Equilibrium constants

112
Equilibrium constants

description

Equilibrium constants. Equilibrium constants. P C c P D d. K =. P A a P B b. [C] c [D] d. K =. [A] a [B] b. Equilibrium constants. P C c P D d. K =. P A a P B b. [C] c [D] d. K =. [A] a [B] b. [solid] = 1. Equilibrium of solids and solutions. - PowerPoint PPT Presentation

Transcript of Equilibrium constants

Page 1: Equilibrium constants

Equilibrium constants

Page 2: Equilibrium constants

Equilibrium constants

K =PC

cPDd

PAaPB

b

K =[A]a[B]b

[C]c[D]d

Page 3: Equilibrium constants

Equilibrium constants

K =PC

cPDd

PAaPB

b

K =[A]a[B]b

[C]c[D]d

[solid] = 1

Page 4: Equilibrium constants

Equilibrium of solids and solutions

Page 5: Equilibrium constants

Equilibrium of solids and solutions

NaCl(s) + H2O(l)

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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.

Page 8: Equilibrium constants
Page 9: Equilibrium constants
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Page 11: Equilibrium constants

Equilibrium of solids and solutions

NaCl(s) + H2O(l) Na+(aq) + Cl-

(aq)

Solution must be saturated for this

equilibrium to take place.

Page 12: Equilibrium constants

Dynamic equilibrium

Page 13: Equilibrium constants

Equilibrium for dissolution-precipitation

reaction:

Page 14: Equilibrium constants

Equilibrium for dissolution-precipitation

reaction:

I2(s) + CCl4(l) I2(CCl4)

Page 15: Equilibrium constants

Equilibrium for dissolution-precipitation

reaction:

K [I2]CCl4

=

I2(s) + CCl4(l) I2(CCl4)

Page 16: Equilibrium constants

Equilibrium for dissolution-precipitation

reaction:

[I2]CCl K4=

I2(s) + CCl4(l) I2(CCl4)

Molarity of saturated solution = K

Page 17: Equilibrium constants

[I2]CCl K4=

K will vary when conditions are changed

based on Le Chatelier’s Principle.

Page 18: Equilibrium constants

[I2]CCl K4=

If solution of a material is exothermic,

increasing the temperature will

decrease K.

A(s) A(sol)

Page 19: Equilibrium constants

[I2]CCl K4=

If solution of a material is endothermic,

increasing the temperature will

increase K.

A(s) A(sol)

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Kinetics (reaction rates) must

be considered.

Page 22: Equilibrium constants

The value of K is not an indicator

of how long it takes to attain

equilibrium.

Page 23: Equilibrium constants

A K = 5 does not guarantee a 5 M

solution in a few minutes.

Page 24: Equilibrium constants

Not all solutions are ‘ideal’

Page 25: Equilibrium constants

Not all solutions are ‘ideal’

Ideal Solution:

Widely separated species

(ions or molecules) that do not

interact.

Page 26: Equilibrium constants

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

(aq)

Page 27: Equilibrium constants

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

(aq)

As the concentration of ions increases,

Cs+ to Cl- distances decrease.

Page 28: Equilibrium constants

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

(aq)

As the concentration of ions increases,

Cs+ to Cl- distances decrease.

Cs+…Cl-Cs+…Cl-

Cl-…Cs+

Page 29: Equilibrium constants

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

(aq)

Cs+…Cl-Cs+…Cl-

Cl-…Cs+

Ion pairing may occur before equilibrium.

Page 30: Equilibrium constants

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

(aq)

Cs+…Cl-Cs+…Cl-

Cl-…Cs+

These solutions are non-ideal.

Ion pairing may occur before equilibrium.

Page 31: Equilibrium constants

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

(aq)

Salts of low solubilities allow the study

of solutions that are essentially ideal.

Page 32: Equilibrium constants

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.

Page 33: Equilibrium constants

Solubility product

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Solubility product

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

(aq)

Page 35: Equilibrium constants

Solubility product

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

(aq)

Same form as equilibrium constant

Page 36: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-]

[AgCl]

Page 37: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-]

1= Ksp

Page 38: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-] = Ksp

Page 39: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-] = Ksp

25oC Ksp = 1.6 x 10-10

Page 40: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-] = Ksp

25oC Ksp = 1.6 x 10-10

[Ag+] = [Cl-] = y

Page 41: Equilibrium constants

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

Page 42: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-] = Ksp

y2 = Ksp = 1.6 x 10-10

= 1.26 x 10-5 M[Ag+] = [Cl-]

Page 43: Equilibrium constants

Solubility product

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

(aq)

[Ag+][Cl-] = Ksp

y2 = Ksp = 1.6 x 10-10

= 1.26 x 10-5 M[Ag+] = [Cl-]

1.8 x 10-3 g/L

Page 44: Equilibrium constants

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

(aq)

Page 45: Equilibrium constants

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

(aq)

[Ba2+][F-]2 = Ksp

Page 46: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

Page 47: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

Page 48: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

For every Fe3+ that goes into solution,

3 OH- go into solution.

Page 49: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

[y][3y]3 = 1.1 x 10-36

Page 50: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[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.

