In a dilute nitric acid solution, Fe^3+ reacts with thiocyanate ion (SCN^-) to f
ID: 967896 • Letter: I
Question
In a dilute nitric acid solution, Fe^3+ reacts with thiocyanate ion (SCN^-) to form a dark-red complex: [Fe(H_2O)_6]^3+ + SCN^- H_2O + [Fe(H_2O)_5NCS]^2+ The equilibrium concentration of [Fe(H_2O)_5NCS]^2+ may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, 1.00 mL of 0.200 M Fe(NO_3)_3 was mixed with 1.00 mL of 1.00 times 10^-3 M KSCN and 8.00 mL of dilute HNO_3. The color of the solution quantitatively indicated that the [Fe(H_2O)_5NCS]^2+ concentration was 7.30 times 10^-5 M. Calculate the formation constant for [Fe(H_2O)_5NCS]^2+.Explanation / Answer
Calculate initial concentration of Fe3+ and SCN-.
nFe(NO3)3 = 0.200 M . (1.00 ml/1000 ml)L
nFe(NO3)3 = 2x10-4 mol
nFe(NO3)3 = nFe3+ = nFe(H2O)63+
[Fe(H2O)5NCS]2+ = 7.30x10-5 M = (2x10-4 mol)/((1.00 ml + 1.00 ml + 8.00 ml)/1000 ml))L*
[Fe(H2O)63+] = 0.02 M
*final volume after mixing
nKSCN = 1x10-3 M . (1.00 ml/1000 ml)L
nKSCN = 1x10-6 mol
nKSCN = nSCN-
[SCN-] = (1x10-6 mol)/((1.00 ml + 1.00 ml + 8.00 ml)/1000 ml))L*
[SCN-] = 1x10-4 M
*final volume after mixing
After mixing and reaching equilibrium:
[Fe(H2O)5NCS]2+ = 7.30x10-5 M
[Fe(H2O)63+] = 0.02 M – 7.30x10-5 M = 0.019927 M
[SCN-] = 1x10-4 M – 7.30x10-5 M = 2.7x10-5 M
Equilibrium equation:
Kf = [Fe(H2O)5NCS]2+/([Fe(H2O)63+].[SCN-])
Kf = 135.68