Question

A student mixes 5.00 mL 4.00 x 10^-3 M Fe(NO₃)₃ with 5.00 mL 4.00 x 10^-3 M KSCN. The student finds that the equilibrium concentration of FeSCN²⁺ is 5.6 x 10^-4M. 

a. Set up the Kc expression for the following equation. 

Fe³⁺ + SCN^-  ↔ FeSCN²⁺

b. What is the total volume of the solution? _______ mL 

c. Fill in the following table for the moles of each species.

d. Find the concentration for the above ions. 

e. Calculate Kc for the reaction.

Answer 1

(a) K_c = [FeSCN²⁺] / ([Fe³⁺] [SCN^-])

(b) Total Volume = 5mL + 5mL = 10 mL = 0.01L

(c) Moles = molarity * Volume

So, Initial Moles of Fe³⁺ = 0.005 L * 4.00 *10^-3 M = 2.0 *10^-5 moles

Initial moles of SCN^- = 0.005 L * 4.00 *10^-3 M = 2.0 *10^-5 Moles

So, table is

	Fe3+(aq)	SCN-(aq)	$\rightleftharpoons$	FeSCN2+
Initial (moles)	2.0 *10-5	2.0 *10-5		0
Change (moles)	-x	-x		+x
Equilibrium (moles)	2.0 *10-5 -x	2.0 *10-5 - x		x

d. At equilibrium, [FeSCN²⁺] = 5.6 *10^-4 M

Volume = 0.010 L

moles of FeSCN²⁺ = Molarity * volume = 5.6 *10^-6 mol = x

moles of Fe³⁺ = (2.0 *10^-5) - (5.6 *10^-6) = 1.44 *10^-5

Volume = 0.010 L

So, [Fe³⁺] = [SCN^-] = 1.44 *10^-5 moles / 0.010 L =1.44 * 10^-3

[Equil] (M)

1.44 * 10-3

1.44 * 10-3

5.6 *10-4

(e) K_c = (5.6 *10^-4) / (1.44 * 10^-3)² = 270.06