Question

3a. Suppose that 5.0 mL of a 0.0030M solution of Fe3+ is mixed with 10.0 mL of a 0.0060 M solution of SCN-. If the final equilibrium Fe(SCN)2+ concentration is measured to be 0.00524M, what are the final equilibrium concentrations for the other two species?

3b. Using the results from question 3a above, calculate the equilibrium constant K for the formation of Fe(SCN)2+ from SCN- and Fe3+.

Answer 1

3a) Fe³⁺ + SCN- FeSCN²⁺

Equilibrium concentration ofFe(SCN)2+=[Fe(SCN)2+]eq=0.00524M

ICE table

	[Fe3+]	[SCN-]	[FeSCN2+]
initial concentration	0.005L*0.003 mol/L=1.5*10^-5mol in molarity, 1.5*10^-5mol/total volume =1.5*10^-5mol/15ml =1.5*10^-5mol/0.015L=0.001 M	0.010L*0.006mol/L=6.0*10^-5mol molarity=.6.0*10^-5mol/0.015L=4.0*10^-3=0.004M	0
change	-x	-x	+x
equilibrium concentration	0.001M-x	0.004M-x	x

But final equilibrium Fe(SCN)2+ concentration is measured to be 0.000524M,=x

equilibrium concentration of Fe3+=0.001-0.000524=0.000476M

equilibrium concentration of SCN-=0.004-0.000524=0.00348M

3b) K=[FeSCN2+]eq/[Fe3+]eq *[SCN-]eq=0.000524M/(0.000476M)(0.00348M)=316.333M^-1

So,