Equilibrium Shift-Kemtec Educational
Purpose:
To observe shifts of chemical equilibriums in a variety of systems.
Part 1: Iron(III) Thiocyanate Complex Equilibrium System
Write a balanced equation for the iron (III) thiocyanate complex system:
FeSCN2+ (aq)
iron(III) thiocyanate
red
-
 Fe3+ (aq)
+ SCN (aq)
iron ion
thiocyanate ion
clear
Procedure:
X 1) Place 1 mL of 0.05 M FeCl3 into a 50 mL beaker. Using a graduated cylinder,
measure a 1 mL portion of 0.05 M KSCN. Record the colour of each solution.
colour of FeCl3: yellow-brown
colour of KSCN: clear
X 2) Add the measured portion of 0.05 M KSCN solution to the beaker containing the 0.05
M FeCl3. Swirl the mixture and record the colour. Add enough distilled water
(approximately 20 mL to the solution to dilute the intense colour to a light amber colour.
3) Half-fill seven 10 X 75 mm test tubes labelled A to G with this solution. Test
tube A serves as a control.
4) For each of the following reactions (steps 5-8) record the results.
5) To test tube B, add 2-3 drops of 0.2 M KCl
6) To test tube C, add 2-3 drops of 0.2 M Fe(NO3)3
7) To test tube D, add 2-3 drops of 0.2 M NH4SCN too weak ??
8) To test tube E, add 2-3 drops of 6.0 M NaOH
9) To test tube F, add 2-3 drops of 0.2 M FeSO4
10) To test tube G, add 2-3 drops of 1.0 M KSCN
When the concentration of the iron (III) thiocyanate ion increases, so does its
deep red colour.
FeSCN2+
 Fe3+ + SCNred
clear
Equilibrium Involving Iron (III) Thiocyanate Ion
1
Reagent
Added
KCl
Fe(NO3)3
NH4SCN
NaOH
FeSO4
Stress
(Ion Added)
----Fe3+
-
SCN
Observation
-
NR (No rxn)
Control tt
Red
----
Red
<--
Brown ppt
of Fe(OH)3
Red
-->
K+, Cl
NO3
NH4+
Na+
-
OH
Fe2+
Spectator Ion
 Fe3+
SO4-2
K+
Direction of
Equilibrium Shift
<--
<--
Red
<-SCN
2) What is the purpose of the control? (Test tube A) (Hint: spectator ions)
To compare the colour change with a solution that has not received any stress.
KSCN
**T**3) Apply Le Châtelier’s principle to explain the results obtained when
6.0 M NaOH was introduced into the iron (III) thiocyanate ion equilibrium system.
There are 2 rxns taking place:
FeSCN2+  Fe3+
red
+
NaOH --> OH
-
+ SCN
clear
+
Na+
Fe(OH)3
brown ppt
As the Fe3+ is removed as a precipitate, the eqm shifts towards the products side
of the iron (III) thiocyanate eqm.
Results: clear soln with a brown ppt
Part 2: Formation of the Chromate/Dichromate Equilibrium System
Write a balanced equation for the chromate/dichromate equilibrium system:
Cr2O72- (aq) + 2OH (aq)  2CrO42-(aq) + H2O (l)
dichromate
chromate
orange
yellow
(spectator ions have been removed; potassium ion; K+)
Procedure:
1) Place 3 drops of potassium chromate solution into two depressions of
a spot plate.
2
2) Place 3 drops of potassium dichromate solution into two depressions
spot plate.
The colour of CrO42- (aq) is:
The colour of Cr2O72- (aq) is:
of a
yellow chromate
orange dichromate
3) Effect of increasing the concentration of H+(aq)
Add dilute HCl solution dropwise to both potassium chromate and potassium dichromate
solution respectively. When a reaction occurs, it appears after the addition of 4-5 drops.
Results:
Cr2O7-2 + 2OH-  2CrO4-2 + H2O
orange
yellow
Results:
Equilibrium Involving Potassium Chromate(K2CrO4):
Reagent
Stress
Spectator Ions
Observation
Added
HCl
H+
yellow to
K+, Clorange
Equilibrium Involving Potassium Dichromate(K2Cr2O7):
Reagent
Stress
Spectator Ions
Observation
Added
HCl
H+
stays same
K+, Cl-
Direction of
Equilibrium Shift
<--
Direction of
Equilibrium Shift
---
4) Effect of increasing the concentration of OH-(aq)
Add 1M NaOH solution dropwise to both potassium chromate and potassium
dichromate solution respectively. When a reaction occurs,
it appears after the addition of 4-5 drops.
Cr2O7-2 + 2OH-  2CrO4-2 + H2O
orange
yellow
Results:
Equilibrium Involving Potassium Chromate(K2CrO4):
Reagent
Stress
Spectator Ions
Observation
Added
NaOH
OHStays yellow
Na+, K+
Direction of
Equilibrium Shift
----
Equilibrium Involving Potassium Dichromate(K2Cr2O7):
3
Reagent
Added
NaOH
Stress
Spectator Ions
Observation
OH-
Na+, K+
Orange to
Yellow
Direction of
Equilibrium Shift
--->
*T* 5) Write a balanced equation which describes the equilibrium
Cr2O7-2(aq) + 2OH-(aq)  2CrO4 -2(aq) + H2O(l)
orange
yellow
*T*6) Explain the reasons for the equilibrium shifts observed with adding HCl and
NaOH to the chromate and dichromate solutions.
with HCl:
Cr2O7-2(aq) + 2OH-(aq)  2CrO4 -2(aq) + H2O(l)
orange
+
yellow
HCl (aq)  H+(aq) + Cl (aq)

