CM4106 Chemical Equilibria & Thermodynamics

Lesson 4Solubility Equilibria

A Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/

Solubility of Common Salts

Courtesy of Chemistry, 10th Edition by Raymond Chang

Fundamentals:

AxBy (s) ⇌ xAy+(aq) + yBx(aq)

• The equilibrium constant for this heterogeneous equilibrium is called the solubility product, Ksp, is written as:

Ksp = [Ay+(aq)]x

eqm [Bx-(aq)]y

eqm

• The solubility product, Ksp, of a sparingly soluble salt is defined as the product of the concentration of the ions (in M) in a saturated solution at a given temperature raised to the power of its coefficient in the equilibrium equation.

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• The solubility of AgCl at 18°C is 1.46 x 10-3 g/L, what is the solubility product of AgCl at 18°C?

Since the solubility of AgCl = 1.46 x 10-3 g / L; [Ag+] = (1.46 x 10-3 g/L) / (143.5 g/mol)

= 1.017 x 10-5 mol/L

Calculating Ksp value from solubility

AgCl (s) ⇌ Ag+ (aq) + Cl (aq) Initial / M - 0 0

Change / M - s + s + s

Eqm / M - s s

Ksp = [Ag+] [Cl] = (1.017 x 10-5)2

= 1.03 x 10-10

The solubility product of AgCl at 18°C is 1.03 x 10-10

• The Ksp for Ag2CO3 is 8.0 x 10-12 M3, calculate its solubility at this temperature.

Calculating Solubility from Ksp

Ag2CO3(s) ⇌ 2Ag+(aq) + CO32(aq)

Initial / M - 0 0

Change / M - s + 2 s + s

Eqm / M - 2 s s

Ksp = [Ag+]2[CO32]

8.0 x 10-12 = [2s]2[s]8.0 x 10-12 = 4 s3

s = 1.26 x 10-4 M= 1.3 x 10-4 M (2 s.f.)

Solubility of Ag2CO3 = 1.3 x 10-4 M

Let s be the solubility of Ag2CO3 in mol / L

Calculate solubility given Ksp

Calculate Ksp given solubility

Quantitative Problems

Relationship between Ksp and Solubility (M)

Predicting Precipitation - Will a PPT form?Knowing the value of Ksp allows us to predict if a ppt will be

formed when two solutions containing ions to form an insoluble salt are mixed.

Step 1: Calculate reaction quotient Qsp

Step 2: Compare Qsp with Ksp

Inference

Q > K Precipitation occurs till Q = Ksp

Q < K No precipitation is seen because the solution is not saturated with the ions hence they remain dissolved in the solution

Q = K A saturated solution is obtained

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Predicting Precipitation (Ksp vs Q)Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 x 10-3 M Na2SO4? (Ksp = 6.3 x 10-7)

PbSO4 (s) ⇌ Pb2+ (aq) + SO42-

(aq)

Q = [Pb2+][SO42-]

= (0.0016) (0.0040)

= 6.4 x 10-6 > Ksp = 6.3 x 10-7

Q > Ksp ppt will form

REMEMBER: TAKE INTO ACCOUNT OF CONCENTRATION OF IONS IN MIXTURE (DILUTION UPON MIXING)

[Pb2+] = (0.10/0.50) x 8.0 x 10-3

= 0.0016 M [SO4

2-] = (0.40/0.50) x 5.0 x 10-3

= 0.0040 M

Factors affecting solubility

1.Common Ion Effect

2.pH of solution

3.Formation of Complexes

1. Common Ion Effect

Solubility of a substance is affected by the presence of other solutes, especially if there is a common ion present

CaF2 (s) ⇌ Ca2+ (aq) + 2F-

(aq)What happens to the solubility of CaF2 if the solution already contains Ca2+ (aq) or F- (aq)?

1. Equilibrium position shifts to the left2. Solubility of CaF2 decreases

Ksp = [Ca2+][F-]2 = (0.010 + s)(2s)2

Since s is small, assume 0.010 + s ≈ 0.010

(CaF2 is a sparingly soluble salt and the solubility is further

suppressed by the presence of common ion effect)

3.9 x 10-11 = (0.010)(2s)2

s = solubility = 3.1 x 10-5 mol/L

1. Common Ion EffectCalculate the molar solubility of CaF2 at 25 oC in 0.010 M Ca(NO3)2 solution

CaF2 (s) ⇌ Ca2+ (aq) + 2F-

(aq)Initial / M - 0.010 0

Change / M - +s + 2s

Eqm / M - 0.010 + s 2s

Assumption is valid, s << 0.010

2. pH of Solution

If a substance has a basic anion, it will be more soluble in an acidic solution.

If a substance has an acidic cation, it will be more soluble in an basic solution.

2. pH of Solution

Calculate the molar solubility of Mg(OH)2 in pure water. What is the pH of the resulting solution?

Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)

Ksp = [Mg2+][OH-]2

= (x)(2x)2

1.8 x 10-11 = 4x3

x = Solubility = 1.651 x 10-4 mol/L = 1.7 x10-4 M (2 s.f.)

[OH-] = 2x = 3.302 x 10-4 M pOH = - lg(3.302 10-4) = 3.48 pH = 10.5 (1 d.p.)

Ksp = 1.8 x 10-11 (25 oC)

Initial / M - 0 0

Change / M - +x +2x

Eqm / M - x 2x

Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)

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2. pH of Solution What is the solubility of Mg(OH)2 in a less alkaline solution buffered at pH 9?

pOH = 5[OH-] = 1.0 x 10-5 M

Ksp = [Mg2+] [OH-]2

= [Mg2+](1.0 x 10-5)2

[Mg2+] = 0.18 M (2 s.f.)

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Initial / M - 0 1.0 x 10-

5

Change / M - +x -

Eqm / M - x 1.0 x 10-

5

Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)

Ksp = 1.8 x 10-11 (25 oC)

With the addition of

acid

3. Complexation

When aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate is observed. As aqueous ammonia is added in excess, the precipitate dissolves to give a deep blue solution.

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3. Complexation1. When aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate is

observed. As aqueous ammonia is added in excess, the precipitate dissolves to give a deep blue solution.

NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH (aq)

When added to Cu2+(aq) solution, Cu2+(aq) + 2OH (aq) Cu(OH)⇌ 2(s) – (1)Hence a blue precipitate is observed. As more NH3(aq) is added, Cu2+(aq) + 4NH3(aq) Cu(NH⇌ 3)4

2+(aq) [Cu2+] drops as Cu(NH3)4

2+(aq) is formed, thus position of equilibrium in equation 1 shifts left to produce more Cu2+(aq), hence Cu(OH)2(s) dissolves.

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[Cu(H2O)6]2+ + 2 OH- (aq) → [Cu(H2O)4(OH)2] (s)

[Cu(H2O)6]2+ + 4 NH3 (aq) → [Cu(NH3)4(H2O)2]2+ (aq)

Ksp values