Question

A student mixes 5.00 mL of 2.00 x 10‐3 M Fe(NO3)3 with 5.00 mL 2.00 x 10‐3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN+2 is 1.40 x 10‐4 M.

a.   What is the initial concentration in solution of the Fe+3 and SCN‐ ?

b.   What is the equilibrium constant for the reaction?

c. What happened to the K+ and the NO3 ‐ ions in this solution?

Answer 1

Let's write the overall reaction:

Fe³⁺ + SCN^- <--------> FeSCN²⁺

The total volume is 10 mL. Now, let's calculate the innitial concentrations of Fe³⁺ and SCN^-

[Fe³⁺] = 2x10^-3 M * (5/10) = 1x10^-3 or simply 0.001 M

[SCN^-] = 2x10^-3 M (5/10) = 1x10^-3 = 0.001 M

For the equilibrium concentration Kc, do an ICE chart:

r: Fe³⁺ + SCN^- <--------> FeSCN²⁺

i: 0.001 0.001 0

e. 0.001-x 0.001-x x

Kc = [FeSCN²⁺] / [Fe³⁺] [SCN^-]

Kc = x / (0.001-x) (0.001-x)

The problem tell us the concentration of equilibrium of FeSCN, which is 1.4x10^-4 M, and this is the value of "x" too so:

Kc = 1.4x10^-4 / (0.001-1.4x10^-4) (0.001-1.4x10^-4)

Kc = 189.29

As for where do NO₃^- and K⁺ go, they become spectator ions in the solution; ie., one associated with KNO₃ and the other two with FeSCN(NO₃)_2.

Hope this helps