+(aq)] = [Br-(aq)] = [AgBr(aq)]

    [AgBr(aq)] is moles of AgBr dissolved in 1 L of solution.
    Assume volume of AgBr is negligible compared to the volume of water, so 1 L of solution = 1 L of water.
    [AgBr(aq)] = n(AgBr(aq)) mol ÷ 1 L
    n(AgBr(aq)) = [AgBr(aq)]

  4. Perform the calculations.

    5. x = √Ksp = √5.0 x 10-13 = 7 × 10-7 mol L-1

    [AgBr(aq)] = 7 × 10-7 mol L-1

    7 × 10-7 moles of AgBr is dissolved in 1 L of solution.

  5. Is your answer plausible?

    6. Use your calculated values for concentration to calculate Ksp (checking your original calculations)

    Ksp = x2 = (7 × 10-7)2 = 5 × 10-15

    Ksp was given as 5 × 10-15 so we have confidence in our calculations.

  6. State your solution to the problem.

    7. The solubility of AgBr at 25oC is 7 x 10-7 mol L-1
    7 x 10-7 moles AgBr will dissolve in 1 L of water at 25oC.

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Example : Calculating the solubility of an ionic compound (MA2)

(based on the StoPGoPS approach to problem solving)

Question: Calculate how much strontium fluoride in moles per litre will dissolve in 1 L of water given Ksp = 2.5 × 10-9 at 25oC.

Response:

  1. What have you been asked to do?

    2. Calculate solubility of strontium flouride in moles per litre

    [SrF2(aq)] = ? mol L-1

  2. What information have you been given in the question?

    3. Ksp = 2.5 × 10-9 at 25oC

  3. What is the relationship between what you know and what you need to find out?

    4. SrF2(s) Sr2+(aq) + 2F-(aq)

    Ksp = [Sr2+(aq)][F-(aq)]2

    Let x = [Sr2+(aq)]

    then [F-(aq)] = 2x (from the balanced chemical equation above)

    Ksp = [x][2x]2 = 4x3

    x = 3√ Ksp
    4

    At equilibrium, [SrF2(aq)] = [Sr2+(aq)] = x mol L-1

  4. Perform the calculations.

    5. x = 3√ Ksp
    4
    x = 3√ 2.5 × 10-9
    4
    x = 8.5 × 10-4

    [SrF2(aq)] = 8.5 × 10-4 mol L-1

  5. Is your answer plausible?

    6. Use the calculated values for concentrations to calculate Ksp (checking our calculations):

    [Sr2+(aq)] = x = 8.5 × 10-4 mol L-1

    [F-(aq)] = 2x = 2 × 8.5 × 10-4 = 1.7 × 10-3 mol L-1

    Ksp = [Sr2+(aq)][F-(aq)]2
          = 8.5 × 10-4 × (1.7 × 10-3)2
          = 2.5 × 10-9

    Ksp was given as 2.5 × 10-9 in the question, so we are confident our answer is correct.

  6. State your solution to the problem.

    7. The solubility of SrF2 at 25oC is 8.5 × 10-4 mol L-1
    8.5 × 10-4 mol L-1 moles SrF2 will dissolve in 1 L of water at 25oC.

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Example : Deciding whether a precipitate will form

(based on the StoPGoPS approach to problem solving)

Question: Will a precipitate form if 25.0 mL of 1.4 × 10-9 mol L-1 NaI and 35.0 mL of 7.9 × 10-7 mol L-1 AgNO3 are mixed?

(Ksp for AgI at 25oC is 8.5 × 10-17)

Response:

  1. What have you been asked to do?

    2. Decide if a precipitate of AgI(s) forms.

  2. What information have you been given in the question?

    3. [NaI(aq)] = c(NaI(aq)) = 1.4 × 10-9 mol L-1

    V(NaI(aq)) = 25.0 mL = 25.0/1000 = 0.0250 L

    [AgNO3(aq)] = c(AgNO3(aq)) = 7.9 × 10-7 mol L-1

    V(AgNO3(aq)) = 35.0 mL = 35.0/1000 = 0.0350 L

    Ksp(AgI) = 8.5 × 10-17

  3. What is the relationship between what you know and what you need to find out?

    4. AgI(s) Ag+(aq) + I-(aq)

    IF ion product > Ksp a precipitate will form:
    [Ag+(aq)][I-(aq)] > Ksp

    [Ag+(aq)] = moles from AgNO3(aq) ÷ total volume of solution
    [Ag+(aq)] = c(AgNO3(aq) × V(AgNO3(aq))/(V(AgNO3(aq)) + V(AgI(aq)))

    [I-(aq)] = moles from NaI(aq) ÷ total volume of solution
    [I-(aq)] = c(NaI(aq) × V(AgI(aq)/(V(AgNO3(aq)) + V(AgI(aq)))

  4. Perform the calculations.

    5. [Ag+(aq)] = c(AgNO3(aq) × V(AgNO3(aq))/(V(AgNO3(aq)) + V(AgI(aq)))
          = (7.9 × 10-7 × 0.0350)/(0.0350 + 0.0250)
          = 4.6 × 10-7 mol L-1

    [I-(aq)] = c(NaI(aq) × V(AgI(aq))/(V(AgNO3(aq)) + V(AgI(aq)))
          = (1.4 × 10-9 × 0.0250)/(0.0350 + 0.0250)
          = 5.8 × 10-10 mol L-1

    ion product = [Ag+(aq)][I-(aq)]
          = [4.6 × 10-7][5.8 × 10-10]
          = 2.7 × 10-16

    ion product > Ksp (that is 2.7 × 10-16 > 8.5 × 10-17)

    Precipitate will form

  5. Is your answer plausible?

    6. Approximations:
    [NaI] = [I-] = 1.4 × 10-9 M before mixing
    [I-] ≈ 0.7 × 10-9 M after mixing (volume roughly doubles so concentration halves)
    [AgNO3] = [Ag+] ≈ 8 × 10-7 M before mixing
    [Ag+] ≈ 4 × 10-7 M after mixing (volume roughly doubles so concentration halves)

    ion product = [Ag+][I-] = [0.7 × 10-9][4 × 10-7] = 2.8 × 10-16
    ion product > Ksp so precipitate forms

  6. State your solution to the problem.

    7. A precipitate of AgI(s) will form.

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