Qual Analysis

CHEM 122L

General Chemistry Laboratory

Revision 3.2

A Qualitative Analysis for Select Cations

To learn about how to Develope of a Qualitative Analysis Scheme.

To learn about Separation of Cations in an Aqueous Solution.

To learn about Precipitation Equilibria.

To learn about Complex Ion Formation.

To learn about Flame Tests for Cations.

In this laboratory exercise we will separate and identify Cations dissolved in an Aqueous system.

Since we will not quantify the amount of each Cation present, but instead merely discern its

presence, such a scheme for separation and identification is referred to as a Qualitative Analysis.

In our particular case, we will be testing for the presence of the following nine Cations:

Ag+, Pb

2+, Cu

2+, Bi

3+, Fe

3+, Mn

2+, Ni

2+, Ba

2+, Na

+

Although this style of “wet chemical” analysis is no longer commonly used to determine the

presence of these Cations, the development of this type of Qual scheme has many other

applications in chemistry. Additionally, this exercise is useful as a study of Aqueous equilibria

involving precipitates and complexes, each of which do have important applications in

chemistry.

Our general approach to separating these Cations is to Group them according to the types of

precipitates they form: Chlorides (Cl-), Sulfides (S

2-), Hydroxides (OH

-), etc. We will proceed

by selectively precipitating the Cations in each Group. Once a Group of Cations is precipitated,

the Cations will be further separated using techniques specific to that Group. Once each Cation

is separated from the others, a confirmatory test will be used to, as the name implies, confirm the

Cation is actually present. These confirmatory tests are typically Cation specific and run the

gamut from the formation of brightly colored complexes to producing distinctly colored flames

in a Bunsen burner.

Let’s consider an example. Suppose we are testing an Aqueous sample for the presence of Pb2+

,

Hg22+

and Ca2+

ions. (Somehow we know no other Cations are present.) We can start to develop

a Qual scheme by testing separate samples of each Cation for precipitation with Chloride (Cl-).

If we do this, we note Chloride precipitates form from Pb2+

and Hg22+

solutions. Thus the

Chloride Group, or Group 1, Cations include these two ions. We further note that heating each

of these precipitates causes the PbCl2 to re-dissolve. This is because PbCl2 is reasonably soluble

in Water at high temperatures, but Hg2Cl2 is not.

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Thus, we have the beginnings of a Qual scheme for this Cation system; a method for separating

the these Cations in a mixture of these Cations. To the mixture of Cations, add HCl to

precipitate the Chloride salts, centrifuge and decant off the supernatant. This separates the Hg22+

and Pb2+

ions from the Ca2+

. Now, add Water to the precipitate and heat. Again, centrifuge and

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decant off the supernatant. Separation of all the Cations is now complete.

The last piece needed to complete this Qual scheme is to add confirmatory tests. Afterall, how

would we ever know the Ca2+

was present in our system above; all we see is a clear liquid. And,

if you think about it, that clear liquid that we claim contains Pb2+

is also just a clear liquid. How

do we know Pb2+

is actually present. Maybe the only Cation present was Hg22+

. And, it would

be nice to know that white Chloride precipitate is actually Hg2Cl2 and not some other Chloride

salt. Confirmatory tests are very Cation specific. They give a “positive” result when the desired

Cation is present and a “negative” result if not, or if only other Cations are present.

The presence of Pb2+

can be confirmed by adding a little Potassium Chromate (K2CrO4). The

Pb2+

precipitates as a golden yellow solid characteristic of PbCrO4.

The presence of Hg22+

is confirmed by adding Aqueous Ammonia, resulting in a dark grey

precipitate of HgNH2Cl and Hg.

The presence of Ca2+

must be confirmed by a Flame Test. Many metal cations give off brightly

colored light when a few drops are added to a burner flame. Ba gives off green light, Li purple

and Ca bright orange. This is because the outer valence electrons are kicked into higher orbitals

by the energy of the flame. When the atom relaxes, a photon whose wavelength is dependent on

the orbital energy spacing is emitted. Typically a Flame Test is performed by dipping a small

loop of Nicrome Wire into the test solution and placing the drop that hangs on the wire in a

Bunsen burner flame. The resulting color is then observed directly.

