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Heterogeneous Equilibrium Review
Review skills for Quiz

Be able to write the K_{sp} expression for the ionization of any salts.

Calculate K_{sp} from molar solubility and vice versa.
Pay attention to the stoichiometry of the salt. You may have to understand
what the salt really is in order to know how they ionize in the solution.

Calculate the Q_{sp} and compare it with tbe K_{sp}
for a solution. Explain if the solution is unsaturated, saturated, or
supersaturated.

Recognize the common ions from various salts, acids, and bases.
Problems involving a concentration of a common ion are called
common ion problems.

Calculate concentration of a common ion in a solution of several salts.

Calculate the ion concentration required to maintain a given
concentration of some other ion.

Calculate the concentration of a precipitating agent that will
result in the maximum separation of two ions.

Calculate the concentration of ions remaining in solution
after the addition of precipitating agent. Look deeper into a problem.

Calculate the pH or pOH (concentration of H^{+} or OH^{})
in order to cause a precipitate of some salts or some sparingly soluble
hydroxide.

Understand how pH will affect a precipitate of sulphide when H_{2}S
is used as a precipitation agent. H_{2}S is a very useful compound
for separation of metal ions.

Calculate the pH at which two metal ions will be separated using
H_{2}S as the precipitating agent.
Selected Precipitation
The strategy for chemical analysis has been discussed in
Separation by Precipitation.
More examples are given here.
Separate, identify, and quantitatively determine the compounds or ions
present in a sample are essential in the study of chemistry.
Separate and determine amounts of different types of ions are important
industrial and scientific applications.
For example, if a solution contains 0.010 M of both Cl^{} and
Cr_{2}O_{4}^{2}, you may want to find a
cation that reacts with them forming compounds of different solubility.
From a table of K_{sp} values, you find the following:
[Ag^{+}] [Cl^{}] = 1.8e10
[Ag^{+}]^{2} [Cr_{2}O_{4}^{2}] = 1.9e12
Does this difference in K_{sp} values afford a separation for Cl^{} and CrO_{4}^{2}?
To answer this question, we calculate the lowest concentrations of Ag^{+}
at which AgCl and Ag_{2}CrO_{4} start to precipitate when you increase
[Ag^{+}] in the solution.
Lets calculate [Ag^{+}] when AgCl starts to form a precipitate:
[Ag^{+}] = 1.8E10 / [Cl^{}]
= 1.8E10 / 0.010
= 1.8E8
If [Ag^{+}] > = 1.8E8 M, then AgCl precipitates.
Now, lets calculate [Ag^{+}] when Ag_{2}CrO_{4} begins to precipitate:
[Ag^{+}]^{2} = (1.9E12) / [Cr_{2}O_{4}^{2}]
= (1.9E12) / 0.010
= 1.9EE10
[Ag^{+}] = 1.4e5 M
If [Ag^{+}] > 1.4e5, then Ag_{2}CrO_{4} will precipitate
AgCl will precipitate at a lower Ag^{+} concentration than Ag_{2}CrO_{4} will.
As you increase the concentration of Ag^{+}, a series of event will
happen as the diagram below shows.
[Ag^{+}]
 1.8e8  1.8e6  1.4e5  1.4e3  0.014


[Cl^{}]
 1.0e2
 1.0e4
 1.3e5
 1.3e7
 1.3e8


[Cr_{2}O_{4}^{2}]
 0.01  0.01
 0.01
 9.7e7
 9.7e9


Reaction  Forming AgCl solid
 Forming Ag_{2}CrO_{4} solid
plus little AgCl solid


What is the [Cl^{}] when Ag_{2}CrO_{4} begins to precipitate?
[Cl^{}] = 1.8E10 / 1.4E5
= 1.3E5 M
When Ag_{2}CrO_{4} starts to precipitate,
[CrO_{4}^{2}] = 0.01, thus the precipitate
at this point has the molar ratio of:
Ag_{2}CrO_{4} 0.010 200
 =  = 
AgCl 0.00005 1
and note that the Cl concentration has been drastically reduced when
Ag_{2}CrO_{4} begins to precipitate.
Separating Metal Ions as Sulfide
Sulfide  K_{sp}


MnS  7e16

FeS  1e19

ZnS  4.5e24

CdS  1.4e28

HgS  3e53

Many metal ions form insoluble compounds with the sulfide ion, S^{2}.
Some examples are given in a table on the right.
In this group, MnS is the most soluble, whereas HgS is the least soluble.
These metal ions can be separated by careful control of sulfide ion
concentration, [S^{2}].
Hydrogen sulfide is a diprotic acid. Its twostage ionizations are
shown in the following equilibrium equations:
H_{2}S = H^{+} + HS^{}  K_{a1} = 1.1E7

HS^{} = H^{+} + S^{2}  K_{a2} = 1E14

H_{2}S = 2 H^{+} + S^{2}  K_{overall} = 1E21

[S^{2}] = 1e21 / 10^{pH*2}
Thus, by careful control of pH the concentration of sulfide ion can be
adjusted to precipitate a group of metal ions. This was the main technique
of qualitative analysis before 1960s. Today, many spectroscopic methods
have been introduced and minute amount of metal ions can be detected using
for example atomic absorption (AA) spectroscopy.
Confidence Building Questions or
Quiz

The K_{sp}'s for BaF_{2} and CaF_{2} are 1.7e6 and 1.7e10 respectively.
If a solution contains 0.100 M Ba^{2+} and Ca^{2+}, what is the
compound in the first trace of precipitate as you increase [F^{}]?
Answer CaF_{2}
Discussion...
The smaller the K_{sp}, the less soluble is the compound.
CaF_{2} is less soluble than BaF_{2}.

