What is the Formation Constant of Iron(III) Thiocyanate?

  • Thread starter Thread starter phizuks
  • Start date Start date
Click For Summary
SUMMARY

The formation constants for the reaction of Iron(III) Thiocyanate (Fe3+ + HSCN <--> H+ + FeSCN2+) are established as follows: log K1 = 3.02, log K2 = 4.64, log K3 = 5.0, log K4 = 6.3, log K5 = 6.2, and log K6 = 6.1. These constants represent the overall equilibrium expressions for the formation of the complex ions, with K5 specifically defined as K5 = [Fe(SCN)5(2-)]/([Fe(3+)][SCN-]^5). The reaction is conducted in a 0.5M nitric acid solution to inhibit the formation of Fe(OH)3, ensuring that SCN- ions remain absent due to the high concentration of H+. The discussion highlights the commonality of this lab experiment among freshman chemistry students.

PREREQUISITES
  • Understanding of chemical equilibrium and formation constants
  • Familiarity with complex ion formation and stoichiometry
  • Knowledge of acid-base chemistry, particularly weak acids
  • Experience with laboratory techniques in analytical chemistry
NEXT STEPS
  • Research the implications of pH on complex ion stability in solutions
  • Study the principles of equilibrium constants in coordination chemistry
  • Explore the role of nitric acid in preventing precipitation reactions
  • Learn about the dissociation of weak acids and their impact on equilibrium
USEFUL FOR

Chemistry students, particularly those in introductory courses, laboratory instructors, and anyone involved in analytical chemistry or studying complex ion equilibria.

phizuks
Messages
2
Reaction score
0
Hello,

Does anyone know the formation constant for the reaction
Fe3+ + HSCN <--> H+ + FeSCN2+

I've looked through my textbook with no luck, and I have been googling for a while. I'm writing a lab writeup right now, and I need it to compare my values to the accepted values.

Thanks.
 
Physics news on Phys.org
Six constants, as there are six steps of complex formation:

log K1 = 3.02
log K2 = 4.64
log K3 = 5.0
log K4 = 6.3
log K5 = 6.2
log K6 = 6.1

(these are overall constants, like:

K5 = [Fe(SCN)5(2-)]/([Fe(3+)][SCN-]^5)

and so on)
 
H

I guess this lab is pretty common for freshman chem. I had the report for this due last week. My lab manual gave the expression [H+][FeSCN2+]/[Fe3+][HSCN] but my professor gave the other one(without H).

The reaction was carried out in a .5M nitric acid solution to prevent this reaction:Fe3+(aq) + 3OH-(aq) --> Fe(OH)3. You don't have SCN- ions in solution b/c of the large amount of H+ and the small amount of HSCN. (HSCN <-->H+ + SCN-, shifted to the left by addition of H+) You have Fe3+ in solution, not Fe(NO3)3 b/c that compound dissociates to a large degree but HSCN is a weak acid.

I think my thoughts here make sense. I guess I'll find out when I get my grade.

PS Nice to see someone from Poland, Borek.:smile: Both of my parents are from Poland and I can speak Polish fluently.
 
On one website, it said that the H+ makes no difference in the equilibrium..but that doesn't make much sense. Someone please explain. Thanks.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
6K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
25K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
5K
  • · Replies 9 ·
Replies
9
Views
13K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K