Page 51: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

[y][3y]3 = 1.1 x 10-36

27y4 = 1.1 x 10-36

Page 52: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

[y][3y]3 = 1.1 x 10-36

27y4 = 1.1 x 10-36

y4 =1.1 x 10-36

27= 4.1 x 10-38

Page 53: Equilibrium constants

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

[y][3y]3 = 1.1 x 10-36

27y4 = 1.1 x 10-36

y4 =1.1 x 10-36

27= 4.1 x 10-38

y = 4.5 x 10-10

Page 54: Equilibrium constants

Ksp CuS = 2 x 10-47

CuS(s) Cu2+(aq) + S2-

(aq)

Page 55: Equilibrium constants

Ksp CuS = 2 x 10-47

CuS(s) Cu2+(aq) + S2-

(aq)

[Cu2+][S2-] = 2 x 10-47

Page 56: Equilibrium constants

Ksp CuS = 2 x 10-47

CuS(s) Cu2+(aq) + S2-

(aq)

[Cu2+][S2-] = 2 x 10-47

y2 = 2 x 10-47

Page 57: Equilibrium constants

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-]

Page 58: Equilibrium constants

Ksp CuS = 2 x 10-47

CuS(s) Cu2+(aq) + S2-

(aq)

y = 4.5 x 10-24 = [Cu2+] = [S2-]

mw CuS = 95.6

Page 59: Equilibrium constants

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

Page 60: Equilibrium constants

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

Page 61: Equilibrium constants

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) =

Page 62: Equilibrium constants

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

Page 63: Equilibrium constants

Example 9-3

Determine Ksp for a salt based

on solubility data.

Page 64: Equilibrium constants

Example 9-3

Determine Ksp for a salt based

on solubility data.

PbCl2 8.67 g/L for a saturated solution

Page 65: Equilibrium constants

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)

Page 66: Equilibrium constants

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

Page 67: Equilibrium constants

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

Page 68: Equilibrium constants

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

Page 69: Equilibrium constants

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=

Page 70: Equilibrium constants

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

278.1 g/mol= 0.031 mol

< 0.1 mol

Page 71: Equilibrium constants

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

278.1 g/mol= 0.031 mol

Ksp = 4(0.031)3

Page 72: Equilibrium constants

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

278.1 g/mol= 0.031 mol

Ksp = 4(0.031)3 =1.2 x 10-4

Page 73: Equilibrium constants

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

278.1 g/mol= 0.031 mol

Ksp = 4(0.031)3 =1.2 x 10-4 Table = 1.6 x 10-5

Page 74: Equilibrium constants

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

278.1 g/mol= 0.031 mol

Ksp = 4(0.031)3 =1.2 x 10-4 Table = 1.6 x 10-5

Non-ideal solution

Page 75: Equilibrium constants

Precipitation from solution

Page 76: Equilibrium constants

Precipitation from solution

Start with solution that is not saturated.

Page 77: Equilibrium constants

Precipitation from solution

Start with solution that is not saturated.

Cause solution to become saturated.

Page 78: Equilibrium constants

Precipitation from solution

Start with solution that is not saturated.

Cause solution to become saturated.

Remove solvent by evaporation.

Page 79: Equilibrium constants

Precipitation from solution

Start with solution that is not saturated.

Cause solution to become saturated.

Remove solvent by evaporation.

Change nature of solvent.

Page 80: Equilibrium constants

Change nature of solvent.

The polarity of a solvent

system can be adjusted.

Page 81: Equilibrium constants

Change nature of solvent.

The polarity of a solvent

system can be adjusted.

NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)

Not saturated

Page 82: Equilibrium constants

Change nature of solvent.

The polarity of a solvent

system can be adjusted.

NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)

Not saturatedAdd EtOH to solution.

Page 83: Equilibrium constants

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.

Page 84: Equilibrium constants

Precipitation of a product from

a reaction.

Will it precipitate?

