# Chapter 16 Precipitation Equilibria

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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

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

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

Example 16.3, (Cont’d)

Example 16.3, (Cont’d)

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:

Precipitation Visualized

Example 16.4

Example 16.4, (Cont'd)

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

Example 16.5, (Cont'd)

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

Example 16.6, (Cont'd)

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

Example 16.7, (Cont'd)

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

Example 16.8, (Cont'd)

Example 16.8, (Cont'd)

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 X 3.6 X 108 = 1 X 10-8

Table 16.2

Visualizing Dissolving by Complex Formation

Example 16.10

Example 16.10, (Cont'd)

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