Ch 16 Lecture 3 Solubility Equilibria

I. Solubility BasicsA. The Solubility Equilibrium

1) Dissolution of an ionic compound is an equilibrium process

a) CaF2 (s) Ca2+ (aq) + 2 F- (aq)

b) Ksp = Solubility Product = [Ca2+][F-]2

2) Remember, neither solids nor pure liquids (water) effect the equilibrium constant

a) Dissolving and Reforming change proportionately to the amount of solid

b) Solvent water is at such a high concentration as not to be effected

3) The Solubility Product is an equilibrium constant, so it has only one value at a given temperature

4) Solubility = the equilibrium position for a given set of conditions

a) There are many different conditions that all must obey Ksp

b) Common ions effect the solubility much as they effect pH

5) Ksp values of some slightly soluble ionic solids

a) Most NO3- salts are soluble

b) Most alkali metal and NH4+ salts are soluble

c) Most Cl-, Br-, and I- salts are soluble (except: Ag+, Pb2+, and Hg22+)

d) Most SO42- salts are soluble (except Ba2+, Pb2+, Hg2

2+, and Ca2+)

e) Most OH- salts are insoluble (except NaOH, KOH)

f) Most S2-, CO32-, CrO4

2-, and PO43- salts are insoluble

Note: These values maydiffer from the ones in your text. Use the valuesfrom your text for all homework problems

6) Example: What is the Ksp of a CuBr solution with a solubility of 2.0 x 10-4 M?

CuBr (s) Cu+ (aq) + Br- (aq)

Ksp = [Cu+][Br-] = [2.0 x 10-4][2.0 x 10-4] = 4 x 10-8

7) Example: Ksp = ? for Bi2S3 with solubility of 1.0 x 10-15 M

8) Example: Find the solubility of Cu(IO3)2 (Ksp = 1.4 x 10-7)

Cu (IO3)2 (s) Cu2+ (aq) + 2 IO3- (aq)

Ksp = [Cu2+][IO3-]2 = 1.4 x 10-7

(x)(2x)2 = 4x3 = 1.4 x 10-7 x = 3.3 x 10-3 M

B. Relative Solubilities

1) If the salts being compared produce the same number of ions, we can compare solubilities by comparing Ksp values

AgI Ksp = 1.5 x 10-16

CuI Ksp = 5.0 x 10-12

CaSO4 Ksp = 6.1 x 10-5

Ksp = [Xn+]1[Yn-]1 for all of these, so we can directly compare them

Solubility of CaSO4 > solubility of CuI > solubility of AgI

2) If the salts being compared produce different numbers of ions, we must calculate the actual solubility values; we can’t use Ksp values to compare.

a) CuS (8.5 x 10-45) > Ag2S (1.6 x 10-49) > Bi2S3 (1.1 x 10-73) by Ksp alone 2 ions 3 ions 5 ions

b) Bi2S3 (1.0 x 10-15) > Ag2S (3.4 x 10-17) > CuS (9.2 x 10-23) in solubility

C. The Common Ion Effect1) Common Ion = any ion in the solid we are trying to dissolve that is present in

solution from another source.

2) What is the solubility of Ag2CrO4 (Ksp = 9 x 10-12) in 0.1 M AgNO3?

Ag2CrO4 (s) 2 Ag+ (aq) + CrO42- (aq)

Init. ---- 0.1 0Equil. ---- 0.1 + 2x x

Ksp = 9 x 10-12 = [0.1 + 2x]2[x] ~~ (0.1)2(x)x = 9 x 10-12 / 0.01 = 9 x 10-10 = solubility

5% rule: 9 x 10-10 / 0.1 < 5%, so the approximation is okSolubility in pure water is 1.3 x 10-4 M. Why does the solubility decrease?

3) Example: Find solubility of CaF2 (Ksp = 4.0 x 10-11) in 0.025 M NaF

D. pH and Solubility

1) Many salts contain hydroxide anion: Mg(OH)2 Mg2+ + 2 OH-

a) High pH means large OH- common ion concentration

b) [OH-] would shift the equilibrium to the left

c) [H+] would shift the equilibrium to the right by using up OH- ions

2) Any Basic Anion will be effected by pH

a) OH-, S2-, CO32-, C2O4

2-, CrO42-, and PO4

3- are all basic anions

b) H+ will increase the solubility of their salts by removing the anions

c) Ag3PO4 (s) 3 Ag+ (aq) + PO43- (aq)

PO43- + H+ HPO4

2-

3) Acidic pH has no effect on non-basic anions or on most cations: Cl-, Br-, NO3-

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

b) H+ doesn’t react with either ion

II. Precipitation and Qualitative AnalysisA. We can use the solubility product to predict precipitation

1) If Q > Ksp, precipitation occurs until Ksp is reached

2) If Q < Ksp, no precipitation will occur

3) Example: 750 ml 0.004 M Ce(NO3)3 is added to 300 ml 0.02 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitate?

