Equilibrium constants

Equilibrium constants PCcPDd K = PAaPBb [C]c[D]d K = [A]a[B]b

Equilibrium constants PCcPDd K = PAaPBb [C]c[D]d K = [A]a[B]b [solid] = 1

Equilibrium of solids and solutions

Equilibrium of solids and solutions NaCl(s) + H2O(l)

Equilibrium of solids and solutions NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) To remove an ion from the crystal lattice, the solvating interactions must be stronger than the lattice interactions.

Equilibrium of solids and solutions NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) Solution must be saturated for this equilibrium to take place.

Dynamic equilibrium

Equilibrium for dissolution-precipitation reaction:

Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4)

Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4) K [I2]CCl = 4

Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4) [I2]CCl K = 4 Molarity of saturated solution = K

K will vary when conditions are changed [I2]CCl K = 4 K will vary when conditions are changed based on Le Chatelier’s Principle.

If solution of a material is exothermic, [I2]CCl K = 4 If solution of a material is exothermic, increasing the temperature will decrease K. A(s) A(sol)

If solution of a material is endothermic, [I2]CCl K = 4 If solution of a material is endothermic, increasing the temperature will increase K. A(s) A(sol)

Kinetics (reaction rates) must be considered.

The value of K is not an indicator of how long it takes to attain equilibrium.

A K = 5 does not guarantee a 5 M solution in a few minutes.

Not all solutions are ‘ideal’

Not all solutions are ‘ideal’ Ideal Solution: Widely separated species (ions or molecules) that do not interact.

CsCl(s) Cs+(aq) + Cl-(aq)

CsCl(s) Cs+(aq) + Cl-(aq) As the concentration of ions increases, Cs+ to Cl- distances decrease.

CsCl(s) Cs+(aq) + Cl-(aq) As the concentration of ions increases, Cs+ to Cl- distances decrease. Cs+…Cl- Cs+…Cl- Cl-…Cs+

CsCl(s) Cs+(aq) + Cl-(aq) Cl-…Cs+ Ion pairing may occur before equilibrium.

CsCl(s) Cs+(aq) + Cl-(aq) Cl-…Cs+ Ion pairing may occur before equilibrium. These solutions are non-ideal.

CsCl(s) Cs+(aq) + Cl-(aq) Salts of low solubilities allow the study of solutions that are essentially ideal.

CsCl(s) Cs+(aq) + Cl-(aq) Salts of low solubilities allow the study of solutions that are essentially ideal. A saturated solution of 0.1 or less is a sign of low solubility in a salt.

Solubility product

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

Solubility product Same form as equilibrium constant AgCl(s) Ag+(aq) + Cl-(aq)

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] [AgCl]

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 1

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = y

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = y y2 = Ksp = 1.6 x 10-10

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp y2 = Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = 1.26 x 10-5 M

Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp y2 = Ksp = 1.6 x 10-10 1.8 x 10-3 g/L [Ag+] = [Cl-] = 1.26 x 10-5 M

BaF2(s) Ba2+(aq) + 2 F-(aq)

BaF2(s) Ba2+(aq) + 2 F-(aq) [Ba2+][F-]2 = Ksp

Fe3+(aq) + 3 OH-(aq) Fe(OH)3

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 For every Fe3+ that goes into solution, 3 OH- go into solution.

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36 If there is another source of OH- (NaOH) that provides a higher [OH-] then that is the value of [OH-] to be used.

