Question

standard solution of FeSCN2+FeSCN2+ is prepared by combining 9.09.0 mL of 0.200.20 M Fe(NO3)3Fe(NO3)3 with 1.01.0 mL of 0.00200.0020 M KSCN.KSCN. The equation for the reaction is as follows. Fe(NO3)3+KSCN↽−−⇀FeSCN2++KNO3+2NO−3 Fe(NO3)3+KSCN↽−−⇀FeSCN2++KNO3+2NO3− What allows us to assume that the reaction goes essentially to completion? The reaction quotient ????Q is greater than Kc.Kc. The concentration of Fe(NO3)3Fe(NO3)3 is much higher than the concentration of KSCN.KSCN. The excess Fe3+Fe3+ prevents the formation of the neutral Fe(SCN)3.Fe(SCN)3. The equlibrium reaction has a very high Kc.Kc. Under the conditions given, Le Châtelier's principle dictates that the reaction shifts to the left. Based on that assumption, what is the equilibrium concentration of FeSCN2+?

Answer 1

The equilibrium concentration of FeSCN²⁺ is 0.0002 M

Step-by-step explanation:

Given data:

9 mL of 0.2 M of Fe(NO)₃

1 mL of 0.002 M of KSCN

Question: What is the equilibrium concentration of FeSCN²⁺

Calculate the initial millimoles of Fe(NO₃)₃:

`n_(Fe(NO3)3) =0.2(mmoles)/(mL) *9mL=1.8mmol`

Initial millimoles of KSCN:

`n_(KSCN) =0.002(mmoles)/(mL) *1=0.002mmoles`

Total volume of solution:

Vtotal = 1 + 9 = 10 mL

Initial concentration of Fe(NO₃)₃:

`[Fe(NO3)3]=(1.8)/(10) =0.18M`

Initial concentration of KSCN:

`[KSCN]=(0.002)/(10) =0.0002M`

The reaction:

SCN⁻ + Fe³⁺ → FeSCN²⁺

I 0.0002 0.18 0

C -0.0002 -0.0002 0.0002

E 0 0.1798 0.0002

As you see the equilibrium concentration of FeSCN²⁺ is 0.0002 M