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# Heterogeneous Equilibrium Review

### Review skills for Quiz

• Be able to write the Ksp expression for the ionization of any salts.
• Calculate Ksp 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 Qsp and compare it with tbe Ksp 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 H2S is used as a precipitation agent. H2S is a very useful compound for separation of metal ions.
• Calculate the pH at which two metal ions will be separated using H2S 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 Cr2O42-, you may want to find a cation that reacts with them forming compounds of different solubility. From a table of Ksp values, you find the following:

[Ag+] [Cl-] = 1.8e-10
[Ag+]2 [Cr2O42-] = 1.9e-12

Does this difference in Ksp values afford a separation for Cl- and CrO42-?

To answer this question, we calculate the lowest concentrations of Ag+ at which AgCl and Ag2CrO4 start to precipitate when you increase [Ag+] in the solution.

Lets calculate [Ag+] when AgCl starts to form a precipitate:

[Ag+] = 1.8E-10 / [Cl-]
= 1.8E-10 / 0.010
= 1.8E-8
If [Ag+] > = 1.8E-8 M, then AgCl precipitates.

Now, lets calculate [Ag+] when Ag2CrO4 begins to precipitate:

[Ag+]2 = (1.9E-12) / [Cr2O42-]
= (1.9E-12) / 0.010
= 1.9EE-10
[Ag+] = 1.4e-5 M
If [Ag+] > 1.4e-5, then Ag2CrO4 will precipitate

AgCl will precipitate at a lower Ag+ concentration than Ag2CrO4 will. As you increase the concentration of Ag+, a series of event will happen as the diagram below shows.

 [Ag+] [Cl-] [Cr2O42-] Reaction 1.8e-8 1.8e-6 1.4e-5 1.4e-3 0.014 1.0e-2 1.0e-4 1.3e-5 1.3e-7 1.3e-8 0.01 0.01 0.01 9.7e-7 9.7e-9 Forming AgCl solid Forming Ag2CrO4 solid plus little AgCl solid
What is the [Cl-] when Ag2CrO4 begins to precipitate?

```  [Cl-] = 1.8E-10 / 1.4E-5
= 1.3E-5 M
```
When Ag2CrO4 starts to precipitate, [CrO42-] = 0.01, thus the precipitate at this point has the molar ratio of:
```  Ag2CrO4      0.010       200
-------  =  -------  =  ------
AgCl       0.00005       1
```
and note that the Cl concentration has been drastically reduced when Ag2CrO4 begins to precipitate.

## Separating Metal Ions as Sulfide

Sulfide Ksp
MnS 7e-16
FeS 1e-19
ZnS 4.5e-24
CdS 1.4e-28
HgS 3e-53
Many metal ions form insoluble compounds with the sulfide ion, S2-. 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, [S2-].

Hydrogen sulfide is a diprotic acid. Its two-stage ionizations are shown in the following equilibrium equations:

 H2S = H+ + HS- Ka1 = 1.1E-7 HS- = H+ + S2- Ka2 = 1E-14 H2S = 2 H+ + S2- Koverall = 1E-21
[S2-] = 1e-21 / 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 Ksp's for BaF2 and CaF2 are 1.7e-6 and 1.7e-10 respectively. If a solution contains 0.100 M Ba2+ and Ca2+, what is the compound in the first trace of precipitate as you increase [F-]?

Discussion...

The smaller the Ksp, the less soluble is the compound. CaF2 is less soluble than BaF2.

• The Ksp's for BaF2 and CaF2 are 1.7e-6 and 1.7e-10 respectively. If a solution contains 0.100 M Ba2+ and Ca2+, what is [Ca2+] when [BaF2] begins to precipitate?

Discussion...

When BaF2 begins to precipitate, [F-]2 = (1.7e-6)/0.100 = 1.7e-5 M, and [F-] = 4.1e-3. Thus [Ca2+] = 1.7e-10/1.7e-5 = ?.

• The Ksp's for BaF2 and CaF2 are 1.7e-6 and 1.7e-10 respectively. If a solution contains 1.7e-3 M F- what is the maximum [Ca2+]?

Discussion...

The maximum [Ca2+] = (1.7e-10)/(1.7e-3)2 = 5.9e-5 M.
If [Ca2+] is higher than this value, precipitation will have reduced its concentration to this value, providing that [F-] is maintained.

• The Ksp's for BaF2 and CaF2 are 1.7E-6 and 1.7E-10 respectively. If a solution contains 4.1E-2 M F- what is the maximum [Ba2+]?

Discussion...

The maximum [Ba2+] = (1.7E-6)/(4.1E-2)2 = ?.

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

• The Ksp's for AgCl, AgBr and AgI are 1.8E-10, 5.2E-13, and 8.5E-17 respectively. Would you expect AgF to be more soluble than AgCl?

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 Ksp's for AgCl, AgBr and AgI are 1.8E-10, 5.2E-13, and 8.5E-17respectively. Do the differences in Ksp values afford a reasonable separation of halide ions by AgNO3?

Discussion...

Theoretically, they can be separated. However, it still takes a skilled chemist to do the job properly.

• The Ksp's for AgCl, AgBr and AgI are 1.8E-10, 5.2E-13, and 8.5E-17 respectively. What is [Br-] when a trace of AgCl precipitates, if the solution contains 0.010 M Cl-?

Discussion...

When a trace of AgCl precipitates, [Ag+] = (1.8E-10) / (0.01) = 1.8E-8 M.
Maximum [Br-] = (5.2E-13) / 1.8E-8 = 2.9E-5.
[Cl-] = 0.010 M
[Br-] = 0.000029 M

• Calculate the equilibrium concentration of Ag+ and Ag(NH3)2+ ions in a solution obtained by adding 0.10 mol AgNO3 to 1.0 L of 2.0 M NH2 solution. Find the formation constant for Ag(NH3)2+ ions from a Handbook.

Answer [Ag(NH3)2+] = 0.10 M; [Ag+] = 2.0e-7 M
Discussion...

Consider the formation of 0.10 Ag(NH3)2+ ions first. Then assume x M of Ag+ ions at equilibrium.

``` Ag+ + 2 NH3 = Ag(NH3)2+
x  1.80+2x   0.10-x
and
0.10-x
---------- = 1.6e7  (Kf 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 NH3 = Ag(NH3)2+ + Cl- You have to learn how to look up Ksp and Kf for the salt and complex.

```AgCl = Ag+ + Cl-,        Ksp = 1.8e-10