Page 85: Equilibrium constants

Determine reaction quotient.

AgNO3 (solution) mixed with

NaCl (solution)

Will AgCl precipitate?

Page 86: Equilibrium constants

Determine reaction quotient.

AgNO3 (solution) mixed with

NaCl (solution)

Will AgCl precipitate?

Q = [Ag+][Cl-]

Q = reaction quotient

Page 87: Equilibrium constants

AgNO3 (solution) mixed with

NaCl (solution)

Q = reaction quotient

Qinit = conditions just as solutions are mixed

Q = [Ag+][Cl-]

Page 88: Equilibrium constants

AgNO3 (solution) mixed with

NaCl (solution)

Q = reaction quotient

Qinit = conditions just as solutions are mixed

If Qinit < Ksp then no AgCl willprecipitate.

Q = [Ag+][Cl-]

[Ag+][Cl-] too low

Page 89: Equilibrium constants

AgNO3 (solution) mixed with

NaCl (solution)

Q = reaction quotient

Qinit = conditions just as solutions are mixed

If Qinit > Ksp then AgCl will precipitate.

Q = [Ag+][Cl-]

Ag+ Cl- too high

Page 90: Equilibrium constants

AgNO3 (solution) mixed with

NaCl (solution)

Q = reaction quotient

Qinit = conditions just as solutions are mixed

If Qinit > Ksp then AgCl will precipitate.

As the reaction continues, precipitation willstop when Q = Ksp

Q = [Ag+][Cl-]

Page 91: Equilibrium constants

precipitate

No precipitate

Page 92: Equilibrium constants

Exercise page 384

TlIO3 555 ml 0.0022 M

NaIO3 445 ml 0.0022 M

Will TlIO3 precipitate atequilibrium?

Page 93: Equilibrium constants

Exercise page 384

TlIO3 555 ml 0.0022 M

NaIO3 445 ml 0.0022 M

Will TlIO3 precipitate atequilibrium?

Ksp = 3.1 x 10-6

Page 94: Equilibrium constants

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) 1.000

= 0.0012 M

Page 95: Equilibrium constants

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

Page 96: Equilibrium constants

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 .555 L x 0.0022 M = 0.0012 moles

[IO3-] = .00099 M + 0.0012 M = 0.0022

Page 97: Equilibrium constants

TlIO3 555 ml 0.0022 M

NaIO3 445 ml 0.0022 M

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-] = (0.0012)(.00099) = 1.2 x 10-6

[Tl+] = 0.0012 M .555 L x 0.0022 M = 0.0012 moles

[IO3-] = .0022 M

For this solution

Page 98: Equilibrium constants

TlIO3 555 ml 0.0022 M

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

For mixedsolutions

Page 99: Equilibrium constants

Common ion effect

Page 100: Equilibrium constants

Common ion effect

Adding more of a cation or anion

that is already in solution will

cause Q to change.

Page 101: Equilibrium constants

Common ion effect

Adding more of a cation or anion

that is already in solution will

cause Q to change.

Q = [A+][B-]

Page 102: Equilibrium constants

AgCl(s)Ag+

(aq) + Cl-(aq)

Q = [Ag+][Cl-]

Add more Cl- (NaCl)

Page 103: Equilibrium constants

Q < Ksp

Q > Ksp

Page 104: Equilibrium constants

AgCl(s)Ag+

(aq) + Cl-(aq)

Q = [Ag+][Cl-]

Add more Cl- (NaCl)

This raises Q above Ksp

Page 105: Equilibrium constants

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.

Page 106: Equilibrium constants

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-]

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

0.050 M KIO3

What is the solubility of TlIO3

in this solution?

Page 107: Equilibrium constants

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-]

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

0.050 M KIO3

Ksp = [Tl+][IO3-]

[Tl+] = y

Page 108: Equilibrium constants

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-]

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

0.050 M KIO3

Ksp = [Tl+][IO3-]

[Tl+] = y [IO3-] = y + 0.050

Page 109: Equilibrium constants

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-]

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

0.050 M KIO3

Ksp = [Tl+][IO3-]

[Tl+] = y [IO3-] = y + 0.050 0.050

Page 110: Equilibrium constants

Ksp = 3.1 x 10-6

Q = [Tl+][IO3-]

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

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

Page 111: Equilibrium constants

Ksp = 3.1 x 10-6

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

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

Page 112: Equilibrium constants

Ksp = 3.1 x 10-6

Exercise page 387

TlIO3(s)Tl+

(aq) + IO3-(aq)

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 [Tl+] = 1.8 x 10-3

No common ion