In preparing solution number I (see experimental procedure), a student measures
ID: 488719 • Letter: I
Question
In preparing solution number I (see experimental procedure), a student measures 5.0 ml. of 2.00 times 10^3 M Fe(NO_3)3. 2.0 ml. of 2.00 times 10^-3 M HSCN. and 3.0 mL of 0.50M HNO_3. (The solvent for the Fe(NO_3)_3 and HSCN solutions is 0.50 M HNO_3). After combining these, what are the concentrations of Fe^3+, HSCN. and H^+ in the mulling solution? (These are the initial concentrations for prelab question #3). Show all work. When at equilibrium, the solution prepared in question #1 has an absorbance of 0.36. What is the Fe(SCN) concentration, assuming that the absorptivity of Fe(SCN)^2+ is 5150 L mol^-1 cm^-1 for 447-nm light? Use the answers from question #1 as your initial concentrations and your answer from question #2 as the equilibrium concentration of Fe(SCN)^2+ (x). and begin filling in the equilibrium table (ICE table). Then determine the equilibrium concentrations of Fe^3+, HSCN. and H^+. Finally, determine the equilibrium constant for the system.Explanation / Answer
(1)
Moles of Fe(NO3)3 = 0.005*0.002 = 10-5 moles
Moles of HSCN = 0.002*0.002 = 4*10-6
Moles of HNO3 = 0.003*0.5 = 0.0015 moles
Final volume = 10 mL = 0.01 L
[Fe3+] = 10-5/0.01 = 10-3 M
[HSCN] = 4*10-6/0.01 = 4*10-4 M
[H+] = 0.0015/0.01 = 15*10-6 M
(2)
Using relation:
A = e*l*c
Here, l = path length = 1 cm
Putting values:
0.36 = 5150*1*c
So,
c = 6.9*10-5 M
(3)
Reaction: Fe3+ + HSCN ---> Fe(SCN)2+ + H+
Initial 10-3 4*10-4 0 15*10-6
Eqb 10-3 - x 4*10-4 - x x 15*10-6 + x
Putting values in Keq, we get:
Keq = [ (x)*(15*10-6 + x) ] / [ (10-3 - x)*(4*10-4 - x) ]
Given: x = 6.9*10-5
Thus,
Keq = 188*10-4