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36 27y4 = 1.1 x 10-36

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36 27y4 = 1.1 x 10-36 1.1 x 10-36 y4 = = 4.1 x 10-38 27

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36 27y4 = 1.1 x 10-36 1.1 x 10-36 y4 = = 4.1 x 10-38 27 y = 4.5 x 10-10

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq)

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) [Cu2+][S2-] = 2 x 10-47

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) [Cu2+][S2-] = 2 x 10-47 y2 = 2 x 10-47

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) [Cu2+][S2-] = 2 x 10-47 y2 = 2 x 10-47 y = 4.5 x 10-24 = [Cu2+] = [S2-]

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) y = 4.5 x 10-24 = [Cu2+] = [S2-] mw CuS = 95.6

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) y = 4.5 x 10-24 = [Cu2+] = [S2-] mw CuS = 95.6 4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) y = 4.5 x 10-24 = [Cu2+] = [S2-] 4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L Atoms/L = moles x Ao

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) y = 4.5 x 10-24 = [Cu2+] = [S2-] 4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) =

Ksp CuS = 2 x 10-47 CuS(s) Cu2+(aq) + S2-(aq) y = 4.5 x 10-24 = [Cu2+] = [S2-] 4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) = 2.7 atoms/L

Example 9-3 Determine Ksp for a salt based on solubility data.

Example 9-3 Determine Ksp for a salt based on solubility data. PbCl2 8.67 g/L for a saturated solution

Example 9-3 Determine Ksp for a salt based on solubility data. PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq)

Example 9-3 Determine Ksp for a salt based on solubility data. PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp (y)(2y)2 = Ksp

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp (y)(2y)2 = Ksp 4y2 = Ksp

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y2 = Ksp 8.67 g PbCl2 = 278.1 g/mol

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y3 = Ksp 8.67 g PbCl2 = 0.031 mol 278.1 g/mol < 0.1 mol

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y3 = Ksp 8.67 g PbCl2 = 0.031 mol 278.1 g/mol Ksp = 4(0.031)3

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y3 = Ksp 8.67 g PbCl = 0.031 mol 278.1 g/mol Ksp = 4(0.031)3 =1.2 x 10-4

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y3 = Ksp 8.67 g PbCl = 0.031 mol 278.1 g/mol Table = 1.6 x 10-5 Ksp = 4(0.031)3 =1.2 x 10-4

Example 9-3 PbCl2 8.67 g/L for a saturated solution PbCl2 Pb2+(aq) + 2 Cl-(aq) [Pb2+][Cl-]2 = Ksp 4y3 = Ksp 8.67 g PbCl = 0.031 mol 278.1 g/mol Table = 1.6 x 10-5 =1.2 x 10-4 Ksp = 4(0.031)3 Non-ideal solution

Precipitation from solution

Precipitation from solution Start with solution that is not saturated.

Precipitation from solution Start with solution that is not saturated. Cause solution to become saturated.

Precipitation from solution Start with solution that is not saturated. Cause solution to become saturated. Remove solvent by evaporation.

Precipitation from solution Start with solution that is not saturated. Cause solution to become saturated. Remove solvent by evaporation. Change nature of solvent.

Change nature of solvent. The polarity of a solvent system can be adjusted.

Change nature of solvent. The polarity of a solvent system can be adjusted. NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) Not saturated

Change nature of solvent. The polarity of a solvent system can be adjusted. NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) Not saturated Add EtOH to solution.

Precipitation from solution Start with solution that is not saturated. Cause solution to become saturated. Remove solvent by evaporation. Change nature of solvent. Cool or warm system.

Precipitation of a product from a reaction. Will it precipitate?

Determine reaction quotient. AgNO3 (solution) mixed with NaCl (solution) Will AgCl precipitate?

Determine reaction quotient. AgNO3 (solution) mixed with NaCl (solution) Will AgCl precipitate? Q = [Ag+][Cl-] Q = reaction quotient

AgNO3 (solution) mixed with NaCl (solution) Q = [Ag+][Cl-] Q = reaction quotient Qinit = conditions just as solutions are mixed

AgNO3 (solution) mixed with NaCl (solution) Q = [Ag+][Cl-] Q = reaction quotient Qinit = conditions just as solutions are mixed If Qinit < Ksp then no AgCl will precipitate. [Ag+][Cl-] too low