With these confirmatory tests in hand, we now have a fully developed Qual scheme for this

system of Cations. In Flow chart form, this is represented as:

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This is the type of scheme you will develop for our list of 9 Cations. You will then use this

scheme to test a solution of known composition, and one of unknown composition, for the

Cations present.

Now, to some of those pesky details. How do we know a precipitate will form? Is selective

precipitation an effective method of separating Cations? How might we re-solubilize our various

precipitates so as to run confirmatory tests, etc.?

In order to predict whether or not a precipitate will form, we need to examine the equilibrium

between the potential solid and its aqueous ions. Typically, this is written as a Solubility

Equilibrium and the Equilibrium Constant is referred to as a Solubility Product, Ksp. For Hg22+

precipitating as a Chloride (Cl-) salt, we have:

Hg2Cl2(aq) Hg22+

(aq) + 2 Cl-(aq) (Eq. 1)

where:

Ksp = [Hg22+

] [Cl-]

2 = 1 x 10

-18 (Eq. 2)

To determine if a precipitate will form we calculate the Reaction Quotient, Q, based on the

experimental conditions and compare the result with the Ksp. If:

Q > Ksp A ppt will form (Eq. 3)

and if:

Q < Ksp A ppt will not form (Eq. 4)

For our Mercurous Chloride (Hg2Cl2) example, suppose our mixture is brought to [Cl

-] = 0.1M by

adding HCl. Further suppose the Hg22+

Cation is at 0.1M. We have:

Q = [Hg22+

] [Cl-]

2 = (0.10) (0.10)

2 = 10

-3

So,

10-3

> 1 x 10-18

and a precipitate will form.

Next, we desire to know if precipitation can be used to selectively precipitate one Cation and not

another. For instance, suppose we have a solution containing Cu2+

and Fe2+

, both at 0.1M. Is it

possible to bring the Sulfide (S2-

) concentration high enough to precipitate 99.999% of the Cu2+

without also precipitating the Fe2+

. The relevant equilibria are:

Ksp = 1 x 10-36

CuS(s) Cu2+

(aq) + S2-

(aq) (Eq. 5)

Ksp = 2 x 10-19

FeS(s) Fe2+

(aq) + S2-

(aq) (Eq. 6)

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First, determine the Sulfide concentration needed to precipitate 99.999% of the Cu2+

; meaning

0.001%, or 0.00001 x 0.1M = 10-6

M, will remain. At this point, using the Equilibrium Constant

Expression, we have a Sulfide Ion concentration of:

Ksp = 1 x 10-36

= [Cu2+

] [S2-

] = (10-6

) [S2-

] (Eq. 7)

or,

[S2-

] = 10-30

M (Eq. 8)

Thus, the Reaction Quotient, Q, for the FeS system under these conditions will equal:

Q = [Fe2+

] [S2-

] = (0.1) (10-30

) = 10-31

(Eq. 9)

Comparing this with the Ksp for FeS, we see:

Q <<< Ksp (Eq. 10)

So, no FeS will precipitate. In other words, precipitation using Sulfide Ion is an effective means

of separating Cu2+

from Fe2+

in an aqueous system.

A final concern is how to re-solubilize, selectively, salts that form co-precipitates. One method

is to change the pH of the solution. Zinc Carbonate is an example of a precipitate that re-

solubilizes as the pH is lowered by adding Acid (H+) to the system:

ZnCO3(aq) + 2 H+(aq) Zn

2+(aq) + H2CO3(aq) (Eq. 11)

Another trick is to form complex ions of the Cation. For example, in an Ammonia (NH3)

solution, Ag+, forms an Ag(NH3)2

+ complex:

AgCl(s) Ag+(aq) + Cl

-(aq) (Eq. 12)

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq) (Eq. 13)

The second of these reactions is referred to as a Formation Reaction, the complex ion is

“formed” from the Cation (Ag+) and its ligands (NH3), and the Equilibrium Constant is a

Formation Constant, Kf.

