In a dilute nitric acid solution, Fe3+ reacts with thiocyanate ion (SCN)^- to fo
ID: 927316 • Letter: I
Question
In a dilute nitric acid solution, Fe3+ reacts with thiocyanate ion (SCN)^- to form a dark-red complex:
[Fe(H2O)6] ^3+ + SCN^- = H2O + [Fe(H2O)5 NCS] ^2+
The equilibrium concentration of [Fe(H2O) NCS] 2+ may be determined by how darkly colred the solution is (measured by a spectrometer). In one such experiment 1.0 mL of 0.20 M Fe(NO3)3 was mixed with 1.0 mL of 3.6x10^-3 KSCN and 8.0mL of dilute HNO3. The color of the solution quantitively indicated that the [Fe(H2O)5 NCS] ^2+ concentration was 7.3x10^-5 M. Calculate the formation constant for [Fe(H2O)5 NCS]^2+.
Explanation / Answer
from teh balanced reaction
K = [Fe(H2O)5NCS]^2+ / [ Fe3+] [SCN-]
we have [Fe(NO3)3] = 0.2M = [Fe3+] = 0.2 M initially with vol 1 ml
final volume is 1+ 1+ 8 = 10 ml
thus [Fe3+] final = M1V1/V2 = 0.2 x 1 /10 = 0.02 M
[SCN-]initial = [KSCN] = 0.0036 M
[SCN-] final = ( 0.0036 x 1 /10) = 0.00036 M
[Fe(H2O)5NCS]^2+ = 7.3 x 10^ -5 M
Now K = ( 7.3 x10^ -5) / ( 0.00036) ( 0.02) = 10