H2O(l)
with NaOH:
[OH-] increases shifting the Cr2O72-/CrO42- eqm to product side and the solution
becomes yellow.
NaOH

-2
Cr2O7 (aq) + 2OH-(aq)  2CrO4 -2(aq) + H2O(l)
orange
+
yellow
Na+
Part 3: Equilibrium Between Solid Barium Chromate and a Saturated Solution of
Its Ions
Write a balanced equation for both the barium chromate and for barium
dichromate equilibrium systems:
BaCrO4(s)  CrO42-(aq) + Ba2+(aq)
baby
yellow ppt
BaCr2O7(aq) --> Cr2O72-(aq) + Ba2+(aq)
orange soln
Procedure:
1a) Add barium nitrate solution dropwise to the cavity that contains potassium
chromate solution. Record the result.
b) Add barium nitrate solution dropwise to the cavity that contains
potassium dichromate solution. Record the result.
Results:
Equilibrium Involving Potasium Chromate (K2CrO4):
Reagent
Stress
Spectator Ions
Observation
Direction of
Added
Equilibrium Shift
+
Ba(NO3)2 Ba2+
baby
yellow
<---K , NO3
4
ppt
Equilibrium Involving Potassium Dichromate (K2Cr2O7):
Reagent
Stress
Spectator Ions
Observation
Added
Ba(NO3)2 Ba2+
soluble orange
K+ , NO3soln
Direction of
Equilibrium Shift
----
2) Compare the solubilities of barium chromate and barium dichromate
and their appearance
BaCr2O7: orange soln(soluble,)
BaCrO4: baby yellow ppt(insoluble)
ppt: solid that deposits at the bottom of a test tube. It looks cloudy or murky.
3) Explain why no precipitate formed in Cr2O72-, while some did with CrO42-?
Keq (BaCrO4): low value (favours reactants, yellow solid-ppt)
BaCr2O7 :100% dissolved, 100% products, orange solution of ions
4a) Add dilute hydrochloric acid dropwise to the cavity that contains
barium chromate solution. Record the result.
b) Add dilute hydrochloric acid dropwise to the cavity that contains
barium dichromate solution. Record the result.
Equilibrium Involving Barium Chromate(BaCrO4):
Reagent
Stress
Spectator Ion
Observation
Added
HCl
H+
ppt dissolves,
Clorange soln
Direction of
Equilibrium Shift
--->
Equilibrium Involving Barium Dichromate(BaCr2O7):
Reagent
Stress
Spectator Ion
Observation
Added
HCl
H+
stays same:
Clorange soln
Direction of
Equilibrium Shift
----
5) How could the changes you observe be reversed? Try it, and record what you
did, and the result.
*Add NaOH*
For CrO4 2-: yellow ppt of BaCrO4
For Cr2O72-: no ppt, orange soln
6) Examine the equations in part 2 and in Part 3. Explain your results.
BaCrO4(s)  CrO42-(aq) + Ba2+-(aq) Keq= 1.2 X 10-10 (favours reactants)
BaCr2O7(aq)  Cr2O72-(aq)- + Ba2+(aq) 100% products
Keq = [products]
5
[reactants]
HCl (aq)