For our system of 9 Cations, we will selectively precipitate them in four Groups. These are:

Group 1

Group 1 Cations precipitate as a Chloride. From the examples above, we note Hg22+

(not a Cation

in our system) is a Group 1 Cation:

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Hg22+

(aq) + 2 Cl-(aq) Hg2Cl2(aq) (Eq. 14)

Group 2

Group 2 Cations do not precipitate as Chlorides but will precipitate upon treatment with Hydrogen

Sulfide (H2S). In an aqueous environment, Hydrogen Sulfide dissociates as a weak acid:

2 H+(aq) + S

2-(aq) (Eq. 15) H2S(aq)

If the solution is already Acidic, the equilibrium will shift left and the concentration of S2-

will

remain fairly low. Thus, only very insoluble Sulfides will precipitate in this Group. Cd2+

, another

example not in our system, is a member of this Group:

Cd2+

(aq) + S2-

(aq) CdS(s) (Eq. 16)

Group 3

These Cations precipitate in an Ammoniacal Solution. Because Ammonia solutions are Basic

(OH-):

NH4+(aq) + OH

-(aq) (Eq. 17) NH3(aq) + H2O

many of the Cations in this Group precipitate as Hydroxides. Cr3+

is an example:

Cr3+

(aq) + 3 OH-(aq) Cr(OH)3(s) (Eq. 18)

Other members of this Group precipitate as the Sulfide. This is because the Basic environment

causes the Hydrogen Sulfide equilibrium to shift toward the Right:

2 H+(aq) + S

2-(aq) (Eq. 19) H2S(aq)

drastically increasing the Sulfide Ion (S2-

) concentration. This causes Sulfides that are otherwise

more soluble (i.e., did not precipitate as a Group 2 Cation.) to suddenly precipitate.

Group 4

These are Cations that do not precipitate. They will remain in solution even after performing the

procedures to precipitate the Group 1, 2, and 3 Cations.

Thus, we will initially treat all 9 of our Cations individually with the Group 1 precipitating

reagent (HCl) to determine which are members of this Group. Once this is determined, we will

then examine methods for separating them and confirming their presence. Having completed

this task, we will proceed with the remaining Cations and categorize, separate and confirm them.

Once this has been completed with the individual Cations, a Qual Scheme for these Cations will

be constructed. And, having done this, we will proceed to the task of analyzing a mixture of

these possible Cations of unknown composition.

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Pre-Lab Questions

Week 1

1. Consult an appropriate source and determine the Concentration of each of the following

species, when in Concentrated form: HCl, H2SO4, and NH3.

2. A 10mL solution of 0.010M HCl is mixed with 20mL of a 0.01M Pb2+

solution, giving a

total volume of 30mL. What are the concentrations of Cl- and Pb

2+ after the mixing? Will

a precipitate of PbCl2 form? (Ksp = 1.7 x 10-5

for PbCl2 at 25oC.)

3. In the Group 2 precipitations, the Sulfide Ion (S2-

) is generated in an Acidic environment.

If the H2S concentration is maintained at 0.1M and the pH = 1, what is the [S2-

]

concentration? (Ka1 = 1 x 10-7

and Ka2 = 1 x 10-13

for H2S.)

4. In the Group 3 precipitations, the pH is raised by adding Ammonia. Suppose it is raised to

pH = 9. What is the [S2-

] concentration under these conditions? Assume the H2S

concentration is again 0.1M.

Week 2

1. Prepare a Flow Chart indicating how you will separate and confirm the presence of each of

the ten Cations in a mixture of these Cations. You will need this Flow Chart in order to

complete the second part of the laboratory exercise. You will not be allowed to start the

second part of the laboratory without this Flow Chart.

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Procedure

During Week 1 you will test the precipitation and confirmation reactions for each

Cation individually. Thus, you will be starting with 9 different samples; each

sample containing a single Cation. The procedural steps below are written for this

style of testing. You will first identify the Cations in a given Group, and then move

on to separating and confirming their presence. Once you identify each Group of

Cations, it is important to run the confirmatory tests on all the Cations in that

Group. This is necessary because you need to show the test confirms the presence

of the target Cation and is negative for other Cations. As you move through these

procedural steps, you should begin to build a flow chart for how the Cations can

be separated from a mixture.

During Week 2 you will follow your flow chart for the separation and identification

of Cations in a mixture. You will do this for two mixtures; a mixture of known

Cation composition and a mixture of unknown composition. It is important that

you realize some of the procedural steps may need to be modified because you will

have only a single sample containing the various Cations and not, as is the case

during Week 1’s analysis, many samples containing a single Cation.

