Question

A student mixes 5.00 mL of 2.00 times 10^-3 M Fe(NO_3)_3 with 5.00 mL of 2.00 times 10^-3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2^+ is 1.40 times 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?

Answer 1

initially:

Vtotal = V1+V2 = 5+5 = 10

mmol of Fe+3 = MV = 5*(2*10^-3) = 0.01

mmol of SCN- = MV = 5*2*10^-3 = 0.01

mmol of FeSCN+2 = 0

the change

mmol of Fe+3 = 0.01 - x

mmol of SCN- =.01 - x

mmol of FeSCN+2 = x

in equilbirium

mmol of Fe+3 = 5*(2*10^-3) = 0.01 - x

mmol of SCN- = 5*2*10^-3 = 0.01 - x

mmol of FeSCN+2 = 0 + x

and we know

mmol of FeSCN+2 = MV = (1.4*10^-4)(10) = 0.0014

so

x = 0.0014

mmol of Fe+3 = 5*(2*10^-3) = 0.01 - 0.0014 = 0.0086

mmol of SCN- = 5*2*10^-3 = 0.01 - 0.0014 = 0.0086

then

a)

initial concentration

[Fe+3] = mmol/V = 0.01 / 10 = 0.001 M

[SCN-] = mmol/V = 0.01 / 10 = 0.001 M

b)

equilbirium constant

Keq = [FESCN+2]/([Fe+3][SCN-])

Keq = (1.4*10^-4)((0.0086/10) * (0.0086/19))

Keq = 1.035*10^-10