22+
H2O + Cr2O7  2CrO4 (aq) +2H (aq)
orange yellow
+
Cl (aq)
+
[H ] increases, shift towards reactant
Part 4: Equilibrium Involving Cobalt (II) Complexes
Write a balanced equation between the hydrated cobalt(III) ion and the hydrated
cobalt (II) chloride complex.
exo
2H2O(l) + Co(H2O)4Cl2(s)  Co(H2O)62+(aq) + 2Cl-(aq)
purple
endo pink
Results:
Colour of Co(H2O)4Cl2 solution: purple
Colour of Co(H2O)62+ solution:
pink (hydrated cobalt complex)
Since chloride is part of a solid on the reactant side and an aqueous ion on the
product side, it is not a spectator. (Cl- is in a different phase)
Step
Reagent
Stress
Observation
Direction of
Equilibrium
Shift
2
HCl
purple

Cl
3
4
5
6
H2O
H2O
Warmed
Cooled
H2O
H2O
T (Heat)
T (Cold)
pink
Light pink
Purple/blue
Pink




Questions:
1) If the hydrated cobalt (II) ion complex were refrigerated, what would you
predict as the colour of the refrigerated solution?
Its colour would be pink. Favours Co(H2O)62+
exo side favoured, system tries to warm up.
2) Look at the equilibrium equation for the reaction involving the hydrated cobalt
(II) ion complex. From your lab results, which reaction is endothermic? Cite
evidence for your answer.
6
T decreases: exo favoured T increases: endo favoured
exo
2H2O(l) + Co(H2O)4Cl2(s)  Co(H2O)62+(aq) + 2Cl-(aq)
purple
endo
pink
Forward rxn: exo since a decrease in T favours pink
Reverse rxn: endo since an increase in T favours purple colour.
3) Predict how the addition of sodium chloride, NaCl, would affect the hydrated
cobalt (II) ion equilibrium? Explain your prediction in terms of Le Châtelier’s
principle.
[Cl ] increases, eqm shifts to reactants side.Soln becomes purple
Follow-Up Questions
1. A student discovers that after doing this experiment, several of the
depression cavities are coated with a yellow precipitate that does not
wash off easily.
Devise a chemical method for removing this material.
Answer:Use an strong acid (HCl, HNO3, H2SO4)to dissolve the ppt.
Most ppts dissolve with strong acids.
Conclusions
1. State the effect on the position of the equilibrium on a change in
concentration of a reactant or of a product:
[reactant] increases
Eqm shifts towards products
[reactant] decreases
Eqm shifts towards reactants
[product] increases
Eqm shifts towards reactants
[product] decreases
Eqm shifts towards products
2. Complete the following table:
Temperature
Forward reaction
Reactants
Products
Increased
Exothermic
towards
Decreased
Exothermic
towards
Increased
Endothermic
towards
Decreased
Endothermic
towards
7