General Precautions

1. Pb2+

, Bi3+

, Cu2+

, Mn2+

, Ni2+

and Ba2+

salts are toxic. Wash your hands after their use.

2. Ag+ will stain your skin.

3. CrO42-

is toxic and will burn your skin.

4. Thioacetamide is toxic and produces toxic H2S gas.

5. HNO3, H2SO4 and HCl are acids and will burn your skin.

Week 1

Confirmation for the Presence of Na+

Because Na+ salts are generally soluble, forming a precipitate of this Cation is difficult.

Additionally, Na+ selective confirmatory reagents are also difficult to come by. Therefore we

will confirm the presence of Na+ using a Flame Test. Sodium (Na) will impart a bright yellow

color to a flame. Since almost all solutions have traces of Na+ present, you must decide if the

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yellow Flame color is due to the presence of Na+ in the original solution, before contaminating

reagents are added, or due to contamination imparted during the Qual Scheme. This will be done

on the basis of the intensity of the color.

1. Using a clean Nicrome Wire loop, perform a Flame Test on each of the original Cation

solutions. Also, for comparison, run a flame test on distilled Water and 0.2M NaCl.

Precipitation of Group 1 Cations

1. Measure out 10 drops of each of the known Cation solutions. Add 4 drops 6M HCl, stir

thoroughly, and then centrifuge. Test for completeness of precipitation by adding 1 drop

6M HCl. If the supernatant is cloudy, stir the solution, add another 2 drops of 6M HCl and

repeat the centrifugation and completeness of precipitation steps. Continue this process

until the supernatant remains clear. The Group 1 Cations will form a Chloride precipitate.

Xn+

(aq) + n Cl-(aq) XCln(s) (Eq. 20)

2. If the Cation did not produce a precipitate, set it aside for the Precipitation of Group 2

Cations analysis.

3. If a precipitate did form, discard the supernatant.

4. Wash each solid by adding 5 drops of Cold Water and stirring. Centrifuge and discard the

supernatant.

5. Add 15 drops of Water to each of the solids and place the test tubes into a hot-water bath.

Stir using a stir rod for ~ 1 minute. Quickly Centrifuge the hot solution, pour the

supernatant into a clean test tube. Repeat this procedure two more times. Retain those

solids that do not dissolve.

6. Confirmation for the Presence of Pb2+

: Add 3 drops of 1M K2CrO4 to the supernatant

containing Pb2+

. PbCrO4, a yellow precipitate, should form; confirming the presence of

Pb2+

.

Pb2+

(aq) + CrO42-

(aq) PbCrO4(s) (Eq. 21)

7. Confirmation for the Presence of Ag+: To the AgCl precipitate from Step 5 that did not

dissolve in Hot Water, add 6 drops of 6M NH3. Centrifuge and decant each supernatant

into a clean test tube. Add 20 drops of 6M HNO3 to the decantate. Stir the solution and

test its acidity with litmus. Continue to add HNO3 until the solution is acidic. A white

cloudiness confirms the presence of Ag+.

AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl

-(aq) (Eq. 22)

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Ag(NH3)2+(aq) + 2 H

+(aq) + Cl

-(aq) AgCl(s) + 2 NH4

+(aq)

(Eq. 23)

At this point you should have confirmed the presence of the Pb2+

and Ag+ ions.


1. In the fume hood, add 10 drops of 1M Thioacetamide (CH3CS(NH2)) to each of the

solutions from the Precipitation of Group 1 Cations that did not form a precipitate in Step 1

of that procedure. In the fume hood, heat each solution in a Hot Water bath for 10

minutes. This should allow the Thioacetamide to decompose into Hydrogen Sulfide (H2S)

and allow the Sulfide precipitates to form in an Acidic environment.

CH3CS(NH2) (aq) + 2 H2O H2S(aq) + NH4CH3CO2(aq) (Eq. 24)

H2S(aq) S2-

(aq) + 2 H+(aq) (Eq. 25)

2 Xn+

(aq) + n S2-

(aq) X2Sn(s) (Eq. 26)

If the Cation did produce a precipitate, Centrifuge the solution and decant the supernatant

into a clean test tube. Save the precipitate and keep track of which supernatant belongs to

which precipitate because if precipitation is not complete, the additional precipitate

collected from the supernatant will have to be combined with the initial precipitate. Test

the supernatant for completeness of precipitation by, again in the fume hood, adding 2

drops of Thioacetamide and allowing it to stand for 1 minute. If more precipitate forms,

add a few more drops of Thioacetamide and heat it in a Hot Water bath for 5 minutes.

