Solubility Product Constants

James Richard Fromm


The form of chemical equilibrium which describes the dissolution of slightly soluble salts was introduced in another section. These equilibria are governed by an equilibrium constant called the solubility product. The following section shows the use of solubility product constants in aqueous solution.

We begin by recalling that one of the forms of chemical equilibrium is that in which a slightly soluble salt dissociates in solution, such as

AgCl(s) rarrow.gif (63 bytes) Ag+(aq) + Cl-(aq).

The equilibrium constant for this dissociation is written as

K = a(Ag+)a(Cl-)/a(AgCl).

Now the activity of AgCl in solid form is constant, so the combined constants are written

Ka(AgCl) = a(Ag+)a(Cl-) = Ksp.

To a good approximation in dilute aqueous solutions this will be

Ksp = [Ag+][Cl-]

The equilibrium constant K<sp is known as the solubility product constant of silver chloride. When molar concentrations are used, as they generally are, this constant is called the molar solubility product. For salts which are not very soluble in water (slightly soluble salts), the value of the molar solubility product is virtually the same as the value of the molal solubility product because pure water has a density which is very close to 1.0 kg/dm3. Molal solubility products are more commonly used in solvents other than water. For other compounds the molar solubility product equation would be written similarly:

BaCl2: Ksp = [Ba2+][Cl-]2

Ag2SO4: Ksp = [Ag+]2[SO42-]

Since this equation was derived on the assumption that solid salt was present, it follows that it can be used if and only if such solid is present.

A table of values of the molar solubility product constants for some selected inorganic solid compounds dissolved in water is accessible here.


Example. Let us calculate the molar solubility of AgCl in water, for which the balanced dissolution equation is AgCl(s) --> Ag+ + Cl-. From this balanced chemical equation, it follows that the molar solubility is equal to [Ag+] = [Cl-]. Since Ksp = 1.76 x 10-10 = [Ag+][Cl-] = [Ag+]2, [Ag+] = 1.33 x 10-5 and the molar solubility of AgCl in water is 1.33 x 10-5 mol/liter.


For salts which produce more than two ions on dissolution the relationship between molar solubility and molar concentration of ion is not quite so simple. In the absence of any other sources of the ions, the dissolution of Ag2SO4 proceeds by the reaction Ag2SO4(s) rarrow.gif (63 bytes) 2Ag+ + SO42-.

Under these conditions, the molar solubility of Ag2SO4 is equal to [SO42-] or [Ag+]/2. In the case of Ba3(PO4)2, the dissolution reaction is

Ba3(PO4)2 rarrow.gif (63 bytes) 3Ba2+ + 2PO43-.

The molar solubility of Ba3(PO4)2 is equal to [PO43-]/2 or [Ba2+]/3; it is not equal to the value of the concentration of any of the constituent ions.


Example. The molar solubility of PbSO4 in water is 1.35 x 10-4 mol/dm3. Let us calculate the value of the molar solubility product. Since Ksp = [Pb2+][SO42-] and the molar solubility is equal to either the molar concentration of Pb2+ or the molar concentration of SO42- since each of them is produced in the ratio of one ion for each lead sulfate dissolved, Ksp = (1.35 x 10-4)2 = 1.82 x 10-8


Example. The molar solubility of CaF2 in water is 3.32 x 10-4 mol/litre. Let us calculate the value of the molar solubility product. The equilibrium is

CaF2(s) larrow.GIF (55 bytes)rarrow.gif (63 bytes) Ca2+(aq) + 2F-(aq) and

Ksp = [Ca2+][F-]2.

On dissolution each mole of CaF2 gives one mole of Ca2+ and two moles of F-. Then [F-] = 2[Ca2+] and the molar solubility is equal to the molar concentration of Ca2+.

Ksp = [Ca2+](2[Ca2+])2; Ksp = (3.32 x 10-4)(6.64 x 10-4)2 = 1.46 x 10-10.


Study Problems

1. The molar solubility of Hg2SO4 in water is 8.9 x 10-4. Calculate the molar solubility product of the salt.

2. The molar solubility product of CaCO3 is 4.96 x 10-9. Calculate the molar solubility of CaCO3 in water.

3. The molar solubility of Cu2S in water is 1.78 x 10-16. Calculate the molar solubility product of the salt.

4. The molar solubility product of MgNH4PO4, the only slightly soluble salt containing ammonium ion, is 2 x 10-13. What is the molar solubility of this salt?

5. A solution which is originally 0.01 molar in carbonate ion and sulfate ion has added to it aqueous calcium chloride. What salt precipitates first? When the second salt just begins to precipitate, what must the concentration of the first salt remaining be?

6. What is the maximum concentration of lead(II) ion which can exist in 2.0 molar sulfate solution? In what form would you expect to find the lead (II) ion in your automobile battery?

7. Calculate the molar solubility of BaSO4 in 0.01 molar Na2SO4 and in 0.01 molar NaCl. Explain the differences between these values and the molar solubility of BaSO4 in distilled water.

8. How many mL of 0.10 molar AgNO3solution would be required to titrate 5.769 g of solid NaCl?

9. A solution of an unknown metal sulfate was diluted to 100 mL, of which 25.00 mL were pipetted into a flask and titrated with 35.61 mL of 0.05 molar barium sulfate solution. Compute the concentration of the metal sulfate solution in the 100 mL of solution prepared. If the mass of anhydrous metal sulfate dissolved to form this solution were 1.075 g, what compound might it have been?

10. Calculate the solubility of barium sulfate in pure water. Explain qualitatively what would happen if a small amount of concentrated sulfuric acid were poured into a saturated aqueous barium sulfate solution.

11. Compute the concentration of lead sulfate which has gone into solution when there is solid lead sulfate sitting on the bottom of the beaker and the beaker contains 0.01 molar aqueous sodium sulfate solution.

12. In the manufacture of pure hydrogen, water gas (CO + H2, in equal quantities) and steam are passed over catalysts and react to give carbon dioxide. Write the equilibrium constant for this reaction. If the carbon dioxide is constantly removed by dissolution in cold water under pressure and the resulting gases are recycled to the catalyst chamber, what will happen? Explain in terms of Le Chatelier's principle.

13. A certain chemist treated a 0.001 molar aqueous solution of chlorine with solid mercury(II) oxide. The reaction produced solid mercury (II) chloride and a solution of hypochlorous acid. Write the equilibrium constant for this reaction. Then discuss, in the light of Le Chatelier's principle, what the effect of (1) pH of the solution, and (2) addition of an agent which would render mercury(II) chloride soluble, would have on the position of equilibrium.

14. A chemist has an aqueous solution which is 0.1 molar in Zn(II) and 0.1 molar in Hg(II). Can he quantitatively precipitate one of these with sulfide ion without precipitating the other? Show calculations to justify your answer.


Copyright 1997 James R. Fromm