Question

Given:
Molar Concentration of Fe(NO3)3 is 0.2
Molar concentration of NaSCN is 0.001
Volume of Fe(NO3)3 is 10.00 ml (in 0.1 M HNO3)
Volume of NaSCN is 2.00 ml (in 0.1 M HNO3)
10.00 ml Fe(NO3)3 + 2.00ml NaSCN + 13.00 ml HNO3= 25 mL

Calculate:
a) Moles of SCN-
b) [SCN-] (25.0 ml)
c) [FeNCS2+]

Answer 1

You can't do this problem unless you know the equilibrium concentration of the complex, which you haven't given.

The initial moles used are

Moles of Fe^-

10 mL =0.01 L

0.01 L x 0.2 mol/L = 2 x 10^-3 mol

Moles of SCN^-

2 mL = 0.002 L

0.002 L X 0.001 mol/L = 2 x 10^-6 mol

They are in a volume of 25 mL, so initial concentrations are 8 x 10^-3 mol of (Fe) and 8 x 10^-6 mol of SCN^-

Suppose the equilibrium concentration of the complex is X mol dm-3 then since 1 mole of the complex is formed from 1 mole of each of the reactants the equilibrium concentration of each reactant [Fe] and [SCN].

Then Kc = [complex]/[Fe][SCN] where the concentrations are the equilibrium values.

If we assume no complex is formed, we can go by the following way.

Moles of SCN^-

2 mL = 0.002 L

0.002 L X 0.001 mol/L = 2 X 10^-6 mol

[SCN^-] (25 mL)

1X10^-6 mol / 0.025 L = 4 x 10^-5 M

[FeNCS²⁺]

In this solution, since there is a huge amount of iron compared to SCN^-, the concentration of FeNCS will equal the concentration of SCN-. So, the answer here is 4 x 10^-5 M.