Centrifuge and decant the supernatant. Combine this additional precipitate with that

initially formed for each appropriate Cation. Do this by adding 1 drop 0.2M NH4NO3

solution and 9 drops Water to the second precipitate, mixing and transferring to the first

precipitate. Mix thoroughly, centrifuge and discard the supernatant. Do this for each

Cation that produces an additional precipitate.

2. If the Cation did not produce a precipitate, keep it for the Precipitation of Group 3 Cations

analysis.

3. Now, in a fume hood, add 10 drops of 6M HNO3 to each precipitate and heat in a Hot

Water bath until the precipitate dissolves. (A light-colored residue of Sulfur may remain.)

If the precipitate has not dissolved in 5 minutes, gently heat over a low flame. Centrifuge

the solution to remove any Sulfur that forms. Dissolution of the precipitate is the result of

H+ combining with Sulfide (S

2-) to reform H2S.

X2Sn(s) + 2n H+(aq) 2 X

n+(aq) + n H2S(aq) (Eq. 27)

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5. Separation of Bi3+

and Confirmation for the Presence of Bi3+

and Cu2+

: In a fume hood,

add dropwise 15M NH3 until each solution is basic to litmus and add 3 more drops. A deep

blue color confirms the presence of Cu2+

.

Cu2+

(aq) + 4 NH3(aq) Cu(NH3)42+

(aq) (Eq. 28)

A white gelatinous precipitate indicates Bi3+

is present.

Bi3+

(aq) + 3 NH3(aq) + 3 H2O Bi(OH)3(s) + 3 NH4+(aq)

(Eq. 29)

This can be confirmed by centrifuging the gelatinous mixture and discarding the

supernatant. Wash the precipitate once with 10 drops of Hot Water. Discard the washings.

Add 6 drops of 6M NaOH and 4 drops of freshly prepared 0.2M SnCl2 to the precipitate

and stir. The formation of a jet-black precipitate confirms the presence of Bi3+

.

2 Bi(OH)3(aq) + 3 Sn(OH)42-

(aq) 2 Bi(s) + 3 Sn(OH)62-

(aq) (Eq. 30)

At this point you should have confirmed the presence of the Cu2+

and Bi3+

ions.


1. Add 2 drops 2M NH4NO3 to each of the solutions from the Precipitation of Group2

Cations that did not form a precipitate in Step 1 of that procedure. Stir. In a fume hood,

add dropwise 15M NH3 until each solution is basic to litmus and add 2 more drops. The

Ammonia produces a Basic (OH-) solution.

NH3(aq) + H2O OH-(aq) + NH4

+(aq) (Eq. 31)

This does two things. It will allow for the precipitation of Group 3 Cations as Hydroxides.

Xn+

(aq) + n OH-(aq) X(OH)n(aq) (Eq. 32)

It also consumes H+ ions, causing the H2S equilibrium, by Le Chatelier’s Principle, to shift

Right, dramatically increasing the concentration of S2-

ion.

H2S(aq) S2-

(aq) + 2 H+(aq) (Eq. 33)

This means Group 3 Cations that form slightly soluble Sulfides will now precipitate.

2 Xn+

(aq) + n S2-

(aq) X2Sn(s) (Eq. 34)

If a precipitate forms, test for completeness of precipitation by Centrifuging the sample

and, in a fume hood, adding an additional drop of 15M NH3.

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2. If the Cation did not produce a precipitate, keep it for the Analysis of Group 4 Cations.

3. In a fume hood, treat each precipitate with 3 drops 16M HNO3 and 9 drops 12M HCl.

Warm in a Hot Water bath until the precipitate dissolves. The Hydroxides dissolve because

of combination with the Acid.

X(OH)n(aq) + n H+(aq) X

n+(aq) + n H2O (Eq. 35)

The Sulfides dissolve as before.