Ksp = [Ce3+][IO3-]3 We need to know concentrations.

Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10 > Ksp Precipitate

4) We can also calculate the equilibrium concentrations after precipitation.

a) Examine the stoichiometry of the precipitation reaction allowed to go to completion

b) Becomes a Ksp problem with a common ion (ion in excess)

M10 x 2.86L 1.050

mol/L) L)(0.004 (0.75][Ce 33

M10 x 71.5L 1.050

mol/L) L)(0.02 (0.30][IO 3-

3

3) Calculate the equilibrium concentrations after precipitation when 100 ml of 0.05 M Pb(NO3)2 is added to 200 ml 0.10 M NaI. Ksp for PbI2 = 1.4 x 10–8.

a) PbI2 (s) Pb2+ (aq) + 2 I- (aq) Ksp = [Pb2+][I-]2

b) [Pb2+] = 1.67 x 10-2, [I-] = 6.67 x 10-2, Q = 7.43 x 10-5 > Ksp precipitate

c) Stoichiometry: Pb2+ + 2 I- PbI2

Initial 5mmol 20 mmol ----

Completion 0 10 mmol ----

d) Equilibrium: some PbI2 redissolves, with I- common ion present

i. [I-] common ion = 10mmol / 300ml = 0.033 M

PbI2 (s) Pb2+ (aq) + 2 I- (aq) Ksp = [Pb2+][I-]2

Initial ---- 0 0.033M

Equilibrium ---- x 0.033 + 2x

i. Ksp = 1.4 x 10–8 = [Pb2 +][I-]2 = (x)(0.033 + 2x)2 ~~(x)(0.033)2

x = 1.3 x 10-5 M = [Pb2 +], [I-] = 0.033 M

4) Example: 150 ml 0.01 M Mg(NO3)2 + 250 ml 0.1 M NaF. Find [Mg2+] and [F-] at equilibrium. Ksp for MgF2 = 6.4 x 10-9

B. Qualitative Analysis

1. Selective Precipitation = addition of an ion that causes only one of a mixture of ions to precipitate

Add NaCl to Ag+ (aq) + Ba2+ (aq) AgCl(s) + Na+ (aq) + Ba2+ (aq)

2. Example: Which ion will precipitate first? I- is added to a solution of 0.0001 M Cu+ and 0.002 M Pb2+? Ksp CuI = 5.3 x 10-12. Ksp PbI2 = 1.4 x 10-8.

a) Use Ksp CuI to find [I-] that will just start precipitation

b) Use Ksp PbI2 to find [I-] that will just start precipitation

c) Whichever has the lowest [I-] will precipitate first

3. Sulfide Ion (S2-) is particularly useful for selective cation precipitation

a) Metal sulfides have very different solubilities

i. FeS Ksp = 2.3 x 10-13

ii. MnS Ksp = 3.7 x 10-19

iii. Mn would precipitate first from an equal mixture of Fe2+ and Mn2+

b) [S2-] can be controlled by pH

i. H2S H+ + HS- Ka1 = 1 x 10-7

ii. HS- H+ + S2- Ka2 = 1 x 10-19

iii. S2- is quite basic. At low pH, there will be very little S2-

iv. At high pH, there is much more S2-

c) We can selectively precipitate metal ions by adding S2- in acidic solution, and then slowly adding base.

CuS Ksp = 8.5 x 10-45

HgS Ksp = 1.6 x 10-54

MnS Ksp = 2.3 x 10-13

NiS Ksp = 3.0 x 10-21

Hg, then Cu, then Ni, then Mnwould precipitate as we raise pH

Solution of Mn2+, Ni2+, Cu2+, Hg2+

Solution ofMn2+, Ni2+

Precip. OfMnS, NiS

Precipitate ofCuS, HgS

Add H2S, pH = 2

Add OH- to pH = 8

4) Qualitative Analysis = scheme to separate and identify mixtures of cations by precipitation

a) Group I Insoluble Chlorides:

Add HCl. AgCl, PbCl2, Hg2Cl2 precipitate

b) Group II Sulfides Insoluble in Acid Solution:

Add H2S. Low [S2-] due to [H+]. HgS, CdS, Bi2S3, CuS, and SnS2 precip.

