William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation

William L MastertonCecile N. Hurleyhttp://academic.cengage.com/chemistry/masterton

Edward J. Neth • University of Connecticut

Chapter 16 Precipitation Equilibria

Outline

1. Precipitate formation: the solubility product constant (Ksp)

2. Dissolving precipitates

Revisiting Precipitation

• In Chapter 4 we learned that there are compounds that do not dissolve in water• These were called insoluble• A reaction that produces an insoluble precipitate

was assumed to go to completion• In reality, even insoluble compounds dissolve to

some extent, usually small• An equilibrium is set up between the precipitate

and its ions• Precipitates can be dissolved by forming complex

ions

Two Types of Equilibria

• AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)• Solid exists in equilibrium with the ions formed

when a small amount of solid dissolves

• AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq)

• Formation of a stable complex ion can cause an otherwise insoluble compound to dissolve

• There are multiple equilibria at work in this example, in similar fashion to the equilibria underlying the function of a buffer (Chapter 14)

Precipitate Formation: Solubility Product Constant, Ksp

• Consider mixing two solutions:

• Sr(NO3)2 (aq)

• K2CrO4 (aq)

• The following net ionic equation describes the reaction:

• Sr2+ (aq) + CrO42- (aq) SrCrO4 (s)

Figure 16.1 – Precipitation of SrCrO4

Ksp Expression

• SrCrO4 (s) ⇌ Sr2+ (aq) + CrO42- (aq)

• The solid establishes an equilibrium with its ions once it forms• We can write an equilibrium expression, leaving out the term for the

solid (recall that its concentration does not change as long as some is present)

• Ksp is called the solubility product constant

]][[ 24

2 CrOSrKsp

Interpreting the Solubility Expression and Ksp

• Ksp has a fixed value at a given temperature

• For strontium chromate, Ksp = 3.6 X 10-5 at 25 °C

• The product of the two concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached

Example 16.1

Ksp and the Equilibrium Concentration of Ions

• Ksp SrCrO4 = [Sr2+][CrO42-] = 3.6 X 10-5

• This means that if we know one ion concentration, the other one can easily be calculated

• If [Sr2+] = 1.0 X 10-4 M, then

• If [CrO42-] = 2.0 X 10-3, then

MX

XCrO 36.0

100.1

106.3][

4

52

4

MXX

XSr 2

3

52 108.1

100.2

106.3][

Example 16.2

Example 16.1, (Cont’d)

Table 16.1

Ksp and Precipitate Formation

• Ksp values can be used to predict whether a precipitate will form when two solutions are mixed• Recall the use of Q, the reaction quotient, from

Chapter 12• We can calculate Q at any time and compare it to

Ksp

• The relative magnitude of Q vs. Ksp will indicate whether or not a precipitate will form

Q and Ksp

• If Q > Ksp, a precipitate will form, decreasing the ion concentrations until equilibrium is established

• If Q < Ksp, the solution is unsaturated; no precipitate will form

• If Q = Ksp, the solution is saturated just to the point of precipitation

Figure 16.2

Example 16.3



Ksp and Water Solubility

• One way to establish a solubility equilibrium• Stir a slightly soluble solid with water• An equilibrium is established between the solid and its ions

• BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)

• If we set the concentration of the ions equal to a variable, s:

MXXKs

sSOBaBaSOK

sSOBa

sp

sp

52

110

224

24

24

2

100.1)101.1(

]][[

][][

Precipitation Visualized

Example 16.4

Example 16.4, (Cont'd)


Calculating Ksp Given Solubility

• Instead of calculating solubility from Ksp, it is possible to calculate Ksp from the solubility

• Recall that solubility may be given in many different sets of units

• Convert the solubility to moles per liter for use in the Ksp expression

Example 16.5


Ksp and the Common Ion Effect

• BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)

• How would you expect the solubility of barium sulfate in water to compare to its solubility in 0.10 M Na2SO4?• Solubility must be less than it is in pure water• Recall LeChâtelier’s Principle

• The presence of the common ion, SO42-, will drive the

equilibrium to the left• Common ions reduce solubility

Visualizing the Common Ion Effect

Example 16.6


Selective Precipitation

• Consider a solution of two cations• One way to separate the cations is to add an

anion that precipitates only one of them• This approach is called selective precipitation

• Related approach• Consider a solution of magnesium and barium

ions

Selective Precipitation, (Cont'd)

• Ksp BaCO3 = 2.6 X 10-9

• Ksp MgCO3 = 6.8 X 10-6

• Carbonate ion is added

• Since BaCO3 is less soluble than MgCO3, BaCO3 precipitates first, leaving magnesium ion in solution

• Differences in solubility can be used to separate cations

Figure 16.3

Figure 16.4 – Selective Precipitation

Example 16.7


Dissolving Precipitates

• Bringing water-insoluble compounds into solution• Adding a strong acid to react with basic anions• Adding an agent that forms a complex ion to react

with a metal cation

Strong Acid

• Zn(OH)2 (s) + 2H+ (aq) Zn2+ (aq) + 2H2O

• This reaction takes place as two equilibria:

• Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)

• 2H+ (aq) + 2OH- (aq) ⇌ 2H2O

• Zn(OH)2 (s) + 2H+ (aq) ⇌ Zn2+ (aq) + 2H2O

• Because the equilibrium constant for the neutralization is so large, the reaction goes essentially to completion

• Note that for the second equilibrium, K = (1/Kw)2 = 1 X 1028

Example 16.8



Insoluble Compounds that Dissolve in Strong Acid

• Virtually all carbonates

• The product if the reaction is H2CO3, a weak acid that decomposes to carbon dioxide• H2CO3 (aq) H2O + CO2 (g)

• Many sulfides

• The product of the reaction is H2S, a gas that is also a weak acid• H2S (aq) ⇌ H+ (aq) + HS- (aq)

Visualizing Selective Dissolving of Precipitates

Example 16.9

Complex Formation

• Ammonia and NaOH can dissolve compounds whose metal cations form complexes with NH3 and OH-

• As with the addition of a strong acid, multiple equilibria are at work:

• Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq) Ksp

• Zn2+ (aq) + 4NH3 (aq) ⇌ Zn(NH3)42+ (aq) Kf

• Net: Zn(OH)2 (s) + 4NH3 (aq) Zn(NH3)42+ (aq) + 2OH- (aq)

• Knet = KspKf = 4 X 10-17 X 3.6 X 108 = 1 X 10-8

Table 16.2

Visualizing Dissolving by Complex Formation

Example 16.10


Example 16.11

Key Concepts

1. Write the Ksp expression for an ionic solid

2. Use the value of Ksp to

A. Calculate the concentration of one ion, knowing the other

B. Determine whether a precipitate will form

C. Calculate the water solubility of a compound

D. Calculate the solubility of a compound in a solution of a common ion

E. Determine which ion will precipitate first

Key Concepts

3. Calculate K for

A. Dissolving a metal hydroxide in a strong acid

B. Dissolving a precipitate in a complexing agent

4. Write balanced, net ionic equations to explain why a precipitate dissolves in

A. Strong acid

B. Ammonia or hydroxide solution

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William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation