Heterogeneous Heterogeneous EquilibriaEquilibria

ObjectivesObjectivesUnderstand and explain the

conditions necessary for precipitation.

Enumerate and explain the conditions for dissolution of precipitates.

Understand and explain the principle of fractional precipitation.

IntroductionIntroductionChemical equilibriums that involve

more than one phase are said to be heterogeneous. A common example of this is the precipitation reaction, which is a formation of an insoluble ionic solid during a chemical reaction between two solutions. Such reaction occurs when certain pairs of oppositely charged ions attract each other strongly that they form precipitate.

With the use of centrifugation for the experimental results and the comparison of the solubility product constant, Ksp, and reaction quotient, Qsp, for the expected results, the concept of heterogeneous equilibrium will be easily understood.

Centrifugation is a process of increasing the effective gravitational force on a test tube so as to more rapidly and completely cause the precipitate to gather on the bottom of the test tube.

ExperimentalExperimental

Label test tubes A to D.

A DCB

Part A. Precipitation

5 drops 0.1 M CaCl2 + 5 drops 1 M NH4OH +

15 drops distilled H2O

5 drops 0.1 M CaCl2 + 5 drops

1 M NaOH + 15 drops

distilled H2O

5 drops 0.1 M CaCl2 + 5 drops 3 M NaOAc +

15 drops distilled H2O

5 drops 0.1 M CaCl2 + 5 drops 3 M NH4OH + 15 drops 3 M

NH4Cl

Centrifuge

Mix reagents. Centrifuge and observe.

Calculate the Ion Product of Ca(OH)2 in each mixture

Part B. Dissolution of Precipitates

10 drops 0.1 M Pb(NO3)2

Dropwise 0.1 M HCl

10 drops 0.1 M AgNO3

Dropwise 0.1 M HCl

1. Place 10 drops each of Pb(NO3)2 and AgNO3 in two separate test tubes. To each one, add HCl dropwise. Shake the test tube after each drop. Continue adding acid until last drop does not result in precipitation.

Centrifuge and observe. Decant supernatant liquid and was precipitate with 5 drops of water. Discard the wash water.

CentrifugeDECANT

SUPERNATANT LIQUID

Wash with 5 drops of

water

DISCARD WASH WATER

2 mL of water

2 mL of water

HEAT BOTH IN A WATERBATH

In each of test tubes, add water and shake the tubes. Heat each one in a water bath. Observe.

Cool to room temperature. Observe and compare to previous observation.

Place 10 drops each of Pb(NO3)2 and AgNO3 ions in two separate test tubes. To each one, add HCl dropwise. Shake after each drop. Continue adding acid until last drop does not result in precipitation.

10 drops 0.1 M Pb(NO3)2

Add 0.1 M HCl dropwise

10 drops AgNO3

Add 0.1 M HCl dropwise

Centrifuge

Wash with 5 drops of water.

Discard supernate.

Discard wash water

Centrifuge and observe. Discard supernatant liquid and was precipitate with 5 drops of water. Discard the wash water.

Into each test tubes, add 2 mL of concentrated NH4OH gradually. Observe.

2 mL concentrated

NH4OH

2 mL concentrated

NH4OH

2. Place Pb(NO3) and Ba(NO3)2 in two separate test tubes. Add K3CrO4. Add HNO3.

10 drops 0.1 M Pb(NO3)2

3 drops of 0.1 M K2CrO4

10 drops 0.1 M Ba(NO3)2

3 drops 1 M HNO3

3 drops of 0.1 M K2CrO4

3 drops 1 M HNO3

Neutralize by adding NaOH or until precipitation occurs. Do not add excess NaOH. Centrifuge and decant supernatant liquid. To each of the precipitates, add 15 drops of 6 M NaOH.

3 drops 6 M NaOH

CentrifugeDECANT AND ADD NaOH

TO PRECIPITATE15 drops of 6 M NaOH

3. Place Ba(NO3)2 in each of two separate test tubes. Add Na2CO3 to the first until precipitation is complete. To the second, add Na2SO3. Centrifuge. Observe result.

10 drops 0.1 M Ba(NO3)2

10 drops 0.1 M Ba(NO3)2

0.1 M Na2CO3

0.1 M Na2SO4

Centrifuge

Centrifuge

Discard supernate. Add concentrated HCl to precipitate.

4. Place Cu(NO3)2 in each of two separate test tubes. Add HOAc dropwise until acidic to litmus, then add thocetamide solution dropwise until precipitation is complete. Centrifuge then decant supernatant liquid.

10 drops 0.1 M Cu(NO3)2

10 drops 0.1 M Cu(NO3)2

3 M HOAc dropwise

3 M HOAc dropwise

Thiocetamide solution

dropwise

Thiocetamide solution

dropwise

Centrifuge

Centrifuge

Add 2 mL of distilled water to each test tube. Stir, centrifuge and discard supernate.

Add 5 drops of HCl to precipitate and HNO3 to the other. Heat both mixtures in water bath. Observe.

5 drops 6 M HCl to

precipitate

5 drops 6 M HNO3

HEAT BOTH IN A WATERBATH

Mix K2CrO4, Na2S, KI and H2O in a test tube. Then add Pb(NO3)2 dropwise.

Part C. Fractional Precipitation

5 drops 0.1 M K2CrO4

5 drops 0.1 M Na2S

5 drops 0.1 M KI

20 drops H2O

Then add Pb(NO3)2 dropwise. Centrifuge after the addition of each drop. Record result. Continue adding until no precipitate is formed.

Pb(NO3)2 dropwise

Centrifuge

ResultsResults

Test Tube Mixtures of:

A No Precipitation.

B White precipitate formed at the bottom, with few suspended and stuck on the walls of the test tube.

C White precipitate formed, more than B.

D No precipitation

Part A. PrecipitatesObserved Data:

Given that the Ksp for Ca(OH)2 is 6.5 x 10-6. We calculate for the Ion Product Constant and prove the existence of precipitate.

Ca2+ + OH- → Ca(OH)2

M1V1 = M2V2(0.1M)(5 drops) = M2(25 drops)

M2 = 0.02 M Ca2+

Test Tube ATest Tube A

NH4OH → NH4+ + OH-

NH4+ + OH- ↔ NH3 + H2O

 M1V1=M2V2(1 M NH4OH)(5 drops) =

M2(25 drops)M2 = 0.2 M NH4OH To get OH-, which is

represented as X,Kb for NH3= [NH4

+][OH-] / [NH3]

1.76 x 10-5 = (x)(x) / 0.2 M[OH-] = 1.88 x 10-3 M 

 Ion Product

Constant = [Ca2+][OH-]

Ca2+ + 2OH- → Ca(OH-)2

 (0.02 M) (1.88 x

10-3 M)2 = 7.07 x 10-8

 IF Ksp > Qsp, then

no precipitation will occur. Correct observation.

Test tube BTest tube BNa+ + OH- → NaOH M1V1 = M2V2(1 M NaOH)(5 drops) = M2(25 drops)M2 = 0.2 M NaOH Ion Product Constant = [Ca2+][OH-]Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(0.2 M)2 = 8 x 10-4

 If Ksp < Qsp, then precipitation will occur.

Correct observation.

Test tube CTest tube CNaOAc → Na+ + OAc M1V1 = M2V2(3 M NaOAc)(5

drops)=M2(25 drops)M2 = 0.6 M NaOAc OAc- + H20 ↔ HOAc + OH-

 Since Kw = KaKb ; where kb =

1.0 x 10-14 / 1.8 x 10-5

5.56 x 10-10 = (x)(x) / 0.06[OH-] = 1.83 x 10-5

Ion Product Constant = [Ca2+][OH-]Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(1.83 x 10-5)2 = 6.69 x 10-12

 If Ksp > Qsp, then precipitation will occur. Erroneous observation.

Test tube DTest tube DNH4OH → NH4

+ + OH-

NH3 + H2O ↔ NH4+ + OH-

 M1V1 = M2V2(3 M)(5 drops) = M2(25 drops)M2 = 0.6 M NH4OH NH4Cl → NH4

+ + Cl-M1V1 = M2V2(4 M)(15 drops) = M2(25

drops)M2 = 2.4 M NH4

+

 Kb of NH3 = (x)(2.4+x) / (0.6)1.8 x 10-5 = 2.4x + x2 / 0.6[OH-] = 4.05 x 10-13

Ion Product Constant = [Ca2+][OH-]Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(4.05 x 10-13) = 8.1 x 10-15

 If Ksp > Qsp, then precipitation will not occur. Correct observation.

