Fe3+ (aq) + SCN-(aq) ⇔FeSCN2+(aq)
One- one ratio between SCN- & FeSCN2+
Moles of FeSCN2+
n=MV
= (0.002) (0.0020)
n= 0.000004 mol SCN-= 0.000004 mol FeSCN2+
Molarity of FeSCN2+
n=MV
0.000004= (M) (0.05)
M=8.0 x 10-5 M
Equation for Graph: y=3256.1x + 0.0049 (y=Absorption, x= [FeSCN2+])
Part Two: Prepare and Test Equilibrium Systems
Table Two
Sample Calculation (Tube 2): Calculating Kc Value
Calculation of [ FeSCN2+] In Equilibrium
y=3256.1x + 0.0049
0.111= 3256.1x + 0.0049
0.1061=3256.1x
x = 3.26 x 10-5
[FeSCN2+] = 3.26 x 10-5
Calculation of [Fe3+] & [SCN-] In Equilibrium
Fe3+ + SCN-⇔FeSCN2+
Kc = [FeSCN2+]
[Fe3+] [SCN-]
[ ] Fe3+ + SCN- → FeSCN2+
Calculation of Kc Value
Kc = [FeSCN2+] = 3.26 x 10-5 .
[Fe3+] [SCN-] [1.97 x 10-3][1.97 x 10-3 ]
Kc= 8.400
Average Kc Value = 15.99
Part Three: Prepare and Analyze Stock Fe Solution from Iron Tablets
Table Three
Sample Calculation (Tube 2): Calculating Mass of Iron In Tablet
y=3256.1x + 0.0049
0.993=3256.1x +0.0049
0.9881=3256.1x
x= 3.03 x 10-4
[FeSCN2+] = 3.03 x 10-4
Kc = [FeSCN2+] Kc= 15.99, [SCN-]= 0.0020 M
[Fe3+] [SCN-]
15.99 = 3.03 x 10-4
[Fe3+][2.0 x 10-3]
[Fe3+]= 9.47 x 10-3
[ ] Fe3+ + SCN- → FeSCN2+
I x 2.0 x 10-3 0
C - 3.03 x 10-4 - 3.03 x 10-4 + 3.03 x 10-4
E =9.47 x 10-3 = 1.70 x 10-3 = 3.03 x 10-3
x= 9.77 x 10-3 M= Concentration of Fe3+ in stock solution
n= MV 2.93 x 10-5 mol Fe3+ x 55.7g = 1.63 x 10-3 g Fe3+
= (9.77 x 10-3) (0.003) 1 mol Fe3+
n=2.93 x 10-5 mol Fe3+
Average Mass of Fe3+ = 2.32 x 10-3 g Fe3+ in Iron Tablets
