Question

You prepare an equilibrium mixture by combining 4.00 mL 0.00129 M Fe3+ with 0.100 mL 0.00582 M SCN-. What is [Fe3+]o for this mixture? Enter the numerical value to 3 significant figures, without units. (You may enter the value using E-format scientific notation, e.g., 1.23×10-4 would be entered as 1.23E-4 -- no spaces, capital E.)

Answer 1

Ans. Step 1: Original solution:

Given, Initial [Fe³⁺] = 0.00129 M                - in its original solution

                        Initial [SCN^-] = 0.00582 M               - in its original solution

Moles of Fe³⁺ in its original solution = Molarity x Volume of solution in liters

                                                = 0.00129 M x 0.004 L                     ; [1 mL = 10^-3 L]

                                                = 5.160 x 10^-6 mol

Moles of SCN^- in its original solution = Molarity x Volume of solution in liters

                                                = 0.00582 M x 0.0001 L                                                                                                                   = 5.820 x 10^-7 mol

Step 2: The mixture

# Final volume of reaction mixture = Vol of original Fe³⁺ soln. + Vol. of original SCN^- soln.

                                                            = 4.000 mL + 0.100 mL

                                                            = 4.100 mL

                                                            = 0.0041

Now,

Initial [Fe³⁺] in mixture = Initial moles of Fe³⁺ / Vol. of reaction mixture in liter

                                    = 5.160 x 10^-6 mol / 0.0041 L

                                    = 1.259 x 10^-3 mol/ L

Hence, initial [Fe³⁺] in reaction mixture = 1.259 x 10^-3 mol/ L

Step 3: The reaction:

Balanced reaction: Fe³⁺(aq) + SCN^-(aq) -------> Fe(SCN)²⁺

Stoichiometry: 1 mol Fe³⁺ reacts with 1 mol SCN^-.

Note that the moles of SCN^- are less than that of Fe³⁺. So, SCN^- is the limiting reactant.

Assuming complete reaction (because equilibrium constant is not mentioned), moles of Fe³⁺ consumed in reaction is equal to the moles of SCN^- (the limiting reactant) present in mixture.

Thus,

            Moles of Fe³⁺ consumed = Moles of SCN^- present in reaction mixture

= 5.820 x 10^-7 mol

Remaining moles of Fe³⁺ = Initial moles of Fe³⁺ - moles of Fe³⁺ consumed in reaction

                                                = 5.16 x 10^-6 mol - 5.820 x 10^-7 mol

                                                = 4.578 x 10^-6 mol

Step 4: [Fe³⁺] at equilibrium

Remaining moles of Fe³⁺ at equilibrium = 4.578 x 10^-6 mol

Final volume of reaction mixture = Vol of original Fe³⁺ soln. + Vol. of original SCN^- soln.

                                                            = 4.000 mL + 0.100 mL

                                                            = 4.100 mL

                                                            = 0.0041 L

Now,

[Fe³⁺] at equilibrium = Moles of Fe³⁺ at equilibrium / Vol. of reaction mixture in liter

                                    = 4.578 x 10^-6 mol / 0.0041 L

                                    = 1.117 x 10^-3 mol/ L                                   ; [1 M = 1 mol/ L]

                                    = 1.117 x 10^-3 M

Hence, [Fe³⁺] at equilibrium = 1.117 x 10^-3 M