X2Sn(s) + 2n H+(aq) 2 X

n+(aq) + n H2S(aq) (Eq. 36)

The Nitric Acid (HNO3) oxidizes the Hydrogen Sulfide (H2S) to keep the Sulfide

precipitate from reforming. Centrifuge the solution to remove any Sulfur that forms.

4. Re-precipitate the Cations of this Group as Hydroxides by adding 6M NaOH until the

solution is strongly Basic. If the precipitate is pasty, add 12 drops of Water. Centrifuge

and decant. To each precipitate add 20 drops of Water and 10 drops of 6M H2SO4. Stir

and heat in the Hot Water bath for 3 minutes or until the precipitate dissolves. Add 12

drops of Water and divide each solution into three parts.

5. Confirmation for the Presence of Fe3+

, Mn2+

and Ni2+

: To the first part for each, add 2

drops 0.2M KSCN. A brick red color indicates the presence of Fe3+

because Ferric

Thiocyanate has formed (Fe(SCN)2+

).

Fe3+

(aq) + SCN-(aq) Fe(SCN)

2+(aq) (Eq. 37)

To the second part, add a solution prepared from 4 drops 3M HNO3 and 4 drops Water.

Mix and add a few grains of Sodium Bismuthate (NaBiO3). Mix thoroughly and

Centrifuge. A pink or purple color is due to the formation of MnO4- and indicates the

presence of Mn2+

originally.

2 Mn2+

(aq) + 5 NaBiO3(s) + 14 H+(aq) 2 MnO4

-(aq) + 5 Bi

3+(aq) +

5 Na+(aq) + 7 H2O

(Eq. 38)

To the third part, add 6M NH3 until the solution is Basic to litmus. Add about 4 drops of

Dimethylglyoxime (DMG). Allow the solution to stand. The DMG complexes with Ni2+

to form a pink precipitate.

Ni2+

(aq) + 6 NH3(aq) Ni(NH3)62+

(aq) (Eq. 39)

Ni(NH3)62+

(aq) + 2 H2DMG(aq) Ni(HDMG)2(s) + 4 NH3(aq) +

2 NH4+(aq)

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(Eq. 40)

At this point you should have confirmed the presence of the Fe2+

, Mn2+

and Ni2+

ions.

Analysis of Group 4 Cations

1. Confirmation for the Presence of Ba2+

: To each of the remaining solutions, add 8 drops of

6M Acetic Acid (CH3CO2H) and 1 drop of 1M K2CrO4. A yellow precipitate indicates the

presence of Ba2+

.

Ba2+

(aq) + CrO42-

(aq) BaCrO4(s) (Eq. 41)

2. Confirmation for the Presence of Na+: This was done as a Flame Test on the original

sample.

At this point you should have confirmed the presence of the Ba2+

and Na+ ions.

At this point you should have a complete Flow Chart for the separation and

confirmation of the presence of any of our 9 Cations.

Week 2

During this week’s lab you will follow your flow chart for the separation and identification of

Cations in a mixture. You will do this for two mixtures; a mixture of “known” Cation

composition and a mixture of unknown composition. It is important that you realize some of the

procedural steps may need to be modified because you will have only a single sample containing

the various Cations and not, as is the case during Week 1’s analysis, many samples containing a

single Cation.

1. Obtain a “known” mixture of cations. Perform your analysis on this sample. Identify the

cations in this mixture. Confirm your results with your laboratory instructor. Re-do any

tests that are not consistent with the “known” results.

2. Obtain an “unknown” mixture of cations. Perform your analysis on this sample.

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Data Analysis

1. Identify the cations in the “unknown” mixture.

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Post Lab Questions

Complete all the questions in the “Titrator Precipitation” appendix.

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Appendix - Titrator Precipitation

In this exercise we will model some of the equilibrium problems encountered during the

execution of the Qual Analysis lab. These problems will revolve around the precipitation

reactions used to classify the various cations into Groups.

We will start with the most basic question, will a given pair of ions form a precipitate at the

concentrations encountered during the implementation of the Qual Scheme? For instance, if we

have a solution that is 0.1M Ag+, will AgCl precipitate if Chloride ion is added? And, if a

precipitate does form, what level of Ag+ will remain in solution?

Ag+(aq) + Cl

-(aq) AgCl(s)

AgCl(s) Ag+(aq) + Cl

-(aq)

This is the type of equilibrium problem we have leveraged in developing our Qual Scheme.

Specifically, we used Cl- precipitation to separate the Group I cations from the others. Basically,

we want certain cations to precipitate in the presence of a given anion, such as Chloride Ion, and

we want the other cations to remain in solution.

Let us start by asking if Ag+ and Cl

- are mixed, will a precipitate form? To answer this question,

we need to calculate the reaction quotient Q for the system. If:

Q > Ksp

then a precipitate will form. Otherwise, it will not.

For Silver Chloride:

Ksp = 1.82 x 10-10

So, for a system that is prepared such that:

[Ag+] = 0.1 M

[Cl-] = 1 x 10

-9 M

we have:

Q = [Ag+] [Cl

-] = (0.1) (1 x 10

-9) = 1 x 10

-10

Since Q < Ksp, no precipitate will form.

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So, our first exercise involves what happens when conc. HCl is added

dropwise to a solution containing Ag+. Will the Ag

+ precipitate, and if so, how

much HCl must be added to completely precipitate the Ag+ as AgCl?

How do we enter this information into Titrator so as to obtain this same answer? Well, first start

by identifying the Components for the system. Recall, these will form a Basis Set from which all

the other Species that make-up the system can be formed. For our system of three Species

(AgCl, Ag+ and Cl

-) we will require two Components. Because we are adding Ag

+ and Cl

- to the

system, it makes sense to include these as Components. They can be entered as:

Component Type Total Conc Guess Log Free M Charge

Ag+ Total 0.1 +1

Cl- Total 1 x 10

-9 -1

Note we are using a “Total” constraint on the concentration as we have Formally added this

amount of each. Although the Ag+ and Cl

- may change forms when precipitating as AgCl, their

Formal concentrations will not change.

The AgCl itself will be formed from these basis components and so we will enter it as a Species:

Species Dissolved/Ppt logK H S

AgCl Precipitate 9.74*

Coefficients: Ag+ +1

Cl- +1

*Equilibrium written as:

Ag+ + Cl

-AgCl logK = - pKsp = 9.74

P a g e | 18

This Species is designated as a potential “Precipitate”. This does not mean that it will

precipitate; only that it has the potential to precipitate. Also, because Titrator requires that we

write all equilibria as Formation reactions for the Species in question, we have reversed the

normal solubility equilibrium.

Finally, the system is solved. Titrator does this rather easily in 2 iterations. The “Free”

concentration of each species is:

[Ag+] = 1.000 x 10

-1M pAg = -1.000

[Cl-] = 1.000 x 10

-9M pCl = -9.000

[AgCl] = 0 M pAgCl = ~

This is as expected for a case where no precipitate forms; the original Ag+ and I

- concentrations

will remain unchanged and the AgCl “concentration” is zero. (Note: As a solid precipitates,

Titrator reports it as having a non-zero “concentration”. This is a formal scheme for indicating

how much precipitate has formed. Keep in mind, the “concentration” of a pure solid cannot

actually change.)

Questions:

1. Enter this system into Titrator. Confirm the above results. Now, let’s Titrate the system

with Chloride Ion. This will mimic the addition of 6M HCl to the system. Click on

“Titrate”. Specify the Titrant composition as:

Titrant Composition

Primary: Cl-

Concentration: 6 M

Set Experimental Volumes

Initial Sol’n Vol: 5 mL

Titrant Vol/Addition: 0.005 mL

Number of Points: 100

Now click “Titrate” and display the AgCl concentration (Molarity) as the 2nd

Species.

What is the volume titrant needed to drop the Ag+ concentration to 0.01M? (You will

have to consult the tabulated results to determine this volume.)

We want to now know if our system contains both Ag+ and Pb

2+, can we

precipitate all the Ag+ before the Pb

2+ starts to precipitate? (The Ksp for

PbCl2 is much larger than that for AgCl and so the AgCl should begin to

precipitate before the PbCl2.) Further, if this is indeed the case, is this an

P a g e | 19

effective means of separating Ag+ from Pb

2+? Keep in mind both Ag

+ and

Pb2+

are both Group I cations.