In the first experiment where 0.1 M HCl was added dropwise to 10 drops of 0.1 M Pb(NO3)2

and 10 drops of 0.1 M AgNO3, it was observed that no precipitate formed in Pb(NO3)2 and it remained clear, while precipitate settled at the bottom of the test tube in the AgNO3 solution.  

When the solutions were added with water and heated in a water bath, the precipitate solidified in the AgNO3 solution and no precipitate still was seen in the Pb(NO3)2 solution.  

When it was cooled to room temperature, it stayed the same and no change happened to the solutions.  

Part B. Dissolution of Precipitates

The reverse happened when NH4OH was added. Pb(NO3) yielded precipitate while the precipitate of AgNO3 was not evident as before.

The reactions involved for the first part were:

Pb(NO3)2 (aq) + HCl (aq) ↔ PbCl2 (s) + HNO3 (aq) AgNO3 (aq) + HCl (aq) ↔ AgCl (s) + HNO3 (aq) The reactions involved when NH4OH was

added were:PbCl2 (s) + 2NH3 (aq) + 2H2O (l) ↔ Pb(OH-)2 (s)

+ 2NH4Cl (aq)AgCl2 (s) + 2NH3 (aq) + 2H2O↔ [Ag(NH3)2]+ (aq)

+ 2Cl- (aq) + 2H2O

In Part B, which concerns the dissolution of precipitates, we have to consider the solubility product constant of the precipitates in each solution.

For Pb(NO3)2 added with HCl, the equation is:

Pb(NO3)2 (aq) + HCl (aq) ↔ PbCl2 (s) + HNO3 (aq)

With PbCl2 = Ksp = 1.6 x 10-5

For Ag(NO3)2 added with HCl, the equation is:

Ag(NO3)2 (aq) + HCl (aq) ↔ AgCl (s) + HNO3 (aq)

With AgCl = Ksp = 1.6 x 10-10

Since, PbCl2 has a higher ksp, it is then more soluble, making the non-evidence of precipitate as compared to AgCl justifiable.

When heated, the precipitate in Ag(NO3)2 looked more solid. However, if we incorporate the Le Chatelier’s Principle, the subjection of heat to the reaction should have created an exothermic process that shifted backward to compensate for the heat.

The precipitates should have lessened in amount and should have dissolved. When cooled, it should have done the opposite.

When added with NH4OH, PbCl2 from the latter set-up would yield precipitate due to the presence of Pb(OH)2 (Ksp = 1.2 x 10-15).

PbCl2 (s) + 2NH3 (aq) + 2H2O ↔ Pb(OH)2 (s) + 2NH4Cl (aq)

AgCl from the latter set-up; however, would yield [Ag(NH3)2]+, an ion which is readily soluble with a Ksp of 1.7 x 107. Because of this, no precipitate would form and the liquid will turn clear.

In the second experiment, when 3 drops of 0.1 M K2CrO4 and 3 drops of 1 M HNO3 was added to 10 drops of 0.1 M Pb(NO3)2 and Ba(NO3)2, it was observed that Pb(NO3)2 was cloudy yellow. The liquid turned yellow with a yellow orange precipitate. Supposedly, it would have no precipitate. Ba(NO3)2 was initially yellow and clear. It remained the same with no precipitate.

 The reactions involved were: Pb(NO3)2 (aq) + K2CrO4 (aq) ↔ PbCrO4 (s) + 2KNO3 (aq)

Ba(NO3)2 (aq) + K2CrO4 (aq) ↔ BaCrO4 (s) + 2KNO3 (aq)

PbCrO4 (s) + HNO3 (aq) ↔ Pb(NO3)2 (aq) + H2CrO4 (aq)

BaCrO4 (s) + HNO3 (aq) ↔ Ba(NO3)2 (aq) + H2CrO4 (aq)

When 6 M NaOH was added, Pb(NO3)2 yielded an orange powdery precipitate while Ba(NO3)2 yielded a white precipitate.

The reactions involved were:Pb(NO3)2 (aq) + 2NaOH (aq) ↔ Pb(OH)2 (s)

+ 2NaNO3 (aq)Ba(NO3)2 (aq) + NaOH (aq) ↔ Ba(OH)2 (s)

+ NaNO3 (aq)

Reagents mixed Observation

Ba(NO3)2 + Na2CO3 white precipitate formed

Ba(NO3)2 + Na2SO4 white precipitate formed more rapidly

The reactions involved were:Ba(NO3)2(aq) + Na2CO3(aq) BaCO3(s) + 2NaNO3 (aq)

Ba(NO3)2(aq) + Na2SO4(aq)  2NaNO3(aq) + BaSO4(s)

In the 3rd set of Part B, BaCO3 and BaSO4

are the insoluble salts of the reactions or the precipitates.

Addition of HCl

Observation

BaCO3 + HCl completely dissolved

BaSO4 + HCl white precipitate is still evident

After adding HCl, BaCO3

dissolved and BaSO4 stayed undissolved.

BaCO3 dissolved because the H+ ion of HCl combined with CO3

2- forming carbonic acid. This causes the equilibrium to shift to the right to replace the supply of CO3

2- that was consumed. On the other hand, BaSO4 did not

dissolve, even though H+ from HCl reacted with SO4

2- to formed HSO4-. This

happened because HSO4- ion will

dissociate since it does not exist as a molecular compound. There would be no shift in equilibrium because the SO4

2- ion is not consumed, making BaSO4 undissolved.

In the 4th set, Cu(NO3)2 was added to thiocetamide to form the precipitate, CuS.

Reagents mixed Observation

CuS+ HCl precipitate dissolved

CuS+ HNO3 precipitate is still evident

Cu(NO3)2 (aq) + CH3CSNH2(aq) <-> CuS(s)

Observations when HCl and HNO3 were added to the mixture.

CuS(s) <-> Cu2+(aq) + S2-(aq)

The experimental results were different from the expected results.

CuS should have dissolved in HNO3 and did not dissolved when HCl was added. This was expected to happen because of the capability of HNO3 to remove S2- ion to elementary sulfur by oxidation favors the forward reaction, so CuS dissolves.

The reaction between CuS and HNO3 can be written as follows:3CuS(s) + 8H++ 2NO3

-(aq)  3Cu 2+(aq) + 4H2O(l)+ 2NO(g) + 3S(s)

ObservationsThe solution is yellow in color. When

Pb(NO3)2 was added, orange and yellow precipitates were formed.

Reactions involved:Pb(NO3)2(aq) + Na2S(aq) PbS(s) +

2NaNO3(aq)Pb(NO3)2(aq) + K2CrO4(aq)

PbCrO4(s) + 2KNO3(aq)Pb(NO3)2(aq) + 2KI(aq) PbI2 (s) +

2KNO3(aq)

Part C. Fractional Precipitation

In part C, precipitates PbS (black), PbCrO4 (yellow), PbI2 (yellow-orange) are expected to form. Using the solubility constants as basis, PbS (4 x 10-26)is expected to precipitate out first since it has the lowest Ksp, second to precipitate out should be PbCrO4 (3 x 10-13) and last is PbI2 is (7.9 x 10-7) which has the highest solubility constant. The experimental results were slightly different with the expected one.

ConclusionConclusionHeterogeneous equilibria involve species in more than

one phase. In precipitation reactions, an insoluble solid called precipitate forms and settles out of the solution.

The comparison of solubility product constant, Ksp, to the reaction quotient, Q, can be used to determine whether a precipitate will form with a given concentration of ions.  If Q < Ksp, forward process is favored. No precipitate occurs.   If Q = Ksp, neither backward nor forward process is favored. Solid and solution are in equilibrium.  If Q > Ksp, reverse process is favored. Precipitation occurs to form more solid.

The factors affecting dissolution of precipitates are temperature and addition of complexing agents. Higher temperature and addition of oxidizing and complexing agent increases the solubility of salts.

In fractional precipitation, the difference in Ksp values of the various reagents is used to identify the order in which salts will precipitate.

Fractional precipitation is a technique in which two or more ions in solution, each capable of being precipitated by the same reagent. It is important that there is a significant difference in the Ksp values of the substances being separated to identify the order in which salts will precipitate.

Ortiz, Anna Katherina P.Ortiz, Anna Katherina P.Paragas, Eula Chrizzelle A. Paragas, Eula Chrizzelle A.

The End.The End.

Thank You!=DThank You!=D