Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp

Introduction andExample 1—Molar

Solubility of an AB type Compound.

Molar Solubilityfrom Ksp

Compounds with low solubility always establish an equilibrium when they are placed in water.

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium.

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium.E.g., when lead(II) sulphate is added to water: 2 2

4(s) (aq) 4(aq)PbSO Pb SO


4(s) (aq) 4(aq)PbSO Pb SO


(aq) 4(aq4( )s) Pb SP ObSO


(aq) 4(aq4( )s) Pb SP ObSO

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium.E.g., when lead(II) sulphate is added to water: 2

4(s) (aq24(aq))PbS SOO Pb

The equilibrium constant expression, Keq, for this is:

2 24(s) (aq) 4(aq)PbSO Pb SO



2 2eq 4Pb SOK


2 2(aq) 4(aq4( )s) Pb SP ObSO

2 2eq 4Pb SOK

Solid



2 2eq 4Pb SOK

2eq

24Pb SOK

For Solubility Equilibrium



2 2eq 4Pb SOK

2sp

24Pb SOK

For Solubility Equilibrium



2 2eq 4Pb SOK

2sp

24Pb SOK

The Solubility Product Constant



2 2eq 4Pb SOK

2sp

24Pb SOK

Product



2 2eq 4Pb SOK

2sp

24Pb SOK

ProductThe Solubility

Product Constant

Solubility depends on the temperature.


Solubility depends on the temperature.


Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution ,


Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution , expressed in moles/L, or molarity (M).


Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution, expressed in moles/L, or molarity (M).A saturated solution is the solution that forms when a compound is at solubility equilibrium.


Molar solubility and

Concentrations of Ions in a

Saturated Solution

The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on:

The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on:• The molar solubility of the compound

The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on:• The molar solubility of the compound• The coefficient of each ion in the

equilibrium equation

For example, consider the compound lead(II) sulphate.

For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is:


2 24(s) (aq) 4(aq)1PbSO 1Pb 1SO

Solubility Equilibrium

Equation




2 24(s) (aq) 4(aq)PbSO Pb SO1 1 1


24(s) (aq)

24(aq)P S1 SO 1b OPb1

× 1/1


2(aq

24(s) 4(aq))Pb1 S P 1O SO1 b

× 1/1

× 1/1


× 1/1

× 1/1

Molar solubility is how much of

this dissolves, in moles/L

= the molar solubility of PbSO4

2(aq

24(aq))4(s)1PbS P 1SOO 1 b

× 1/1

× 1/1



2Pb

2(aq

24(aq))4(s)1PbS P 1SOO 1 b

× 1/1

× 1/1




2Pb


24SO

2 2(aq) 4(aq)4(s)1PbSO Pb O1 S1

× 1/1

× 1/1

One of each ion so PbSO4 is called an “AB” type compound.

22(s) (aq) (aq)PbBr Pb 2B1 r1

Lead(II) bromide

22(s) (aq) (aq)PbBr Pb 2B1 r1

Solubility Equilibrium

Equation

22(s) (aq (aq))P 2BrbBr Pb1 1

× 1/1



= the molar solubility of PbBr2

2Pb

2(aq)2(s) (aq)PbBr 1 Br2P1 b

× 1/1

× 2/1



= the molar solubility of PbBr2

2Pb

= 2 × the molar solubility of PbBr2

Br

2(aq) (aq) )2(sPbBr1 Pb Br1 2

× 1/1

× 2/1

1 of one ion and 2 of the other ion so PbBr2 is called an “AB2” type compound.

Finding the Molar Solubility of an AB

Type Compound from Ksp

Find the molar solubility of PbSO4 at 25°C.Its Ksp at 25°C is 1.8 × 10–8.

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s

1

Find the molar solubility of PbSO4. Its Ksp at 25°C is 1.8 × 10–8.


Let molar solubilityof PbSO4 = s

s s

The molar solubility of PbSO4 = 1.3 × 10–4 M

2 2sp 4Pb SOK

Solubility Equilibrium Equation

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s

1



s s


2 2sp 4Pb SOK


2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s

1



s s


2 2sp 4Pb SOK


× 1/1 1/1 ratio

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s

1





2 2sp 4Pb SOK

1/1 ratios s× 1/1

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s

1

2




s s


2 2sp 4Pb SOK Ksp Expression

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


2 2sp 4Pb SOK

Use this to find the molar solubility

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


s

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


s

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


s s

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s s




s s


So Ksp is equal to s time s

s s

1

2

3

2 2sp 4

sp

8

4 4

2sp

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

s K

K

s s

s




s s


1

2

3

2 2sp 4

sp

8

4 4

2sp

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

s K

K

s s

s




s s


Take the square root of both sides

1

2

3

2 2sp 4

2sp

8

4

sp

4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

s K

K

K

s s

s




s s


1

2

3

2 2sp 4

2sp

8

4

sp

4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

s K

K

K

s s

s




s s


Substitute 1.8 × 10–8 in for Ksp

1

2

3

2 2sp 4

2sp

sp

4 4

81

Pb SO

1.34 10 M

.8 10

1.3 10 M

s

s

K

K

K

s

s

s




s s


1

2

3

2 2sp 4

2sp

s

8

44

p

Pb SO

1.8 10

M 1.3 1.3 0 M1 4 10

K

K s

s K

s

s s




s s


1

2

3

2 2sp 4

2sp

s

8

44

p

Pb SO

1.8 10

1.3 11. M 03 0 M4 1

K

K s

s K

s

s s




s s


M is the unit for molar concentration

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s

s s




s s


2 significant

figures

1

2

3

2 2sp 4

2sp

s

8

44

p

Pb SO

1.8 10

1.3 11. M 03 0 M4 1

K

K s

s K

s

s s




s s


round to 2 significant figures

1

2

3

2 2sp 4

2sp

s

8

4

p

4

Pb SO

1.8 10

1.3 11. M 03 0 M4 1

K

K s

s K

s

s s




s s


2 signi

ficant fi

gures

1

2

3

2 2sp 4

2sp

sp

8

4 4

Pb SO

1.8 10

1.34 10 M 1.3 10 M

K

K s

s K

s

s s




s s

The molar solubility of PbSO4 at 25°C = 1.3 × 10–4 M

Documents

Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp