In a dilate nitric acid solution, Fe^3+ reacts with thiocyanate ion(SCN^-) to fo
ID: 539425 • Letter: I
Question
In a dilate 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 dark the colored solution is (measured by a spectrometer). In one such experiment, 1.0 mL of 0.20 M Fe(NO_3)_3 was mixed with 1.0 mL of 4.9 times 10^-3 M KSCN and 8.0 mL of dilute HNO_3. The color of the solution quantitatively indicated that the [Fe(H_2O)_6NCS]^2+ concentration was 7.3 times 10^-5 M. Calculate the formation constant for [Fe(H_2O)_5NCS]^2+. K_f =Explanation / Answer
K = [Fe(H2O)5NCS]+2 / [SCN-][Fe(H2O)6]+3
initially
Vtotal = 1+1+8 = 10 mL
[Fe(H2O)6]+3 = M1*V1/VT = 1*0.2/10 = 0.02
[SCN-] = M2*V2/VT = 1*(4.9*10^-3)/10 = 4.9*10^-4
[Fe(H2O)5NCS]+2] = 0
after reaction
[Fe(H2O)6]+3 = 0.02 - x
[SCN-] = M2*V2/VT = 4.9*10^-4 - x
[Fe(H2O)5NCS]+2] = + x
and we kow that
[Fe(H2O)5NCS]+2] = + x = 7.3*10^-5
so
[Fe(H2O)6]+3 = 0.02 - 7.3*10^-5 = 0.019927
[SCN-] = M2*V2/VT = 4.9*10^-4 - 7.3*10^-5 = 0.000417
substitute in Kf
K = [Fe(H2O)5NCS]+2 / [SCN-][Fe(H2O)6]+3
K = (7.3*10^-5) /(0.019927*0.000417)
Kf= 8.7850