Solubility Equilibria

· Write a balanced chemical equation to represent equilibrium in a saturated solution. · Write a solubility product expression. · Answer questions about Ksp and various missing

concentrations using I.C.E. tables.

C6H12O6(s) C6H12O6(aq)

There are 3 actions that affect solubility:

1. Nature of the solute and solvent“like dissolves like”

Polar / ionic solute dissolve in polar solvent.Non-polar dissolve in non-polar.

Even the most insoluble ionic solids are actually soluble in water to a limited extent

2. Temperature

Solids in liquids: ↑ temperature - ↑ solubility.Gases in liquids: ↑ in temperature - ↓ solubility.

3. Pressure

Does not affect the solubility of (s)/(l).(g): ↑ pressure ↑ solubility.

AaBb(s) aA+(aq) + bB¯(aq)

Ksp, called the solubility product constant.

Ksp = [A+]a[B-]b

Product of ion concentrations in a saturated solution.

Kc =

[A+]a[B-]b [AaBb]

Write the dissociation and the product constant equation for the solubility of calcium hydroxide.

Ksp = [Ca2+][OH-]2

Ca(OH)2 (s)

Pb3(PO4)2(s) 3 Pb2+(aq) + 2 PO4

3-(aq)

Ksp = [Pb2+]3[PO43-]2

Write a solubility product expression for Pb3(PO4)2.

Ca2+(aq) + OH-

(aq)2

At equilibrium, the [Ag+] = 1.3 x 10-5 M and the [Cl-] = 1.3 x 10-5 M, what is the Ksp of silver chloride?

Ksp = [Ag+][Cl-]

Ksp =(1.3 x 10-5)(1.3 x 10-5)Ksp = 1.7 x 10-10

AgCl (s) Ag+(aq) + Cl-

(aq)

*NOTE: Ksp has no units.

SolubilityAnd

I.C.E. Tables(Yeah!)

Solubility - maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature.

Calculate Ksp of lead (II) chloride if a 1.0 L saturated solution has of lead ions.

I --- 0 0C --- +x +2xE ---

Ksp = [Pb+2][Cl -]2

Ksp = [1.62 x 10 -2][ 3. 24 x 10 -2]2

Ksp = 1.70 x 10 -5

PbCl2(s) Pb2+(aq) + 2 Cl-

(aq)

1.62 x 10-2 M

2(1.62 x 10-2)

1.62 x 10-2 M

The solubility of PbF2 is . What is the value of the solubility product constant?

PbF2(s) Pb2+(aq) + 2 F¯(aq)

0.466 g1 L 245.2 g

1 mol= 1.90 x 10-3 M PbF2

0.466 g/L

Pb – 207 + 2 (19) = 245g/mol

Ksp = [Pb2+][F-]2

Ksp = [Pb2+][F-]2

Ksp = (1.90 x 10-3)(3.80 x 10-3)2

Ksp = 2.74 x 10-8

PbF2(s) Pb2+(aq) + 2 F¯(aq)

[E] 0 1.9 x 10-3 M

[I] 1.9 x 10-3 mol/L 0 0

[C] - x + x + 2x

3.8 x 10-3 MSaturated – all solid reactant dissociates.

Calculate Ksp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride.

I 0.0159 0 0C -x +x +2xE 0 0.0159 M 0.0318 M

Ksp = [Pb2+][Cl-]2

0.2207 g278.1g1 mol = 0.0159 M PbCl20.05 L

PbCl2(s) Pb2+(aq) + 2 Cl-

(aq)

Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

Ksp of magnesium hydroxide is 8.9 x 10-12. What are the [equilibrium] of ions in saturated solution?

I --- 0 0C --- +x +2xE --- x 2x

Mg(OH)2 (s) Mg2+(aq)

+ 2 OH-(aq)

Ksp = [Mg2+][OH-]2

8.9 x 10-12 = [x][2x]2

8.9 x 10-12 = [x]4x2

8.9 x 10-12 = 4x3

[Mg2+] = x = 1.3 x 10-4 mol/L

[OH-] = 2x = 2.6 x 10-4 mol/L

8.9 x 10-12 = 4x3

44

2.23 x 10-12 = x33√ 3√

1.3 x 10-4 = x

Estimate the solubility in g/L of Ag2CrO4 if the Ksp is 1.1 x 10-12.

I --- 0 0C --- +2x +xE ---

Ksp = [Ag+]2[CrO42-]

1.1 x 10 -12 = [2x]2[x] 1.1 x 10 -12 = 4x3

x = 6.50 x 10 -5 M

Ag2CrO4(s) 2 Ag+(aq) + CrO4

2-(aq)

1.30 x 10 -4 M 6.50 x 10 -5 M

[Ag2CrO4]i = 6.50 x 10 -5 moles/1L

Ag2CrO4(s) 2 Ag+(aq) + CrO4

2-(aq)

6.5 x 10-5mol 330 g1 mol = 0.022 g/L

1 L

E --- 1.30 x 10 -4 M 6.50 x 10 -5 M

1 1

Precipitation

Compare value of Q, with given Ksp to determine if an aqueous solution is saturated or unsaturated.

Q = Ksp Saturated solution, no precipitate.

Q >Ksp Precipitate forms (“oversaturated”)

Q < Ksp Solution is unsaturated.

Qsp = [A+]a[B¯]b

PbF2(s) Pb2+(aq) + 2 F¯(aq)

Pb – 207 + 2 (19) = 245g/mol

Ksp of lead (II) fluoride is 1.6 x 10-5. If 0.57 g are mixed with 1500 mL of water, is solution saturated?

0.57 g1 L 245.2 g

1 mol= 2.32 x 10-3 mol/L

Qsp = (2.32 x 10-3)(4.64 x 10-3)2

Qsp = 5.0 x 10-8Q >Ksp Precipitate forms

Ksp = [Pb2+][F-]2

0.01 M NaCl 0.02 M Pb(NO3)2

Predict if there is a precipitate of PbCl2 if 100 mL each of of and are added together. Ksp of PbCl2 = 1.7 x 10 -5

PbF2(s) Pb2+(aq) + 2 Cl¯(aq)

Qsp = [Pb2+][Cl-]2

0.01 M NaCl 0.02 M Pb(NO3)2

C1V1 = C2V2

(0.01 M)(0.1 L) = (C2)(0.2 L) = 0.005 M Cl-

(0.02 M)(0.1 L) = (C2)(0.2 L) = 0.01 M Pb2+

Qsp = [0.01][0.005]2 = 2.5 x 10 -7

Qsp< Ksp A precipitate will not form.

Because you are adding volumes, you must account for the new diluted concentrations.

If 20.0 mL of 0.0010 M silver nitrate is mixed with 20.0 mL of 3.0 x 10-5 M potassium bromide, does silver bromide (Ksp = 5.0 x 10-13) precipitate? Assume

the volumes are additive.

AgBr(s) Ag+(aq) + Br¯(aq)

Ksp = [Ag+][Br-]

(0.001 M)(0.02 L) = (C2)(0.04 L) = 5.0 x 10-4 M Ag+

(2e-5 M)(0.02 L) = (C2)(0.04 L) = 1.5 x 10-5 M Br-

Qsp = [5.0 x 10-4][1.5 x 10-5] = 7.5 x 10 -9

Qsp> Ksp A precipitate will form.

· Substances which are insoluble are actually slightly soluble.

· The solubility product, Ksp, describes the product of ion concentrations in saturated solutions.

· Solubility can be determined from the solubility product.