The K_{sp}'s for BaF_{2} and CaF_{2} are 1.7e6 and 1.7e10 respectively.
If a solution contains 0.100 M Ba^{2+} and Ca^{2+}, what is [Ca^{2+}]
when [BaF_{2}] begins to precipitate?
Answer [Ca^{2+}] = 1e5 M
Discussion...
When BaF_{2} begins to precipitate, [F^{}]^{2}
= (1.7e6)/0.100 = 1.7e5 M, and
[F^{}] = 4.1e3. Thus [Ca^{2+}] = 1.7e10/1.7e5 = ?.

The K_{sp}'s for BaF_{2} and CaF_{2} are 1.7e6 and 1.7e10 respectively.
If a solution contains 1.7e3 M F^{} what is the maximum [Ca^{2+}]?
Answer 5.9e5 M
Discussion...
The maximum [Ca^{2+}] = (1.7e10)/(1.7e3)^{2} = 5.9e5 M.
If [Ca^{2+}] is higher than this value, precipitation will have
reduced its concentration to this value, providing that [F^{}]
is maintained.

The K_{sp}'s for BaF_{2} and CaF_{2} are
1.7E6 and 1.7E10 respectively.
If a solution contains 4.1E2 M F^{} what is the maximum
[Ba^{2+}]?
Answer 1e3 M
Discussion...
The maximum [Ba^{2+}] = (1.7E6)/(4.1E2)^{2} = ?.

All three questions are interrelated. Have you gotten the idea of
selective precipitation?
Answer Yes

The K_{sp}'s for AgCl, AgBr and AgI are 1.8E10, 5.2E13, and
8.5E17 respectively. Would you expect AgF to be more soluble than AgCl?
Answer Yes
Discussion...
The group 7 elements are F, Cl, Br, I, and At in order of increasing
atomic number. Deduction is an important method of study but a scientist
keeps an open mind.

The K_{sp}'s for AgCl, AgBr and AgI are 1.8E10, 5.2E13, and
8.5E17respectively. Do the differences in K_{sp} values
afford a reasonable separation of halide ions by AgNO_{3}?
Answer Yes
Discussion...
Theoretically, they can be separated. However, it still takes a
skilled chemist to do the job properly.

The K_{sp}'s for AgCl, AgBr and AgI are 1.8E10, 5.2E13,
and 8.5E17 respectively. What is [Br^{}] when a trace of AgCl
precipitates, if the solution contains 0.010 M Cl^{}?
Answer 2.9e5 M
Discussion...
When a trace of AgCl precipitates, [Ag^{+}] = (1.8E10) / (0.01)
= 1.8E8 M.
Maximum [Br^{}] = (5.2E13) / 1.8E8 = 2.9E5.
[Cl^{}] = 0.010 M
[Br^{}] = 0.000029 M

Calculate the equilibrium concentration of Ag^{+} and
Ag(NH_{3})_{2}^{+} ions
in a solution obtained by adding 0.10 mol AgNO_{3} to 1.0 L of
2.0 M NH_{2} solution. Find the formation constant for
Ag(NH_{3})_{2}^{+} ions from a Handbook.
Answer [Ag(NH_{3})_{2}^{+}] = 0.10 M;
[Ag^{+}] = 2.0e7 M
Discussion...
Consider the formation of 0.10 Ag(NH_{3})_{2}^{+}
ions first. Then assume x M of Ag^{+} ions at equilibrium.
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+}
x 1.80+2x 0.10x
and
0.10x
 = 1.6e7 (K_{f} from a handbook)
(1.8+2x)^{2} x
Approximation can be used to solve for x. Make sure you know how to set
the formula up.

Calculate the equilibrium constant for the reaction
AgCl + 2 NH_{3} = Ag(NH_{3})_{2}^{+} + Cl^{}
You have to learn how to look up K_{sp} and
K_{f} for the salt and complex.
Answer 2.9e3 M^{1}
Discussion...
Do you know where to look for K_{sp} and K_{f}?
AgCl = Ag^{+} + Cl^{}, K_{sp} = 1.8e10
Ag^{+} + 2 NH_{3} = Ag(NH_{3})_{2}^{+}, K_{f} = 1.6e7.

AgCl + 2 NH_{3} = Ag(NH_{3})_{2}^{+} + Cl^{}, K = ?
Quiz
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