CHAPTER 18: CHEMICAL EQUILIBRIUM

Section 4: Solubility Equilibrium

ObjectivesExplain what is meant by solubility

product constants, and calculate their values.

Calculate solubilities using solubility product constants.

Carry out calculations to predict whether precipitates will form when solutions are combined.

Solubility ProductA saturated solution contains the

maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance.

A saturated solution is not necessarily a concentrated solution.

The equilibrium principles developed in this chapter apply to all saturated solutions of sparingly soluble salts.

Solubility Product, continuedThe heterogeneous equilibrium system in a

saturated solution of silver chloride containing an excess of the solid salt is represented by

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

The solubility product constant, Ksp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.

Solubility Product, continuedThe equation for the solubility equilibrium

expression for the dissolution reaction of AgCl is

spK -[Ag ][Cl ]

The equilibrium expression is written without including the solid species.

The numerical value of Ksp can be determined from solubility data.

Solubility Product, continuedFor a saturated solution of CaF2, the

equilibrium equation is

spK 2 – 2[Ca ][F ]

2 –

2CaF ( ) Ca ( ) 2F ( )s aq aq

The expression for the solubility product constant is

The solubility of CaF2 is is 8.6 10−3/100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1 10−3 mol/L.

Determining Ksp for Reactions at Chemical Equilibrium

Solubility Product, continuedCaF2 dissociates to yield twice as many F−

ions as Ca2+ ions.

[Ca2+] = 1.1 10−3 mol/L [F− ] = 2.2 10−3 mol/L

spK 2 2[Ca ][F ]

spK –3 3 2(1.1 10 )(2.2 10 )

Ksp = 5.3 10-9

Calculations of Ksp ordinarily should be limited totwo significant figures.

Solubility Product Constants at 25°C

Solubility Product, continuedThe solubility product constant is an

equilibrium constant representing the product of the molar concentrations of its ions in a saturated solution.

The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent.

It has only one value for a given solid at a given temperature.

It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.

Solubility Product, continued

Sample Problem BCalculate the solubility product constant,

Ksp ,forcopper(I) chloride, CuCl, given that the

solubilityof this compound at 25°C is 1.08 10–2

g/100. gH2O.

Solubility Product, continuedSample Problem B Solution

2

2 2

1 g H Og CuCl 1000 mL 1molCuClsolubility in mol /L

100. g H O 1mL H O 1L g CuCl

–CuCl( ) Cu ( ) Cl ( )s aq aq

Ksp=[Cu+][Cl–]

Unknown: Ksp

Given: solubility of CuCl = 1.08 10−2 g CuCl/100. g H2O

Solution:

[Cu+] = [Cl–] = solubility in mol/L

Solubility Product, continuedSample Problem B Solution, continued

–22

2 2

solubility in mol /L

1 g H O1.08 10 g CuCl 1000 mL 1molCuCl

100. g H O 1mL H O 1L 99.0 g CuCl

1.09 10-3 mol/L CuCl

Ksp = (1.09 10-3)(1.09 10-3) =

1.19 10-6

[Cu+] = [Cl–]=1.09 10-3 mol/L

Calculating SolubilitiesThe solubility product constant can be

used to determine the solubility of a sparingly soluble salt.

2 2–

3 3BaCO ( ) Ba ( ) CO ( )s aq aq

spK 2 2– –93[Ba ][CO ] 5.1 10

x2 2–3[Ba ] [CO ]

2spK x x x2 2– –9

3[Ba ][CO ] ( )( ) 5.1 10

x –9 –55.1 10 7.1 10

The molar solubility of BaCO3 is 7.1 10−5 mol/L.

How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?

Calculating Solubilities, continued

Sample Problem CCalculate the solubility of silver bromide,

AgBr, in mol/L, using the Ksp value for this

compound.

Calculating Solubilities, continuedSample Problem C Solution

Given: Ksp = 5.0 10−13

Unknown: solubility of AgBr

Solution:

[Ag+] = [Br−], so let [Ag+] = x and [Br−] = x

s aq) + aqAgBr( ) Ag ( Br ( )

spK –[Ag ][Br ]

Calculating Solubilities, continuedSample Problem C Solution, continued

spK x x x– 2[Ag ][Br ] ( )( )

x2 –135.0 10

x –135.0 10

7

–13solubility

7

of AgBr 5.

.1 1

0 10

0 mol /L

Precipitation CalculationsThe equilibrium condition does not

require that the two ion concentrations be equal. Equilibrium will still be established so that the ion product does not exceed the value of Ksp for the system.

If the ion product is less than the value of Ksp at a particular temperature, the solution is unsaturated.

If the ion product is greater than the value for Ksp, solid precipitates.

Precipitation Calculations, continuedUnequal quantities of BaCl2 and Na2CO3

are dissolved in water and the solutions are mixed.

If the ion product exceeds the Ksp of BaCO3, a precipitate of BaCO3 forms.

Precipitation continues until the ion concentrations decrease to the point at which equals the Ksp.

2 2-3[Ba ][CO ]

2 2-3[Ba ][CO ]

The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.

Precipitation Calculations, continued

Sample Problem DWill a precipitate form if 20.0 mL of 0.010

M BaCl2 is mixed with 20.0 mL of 0.0050 M

Na2SO4?

Precipitation Calculations, continued

Sample Problem D SolutionGiven: concentration of BaCl2 = 0.010 M

volume of BaCl2 = 20.0 mL concentration of Na2SO4 = 0.0050

M volume of Na2SO4 = 20.0 mL

Unknown: whether a precipitate forms

Solution: The two possible new pairings of ions are NaCl and BaSO4. BaSO4 is a sparingly soluble salt.

Precipitation Calculations, continuedSample Problem D Solution, continued

2 2–

4 4BaSO ( ) Ba ( ) SO ( )s aq aq

spK 2 2– –104[Ba ][SO ] 1.1 10

220.010 mol Ba

0.020 L 0.000 20 mol Ba1 L

2

2–2–4

–4

4

0.0050 mol SO0.020 L 0.000 10 mol SO

1 L

mol SO ion :

mol Ba2+ ion:

Precipitation Calculations, continuedSample Problem D Solution, continued

total volume of solution:

0.020 L + 0.020 L = 0.040 Lconcentration Ba2+ ion in combined

solution:

2–3 20.000 20 mol Ba

5.0 10 mol /L Ba0.040 L

2–

–3

2–

4

4

2–4

0

co

.000 10 mol SO2.5 10 mol /L SO

0.040 L

ncentration SO ion in combined solution :

Precipitation Calculations, continuedSample Problem D Solution, continued

The ion product:2 2– –3 –3 –5

4[Ba ][SO ] (5.0 10 )(2.5 10 ) 1.2 10

sp

K

–5 –10

2 2–4

1.2 10 1.1 10

[Ba ][SO ]

Precipitation occurs.

Limitations on the Use of KspThe solubility product principle can be

very useful when applied to solutions of sparingly soluble substances.

It cannot be applied very successfully to solutions of moderately soluble or very soluble substances.

The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together.

Sometimes it is necessary to consider two equilibria simultaneously.

Equilibrium Calculations