Solubility Equilibrium

Solubility Product Constant

• Ionic compounds (salts) differ in their solubilities

• Most “insoluble” salts will actually dissolve to some extent in water• Better said to be slightly, or sparingly, soluble in water

• The equilibrium between solids and ions is different from the equilibrium between gases

• The equilibrium between solids and ions is a “phase” equilibrium

NaCl(s)

Na+(aq) + Cl-(aq)

Ksp deals with a phase equilibrium: (s) (aq)

Solubility Product

Consider: NaCl Na+(aq) + Cl-(aq)

NaCl(s)

Na+(aq) + Cl-(aq)

Ksp = [Na+(aq)] [Cl-(aq)]

K =[Na+(aq)] [Cl-(aq)]

[NaCl(s)]

[NaCl(s)] doesn’t change (it’s a constant)

K • [NaCl(s)] = [Na+(aq)] [Cl-(aq)]

Solubility Product

• A solid always dissolves until no more can dissolve - called molar solubility (mol/L or M)

• An equilibrium is established when the amount dissolving equals the amount precipitating

• This can only be true if there is some solid (we can usually see if there is an equilibrium)

• A solution with solid remaining (i.e. in equilibrium) is called “saturated”

Note: adding more solid will not affect equilibrium

Solubility Product

Solubility Product

Ksp = [7.1x10–7][7.1x10–7] = 5.0 x 10–13

Ksp = [Ag+(aq)][Br –(aq)]

1.00 L of water dissolves 7.1x10-7 mol of AgBr. What is the Ksp?

AgBr(s) Ag+(aq) + Br–(aq)

Solubility Product

Ksp = [1.7x10-3][0.1034]2 = 1.7 x 10–5

Ksp = [Pb2+(aq)][Cl–(aq)]2

The molar solubility of PbCl2 is 1.7x10-3 M in a

0.10 M NaCl solution. What is Ksp?

NaCl(s) Na+(aq) + Cl–(aq) PbCl2(s) Pb2+(aq) + 2Cl–(aq)

(Cl- from NaCl must be added to Cl- from PbCl2)

Solubility Product

Solubility Product• Find the concentration of ions present in calcium

fluoride (in water) and the molar solubility.

CaF2(s) Ca+2 + 2 F-

-x +x +2x

Ksp = [Ca+2] [F-]2

(from table) 2 x 10-10 = [x] [2x]2

3.68 x 10 -4 = x

x = [Ca+2 ] = 3.68 x 10-4 2x = [F-] = 7.37 x 10-4

Solubility of CaF2 = 3.68 x 10 -4

Solubility Product

Find the molar solubility of silver chloride (in water).

AgCl (s) Ag+ + Cl - Ksp = 1.6 x 10-10

Copper (II) azide has Ksp = 6.3 x 10–10. Find the solubility of Cu(N3)2 in water, in g/L.

Solubility Product

At any given time, the ion product Q = [ Ba2+ ] [SO4

2– ]

If Q > Ksp... ppt

If Q < Ksp... no ppt

If Q = Ksp... equilibrium

selective precipitation: using the different solubilities of ions to separate them

Solubility Product

Will a precipitate form from mixing 50.0 mL of 8.0 x 10–3 M Pb(NO3)2(aq) and 50.0 mL of 5.0 x 10–3 M Na2SO4(aq)? Ksp = 1.6 x 10-8

Solubility Product

The Ksp of CuS is 2.5 x 10-41 at 18°C. If 2.0 x 10-6

mole of Cu(NO3)2 and 2.0 x 10-6 mole of Na2S are mixed in enough water to give a total volume of 10.0 liters, will a precipitate form?

Solubility Product

Calculate the hydrogen ion concentration of a 0.25 M solution of hypochlorous acid, HOCl, for which Ka = 3.5 x 10-8.

Solubility Product