Precipitation Equilibrium
Precipitation reactions reach a position of
equilibrium – even the most insoluble
electrolyte dissolves to at least a slight
extent, establishing an equilibrium with it’s
ions in solution.
 For example, a solution of
lead II chloride.

Ion-Product Equilibrium Systems

There are 2 types of precipitation equilibria:
(1)
between a precipitate and its ions.
A precipitate (ppt.) forms when a cation from one solution
combines with another forming an insoluble ionic solid.
For example:
Sr(NO3)2 (aq) + K2CrO4 (aq) ⇌ 2KNO3 (aq)
+ SrCrO4 (s)
(2) between a precipitate and the species used to
dissolve it.
AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq)
The Solubility Product Constant



The Ksp expression – this is the equilibrium
constant expression for the dissolving of a
solid.
The smaller the Ksp value the less soluble the
precipitate.
And like the any equilibrium constant value,
the Ksp value is at a fixed temperature.
Writing Ksp Expression:

Write the ion-product expression for each
compound:





(a) magnesium carbonate
(b) iron(II) hydroxide
(c) calcium phosphate
(d) silver chromate
(e) silver sulfide
*(be careful with sulfide ions, they are so reactive with water that they will produce HS
OH-)
-
and
Calculating Ion Concentration

Calcium phosphate is a water-insoluble
mineral, large quantities of which are used to
make commercial fertilizers. Taking it’s
equilibrium constant value of 1 x 10-33,
calculate:

The concentration of the phosphate ion in
equilibrium with the solid if the [Ca2+] = 1 x 10-9 M


Ans = 1 x 10-3 M
The concentration of the calcium ions in
equilibrium with the solid if the [PO43-] = 1 x 10-5 M

Ans = 2 x 10-8 M
Determining Precipitate Formation:

Ksp values can be used to predict when a
precipitate will form or not. To do this we
work with the reaction quotient once more:



If Q > Ksp
- the reaction will shift to the left so
ppt. forms
If Q < Ksp
- the reaction will shift to the right so
no ppt. forms
If Q = Ksp
- then the solution is saturated with
ions at the point of ppt.
Determining if a ppt. will form…

Sodium chromate is added to a solution in which the
original concentration of Sr2+ is 0.0060 M

(a) assuming the [Sr2+] stays constant, will a precipitate of
strontium chromate form when the chromate ion
concentration becomes 0.0030 M?


ans: no precipitate will form, Q < K
(b) will a precipitate of strontium chromate form if 0.200 L of
0.0060 M of strontium nitrate solution is mixed with 0.800 L
of 0.040 M potassium chromate?

ans: a precipitate will just barely form, Q > K (just slightly)
Ksp & Water Solubility:


One way to establish equilibrium between a
slightly soluble solid is to stir the solid with
water to form a saturated solution.
Solubility of the solid, in moles per liter, is
related to the solubility product constant.

For example, determine the solubility of barium
sulfate.

ans: s = 1.0 x 10-5 M
Example: Determine Solubility

Calculate the solubility of barium fluoride in
moles/liter and grams/liter.

ans: 3.6 x 10-3 M & 0.63 g/L
Ksp & The Common Ion Effect


The solubility of an insoluble substance
decreases when adding a solution with a
common ion in comparison to adding water
(similar to Le Chatelier’s Principle).
Which will have the higher solubility:


barium sulfate in water OR barium sulfate in a
0.1M solution of sodium sulfate
Why?
Applying the Common-Ion Effect

Taking the equilibrium constant value of
barium sulfate into account, estimate its
solubility in a 0.10 M solution of sodium
sulfate (hint: you will need an equilibrium
table).

ans: 1.1 x 10-9 M
Selective Precipitation


A way to separate 2 cations in water solution is to
add an ion that precipitates only one of the cations.
For example:

A flask contains a solution 0.10 M Cl- and 0.010 M CrO42-.
When AgNO3 is added:


(a) which anion, chloride or chromate, precipitates first?
ans: The chloride ion will ppt. first since it requires the
lowest silver ion concentration to ppt.
(b) what percentage of the first anion has been precipitated
when the second anion starts to precipitate?
ans: 99.98% of chloride ions precipitated.
Dissolving Precipitates

Water insoluble ionic solids can be brought
into solution by adding a reagent to react with
either the anion or the cation. The 2 most
useful reagents for this are:


1. Strong Acids – the H+ will react with the basic
anions
2. NH3 or OH- to react with the metal cations
Dissolving Zinc Hydroxide


Write the equation that represents the
dissolving of zinc hydroxide by a strong acid
(hint: two-step process).
Determine the equilibrium constant for this
reaction (hint: rule of multiple equilibria). The
Kw for 1 mole of water is 1 x 10-14.
Strong Acids

Strong acids can be used to dissolve water-insoluble
salts in which the anion is a weak base:



Almost all carbonates
Many sulfides
Example: Write balanced equations to explain why
each of the following precipitates dissolve in a
strong acid (assume the acid is in excess):



Aluminum hydroxide
Calcium carbonate
Cobalt II sulfide
Complex Ion Formation


Ammonia and hydroxides are commonly
used to dissolve precipitates containing a
cation that forms a stable complex with NH3
or OH- (reference table 16.2 – page 434).
Write the reaction by which zinc hydroxide
dissolves in ammoina and solve for the
equilibrium constant of this reaction using:

Step 1: Zn(OH)2 (s) ⇌ Zn2+ + 2OH4.0 x 10-17
Ksp =

Step 2: Zn2+ + 4NH3 (aq) ⇌ Zn(NH3)42+ + 2OH3.6x108
Kf =
Kf = equilibrium constant for the formation of complex ions (page 416)
Dissolving AgCl with Ammonia

Consider the reaction by which silver chloride
dissolves in ammonia:

Taking the Ksp AgCl = 1.8 x 10-10 and the Kf
Ag(NH3)2+ = 1.7 x 107, calculate the K for this
reaction.


ans: 3.1 x 10-3
Calculate the number of moles of AgCl that
dissolves in one liter of 6.0 M ammonia.

ans: 0.30 moles/liter