AgNO3 (solution) mixed with NaCl (solution) Q = [Ag+][Cl-] Q = reaction quotient Qinit = conditions just as solutions are mixed If Qinit > Ksp then AgCl will precipitate. Ag+ Cl- too high

AgNO3 (solution) mixed with NaCl (solution) Q = [Ag+][Cl-] Q = reaction quotient Qinit = conditions just as solutions are mixed If Qinit > Ksp then AgCl will precipitate. As the reaction continues, precipitation will stop when Q = Ksp

precipitate No precipitate

Exercise page 384 TlIO3 555 ml 0.0022 M NaIO3 445 ml 0.0022 M Will TlIO3 precipitate at equilibrium?

Exercise page 384 TlIO3 555 ml 0.0022 M NaIO3 445 ml 0.0022 M Will TlIO3 precipitate at equilibrium? Ksp = 3.1 x 10-6

TlIO3 555 ml 0.0022 M NaIO3 445 ml 0.0022 M Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] [Tl3+] = .555(0.0022) = 0.0012 M 1.000

TlIO3 555 ml 0.0022 M NaIO3 445 ml 0.0022 M Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] [Tl+] = 0.0012 M

TlIO3 555 ml 0.0022 M NaIO3 445 ml 0.0022 M Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] .555 L x 0.0022 M = 0.0012 moles [Tl+] = 0.0012 M [IO3-] = .00099 M + 0.0012 M = 0.0022

TlIO3 555 ml 0.0022 M For this solution NaIO3 445 ml 0.0022 M Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] = (0.0012)(.00099) = 1.2 x 10-6 .555 L x 0.0022 M = 0.0012 moles [Tl+] = 0.0012 M [IO3-] = .0022 M

TlIO3 555 ml 0.0022 M For mixed solutions NaIO3 445 ml 0.0022 M Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] = (0.0012)(.0022) = 2.6 x 10-6 Q < Ksp No precipitate

Common ion effect

Common ion effect Adding more of a cation or anion that is already in solution will cause Q to change.

Common ion effect Adding more of a cation or anion that is already in solution will cause Q to change. Q = [A+][B-]

Ag+(aq) + Cl-(aq) AgCl(s) Q = [Ag+][Cl-] Add more Cl- (NaCl)

Q > Ksp Q < Ksp

Ag+(aq) + Cl-(aq) AgCl(s) Q = [Ag+][Cl-] Add more Cl- (NaCl) This raises Q above Ksp

Common ion effect If a solution and a solid salt to be dissolved in it have an ion in common, the solubility of the salt is depressed.

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 Q = [Tl+][IO3-] 0.050 M KIO3 What is the solubility of TlIO3 in this solution?

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Q = [Tl+][IO3-] Ksp = [Tl+][IO3-] [Tl+] = y

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Q = [Tl+][IO3-] Ksp = [Tl+][IO3-] [Tl+] = y [IO3-] = y + 0.050

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Q = [Tl+][IO3-] Ksp = [Tl+][IO3-] [Tl+] = y [IO3-] = y + 0.050  0.050

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Q = [Tl+][IO3-] Ksp = [Tl+][IO3-] [Tl+] = y [IO3-] = y + 0.050  0.050 3.1 x 10-6 = (y)(0.050) = 0.050y

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Ksp = [Tl+][IO3-] [Tl+] = y [IO3-] = y + 0.050  0.050 3.1 x 10-6 = (y)(0.050) = 0.050y y = 6.2 x 10-5 mol/L

Exercise page 387 Tl+(aq) + IO3-(aq) TlIO3(s) Ksp = 3.1 x 10-6 0.050 M KIO3 Ksp = [Tl+][IO3-] [Tl+] = y [IO3-] = y + 0.050  0.050 3.1 x 10-6 = (y)(0.050) = 0.050y No common ion y = 6.2 x 10-5 mol/L [Tl+] = 1.8 x 10-3