Question

Assume that at some temperature, Kc= 100 for the reaction: Fe3+ (aq) + SCN- (aq) ⇌ FeSCN2+. Calculate the equilibrium concentration of FeSCN2+ if 100 mL of 0.24 M Fe(NO3)3 are mixed with 100 mL of .24 M KSCN. Show your work. Specifically, how do I determine the initial amounts? Is it just .24 or .24/100mL? And do I need to convert to Liters or anything?

Answer 1

Given, the equilibrium reaction,

Fe³⁺(aq) + SCN^-(aq) FeSCN²⁺(aq)

Also given,

Concentration of Fe(NO₃)₃ Or Fe³⁺ solution= 0.24 M

Volume of Fe(NO₃)₃ Or Fe³⁺ solution = 100 mL x ( 1L/1000 mL) = 0.1 L

Concentration of KSCN Or SCN^- solution = 0.24 M

Volume of KSCN Or SCN^-​​​​​​​ solution = 100 mL x ( 1L/1000 mL) = 0.1 L

K_c = 100

Calculating the number of moles of Fe³⁺ and SCN^- ion solutions from the given concentrations and volumes,

= 0.24 M x 0.1 L = 0.024 mol of Fe³⁺

Similarly,

= 0.24 M x 0.1 L = 0.024 mol of SCN^-

When both the solutions are mixed,

Total volume = 0.1 L + 0.1 L = 0.2 L

New concentrations of Fe³⁺ and SCN^- ion solutions are,

= 0.024 mol of Fe³⁺ / 0.2 L = 0.12 M solution of Fe³⁺

= 0.024 mol of SCN^- / 0.2 L = 0.12 M solution of SCN^-

Now, Drawing an ICE chart,

	Fe3+(aq)	SCN-(aq)	FeSCN2+(aq)
I(M)	0.12	0.12	0
C(M)	-x	-x	+x
E(M)	0.12-x	0.12-x	x

Now, K_c expression is,

K_c = [FeSCN²⁺] / [Fe³⁺][SCN^-]

100 = [x] / [0.12-x][0.12-x]

100x² -25x + 1.44 = 0

Solving the quadratic equation,

x = 0.09

Thus, equilibrium concentration of FeSCN²⁺ = x = 0.09 M