2. Now add a second cation to the system; Pb2+

. It’s solubility equilibrium is:

PbCl2(s) Pb2+

(aq) + 2 Cl-(aq)

with a Ksp = 1.7 x 10-5

. Titrate the system with Cl- as before.

What volume titrant must be added in order to precipitate the PbCl2? Use the information

provided by Titrator to determine QAgCl and QPbCl2 at this volume. Determine the same

values at the previous volume increment.

3. Has all the Ag+ effectively precipitated out of solution before the PbCl2 begins to

precipitate? In other words, could these precipitation reactions be used in a Qual Scheme

for Ag+ and Pb

2+? Explain. In practical terms, will Chloride precipitation provide an

effective means of separating Ag+ and Pb

2+ in solution?

Next, consider the precipitation reactions involved in separating the Group II and Group III

cations. H2S is added to the mixture and the pH is increased by adding Ammonia. Group II

cations precipitate at low pH’s, whereas Group III cations require a high pH before they

precipitate. Further, the Group III cations could precipitate as either the Hydroxide or the

Sulfide salts.

Let us start simply with just the Hydrogen Sulfide equilibrium in Water. This involves the

following two reactions:

P a g e | 20

H2S(aq) H+(aq) + HS

-(aq)

Ka1 = 10-7

H2S(aq) 2 H+(aq) + S

2-(aq)

K = Ka1 x Ka2 = 10-7

x 10-13

= 10-20

Additionally, we need to couple in the autoionization of Water:

H2O H+(aq) + OH

-(aq)

Kw = 10-14

We will take the H2S concentration as fixed at 0.1M as this component is continuously

regenerated by the Thioacetamide that is also present. (Note: We are ignoring the fact that the

H2S in solution will also equilibrate with gaseous H2S, as we noted was present by the rotten egg

smell in the laboratory.)

A Titrator definition for this system is:

Component Type Total M Guess Log Free M Charge

H2S Free 0.1 0 0

H2O Free 55.5 0 0

H+ Total 0 -7 +1

Species Dissolved/Ppt logK H S

OH- Dissolved -14

Coefficients: H+ -1

H2O +1 HS

- Dissolved -7

Coefficients: H+ -1

H2S +1

S2-

Dissolved -20 Coefficients: H

+ -2

H2S +1

Our question now is, given a system containing Cu2+

and Mn2+

cations and

acidifed Sulfide, will a precipitate form?

P a g e | 21

Questions:

4. Enter this system into Titrator and determine the concentrations of [S2-

] and [OH-].

Suppose the system contains the cations Cu2+

(Group II) and Mn2+

(Group III), each at

0.1M concentration. Given the following Solubility Products:

Solid Ksp

CuS 8 x 10-37

Cu(OH)2 4.8 x 10-20

MnS 3 x 10-11

Mn(OH)2 2 x 10-13

determine the reaction quotient Q for each possible precipitation reaction. Will any of the

solids precipitate? (You do not need to enter the Cu and Mn species into Titrator. Simply

use the results from Titrator for the H2S system and calculate the Q’s by hand.)

In order to raise the pH and increase [S2-

], add a little Ammonia to the system:

NH3(aq) + H2O NH4+(aq) + OH

-(aq)

Kb = 1.75 x 10-5

In order to add this to our Titrator definition, we need to re-write it as (why?):

NH3(aq) + H+(aq) NH4

+(aq)

K = (1.75 x 10-5

) (10+14

) = 1.75 x 10+9

Now we intend to make the system basic by adding Ammonia. Will anything

precipitate? Keep in mind this is the operation you performed in

precipitating Group III cations.

P a g e | 22

Questions:

5. Add this to the above system. Start with a “Total” Ammonia concentration of 10-20

. Solve

the system and then Titrate it with Ammonia:

Titrant Composition

Primary: NH3

Concentration: 1 M

Set Experimental Volumes

Initial Sol’n Vol: 5 mL

Titrant Vol/Addition: 0.001 mL

Number of Points: 100

What volume NH3 must be added before MnS precipitates out? Will the Hydroxide also

precipitate at this point? Explain using a quantitative arguement. (Determine by hand

calculation what S2-

concentration is required to precipitate the Mn2+

. Then check the

tabulated Titrator results to see what volume NH3 is required to produce this

concentration.)

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Qual Analysis