c) Group III Sulfides Insoluble in Basic Solution:

Add OH-. CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, and Al(OH)3 precipitate.

d) Group IV Insoluble Carbonates:

Add CO32-. BaCO3, CaCO3, and MgCO3 precipitate.

e) Group V Alkali Metal and Ammonium

These ions (Na+, K+, NH4+) are soluble with common ions

f) Further tests would tell us which specific ions in each group are present

C. Complex Ion Equilibria1) A complex ion = Lewis Acid—Lewis Base complex of a metal ion

a) Ligand = generic name for the Lewis Base bonded to a metal ion

b) H2O, NH3, Cl-, CN- all have lone pairs to donate; can be ligands

2) Coordination Number (CN) = the number of ligands bonded to a metal ion

a) CN = 6 is common: Co(H2O)62+, Ni(NH3)6

2+

b) CN = 4 CoCl42-, Cu(NH3)4

2+

c) CN = 2 Ag(NH3)2+

d) Coordination number depends on size and properties of the metal ion

3) Formation (or Stability) Constantsa) Ligand addition to the metal ion is stepwise

Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 Ag(NH3)2+ K2 = 8.2 x 103

Ag+ + 2NH3 Ag(NH3)2+ Kf = 1.7 x 107

b) Usually, [Ligand] >> [Mn+] to force the equilibria to the right

c) What are the equilibrium conditions of 100 ml 2.0 M NH3 added to 100 ml of 0.001 M AgNO3?

i. K1 x K2 = Kf is large, favoring complete reaction

ii. [NH3] is much larger than [Ag+], so assume constant

iii. Do stoichiometry of the reaction:

Ag+ + 2 NH3 Ag(NH3)2

+

Initial 5 x 10-4 M 1 M 0

Equilib. 0 1 M 5 x 10-4 M

iv. Then use equilibrium expressions to find concentrations

M10 x 1.6)(1.0)10 x (8.2

5x10

][NHK

])[Ag(NH])Ag(NH[

]][NH)[Ag(NH

])[Ag(NH10 x 8.2K

83

4

32

233

33

2332

M10 x 2.9][NHK

])[Ag(NH]Ag[

]][NH[Ag

])[Ag(NH10 x 1.2K

11

31

3

3

331

4) Example: Find [Ag+], [Ag(S2O3)-], [Ag(S2O3)23-] for 150 ml 0.001 M AgNO3 +

200 ml 5.0 M Na2S2O3. K1 = 7.4 x 108, K2 = 3.9 x 104.

5) Complex Ions and Solubility

a) How do we dissolve AgCl(s)? Ksp = 1.6 x 10-10

Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 Ag(NH3)2+ K2 = 8.2 x 103

b) Adding NH3 to a suspension of AgCl, forces more of it to dissolve. The NH3, removes Ag+ from solution, forcing the equilibrium to the right.

c) [Ag+]T = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]

d) The overall reaction is the sum of three individual reactions:

AgCl Ag+ + Cl- Ksp = 1.6 x 10-10

Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 Ag(NH3)2+ K2 = 8.2 x 103

AgCl + 2 NH3 Ag(NH3)2+ + Cl- K = ???

K = Ksp x K1 x K2 = (1.6 x 10-10)(2.1 x 103)(8.2 x 103) = 2.8 x 10-3

e) Example: Find the solubility of AgCl(s) in 10 M NH3.

AgCl + 2 NH3 Ag(NH3)2+ + Cl-

Initial ----- 10 0 0

Equil. ----- 10 – 2x x x

Solve using the expression for K as in a normal equilibrium problem.

6) Strategies for Dissolving Insoluble Solids

a) If the anion is basic, add acid

b) If the cation will form a complex, add ligand

c) Heat often increases solubility (temp. effects all equil. constants)

d) Example: HgS Hg2+ + S2- Ksp = 1 x 10-54

i. S2- is a basic anion, so add HCl

S2- + H+ HS-

ii. Hg2+ will form a chloride complex

Hg2+ + 4 Cl- HgCl42-

7) The qualitative analysis of the Group I cations illustrates complex ion equilibria

Solution of Ag+, Hg2

2+, Pb2+

Solution of Pb2+Precipitate ofAgCl, Hg2Cl2

Precipitate ofPbCrO4

Solution ofAg(NH3)2+, Cl-

Precip. of AgCl

Precipitate ofHg, HgNH2Cl

Precipitation of AgCl, Hg2Cl2, PbCl2

Add cold HCl

Heat

Add CrO42-

Add H+

Add NH3