Instructions The reaction of Fe 3+ (aq) + SCN - (aq) = [FeSCN] 2+ (aq) is a reve
ID: 1026975 • Letter: I
Question
Instructions
The reaction of Fe3+ (aq) + SCN- (aq) = [FeSCN]2+ (aq) is a reversible reaction. The colors of the iron (from Fe(NO3)3) and KSCN are colorless (or have a slightly yellow tinge), and the product, [FeSCN]2+, is reddish brown. You also need to know that Fe3+ (aq) and HPO42- (aq) combine to form a colorless [FeHPO4]+ (aq) complex and that Fe3+ (aq) and OH- (aq) combine to form a very insoluble Fe(OH)3.
Prepare the main solution) Add 20 mL of distilled water from a graduated cylinder to a 100-mL beaker. Next add 20 drops of 0.1 M Fe(NO3)3 and 20 drops of 0.1 M KSCN to the beaker. The reddish color is due to the formation of the [Fe(SCN)]2+ complex ion.
Test tube 1) Add 20 drops of 0.1 M FeCl3 to the 3mL of the main solution (Color: Darkened reddish brown)
Test tube 2) Add 20 drops of 0.1 M KSCN to the 3mL of the main solution (Color: Reddish brown)
Test tube 3) Add 5-10 small crystals of Na2HPO4 to the 3mL of the main solution. (Color: clear)
Test tube 4) Add drops of 4 M NaOH until a color change is observed to the 3mL of the main solution. (Color: Yellowish brown)
Problem:
Indicate how the level of each reactant and product ion was affected (use for increased, for decreased, NC for no change). For each of the test tubes, also indicate the direction of shift (left , right , or no change (NC)) of the ea for the reaction. Below the table, using Le Chatelier's principle and the experimental evidence, explain your answers. Direction of Kn Shift Change in Ion Concentration Test Tube Fe3+ SCN Fe(SCN) 4Explanation / Answer
The equilibrium reaction is given as
Fe3+ (aq) + SCN- (aq) --------> [Fe(SCN)]2+ (aq)
The equilibrium constant for the reaction is given as
Keq = [Fe(SCN)2+]/[Fe3+][SCN-] ……(1)
We evaluate the effects of the changes as below.
Test Tube
Change in ion concentration
Direction of Keq shift
Explanation
Fe3+
SCN-
[Fe(SCN)]2+
1
decreases
decreases
increases
Right or product side
FeCl3 ionizes completely to produce Fe3+. Therefore, [Fe3+] increases. However, Keq, being an equilibrium constant, must remain constant at a particular temperature. To keep Keq constant, more [Fe(SCN)]2+ is produced, thereby, increasing the reddish brown color of the solution.
2
decreases
decreases
increases
Right or product side
KSCN ionizes completely to produce SCN-. Therefore, [SCN-] increases. However, Keq, being an equilibrium constant, must remain constant at a particular temperature. To keep Keq constant, more [Fe(SCN)]2+ is produced, thereby, increasing the reddish brown color of the solution.
3
increases
increases
decreases
Left or reactant side
NaH2PO4 ionizes completely to produce H2PO4- which combines with Fe3+ to produce colorless [FeHPO4]+. This reduces [Fe3+] at equilibrium. To keep Keq constant, the equilibrium shifts to the left, producing more Fe3+ which further combines with HPO42- to give a colorless solution.
4
increases
increases
decreases
Left or reactant side
NaOH ionizes completely to produce OH- which combines with Fe3+ to produce insoluble brownish Fe(OH)3. This reduces [Fe3+] at equilibrium. To keep Keq constant, the equilibrium shifts to the left, producing more Fe3+ which further combines with OH- to give brown Fe(OH)3.
Test Tube
Change in ion concentration
Direction of Keq shift
Explanation
Fe3+
SCN-
[Fe(SCN)]2+
1
decreases
decreases
increases
Right or product side
FeCl3 ionizes completely to produce Fe3+. Therefore, [Fe3+] increases. However, Keq, being an equilibrium constant, must remain constant at a particular temperature. To keep Keq constant, more [Fe(SCN)]2+ is produced, thereby, increasing the reddish brown color of the solution.
2
decreases
decreases
increases
Right or product side
KSCN ionizes completely to produce SCN-. Therefore, [SCN-] increases. However, Keq, being an equilibrium constant, must remain constant at a particular temperature. To keep Keq constant, more [Fe(SCN)]2+ is produced, thereby, increasing the reddish brown color of the solution.
3
increases
increases
decreases
Left or reactant side
NaH2PO4 ionizes completely to produce H2PO4- which combines with Fe3+ to produce colorless [FeHPO4]+. This reduces [Fe3+] at equilibrium. To keep Keq constant, the equilibrium shifts to the left, producing more Fe3+ which further combines with HPO42- to give a colorless solution.
4
increases
increases
decreases
Left or reactant side
NaOH ionizes completely to produce OH- which combines with Fe3+ to produce insoluble brownish Fe(OH)3. This reduces [Fe3+] at equilibrium. To keep Keq constant, the equilibrium shifts to the left, producing more Fe3+ which further combines with OH- to give brown Fe(OH)3.