Final Concentration

Question

If 3.2 mL of 0.200 M Fe³⁺ solution is mixed with 2 mL of 0.00200M SCN^- solution, and HNO₃ is used to bring the final volume to 10 mL. What is the final concentration in M of the Fe³⁺  ion in the 10 mL test tube?

Answer 1

Fe³⁺ + SCN^- ==> Fe(SCN)²⁺ ... balanced equation

mols Fe³⁺ = 3.2 ml x 1 L / 1000 ml x 0.200 mol / L = 6.40x10^-4 moles Fe³⁺

mols SCN^- = 2 ml x 1 L / 1000 ml x 0.00200 mol / L = 4.00x10^-5 moles SCN^-

Limiting reactant = SCN^- since mol ratio in balanced equation is 1:1

Use mols SCN^- to determine final concentration of Fe³⁺

mols Fe³⁺ used up = 4.00x10^-5 mol SCN^- x 1 mol Fe³⁺ / mol SCN^- = 4.00x10-5 mol Fe³⁺ used

mols Fe³⁺ remaining = 6.40x10^-4 mol - 4.00x10^-5 mol = 6.00x10^-4 mol Fe³⁺ remaining

Final [Fe³⁺] = 6.00x10^-4 mol / 10 ml x 1 L / 1000 ml = 0.0600 mols / L = 0.0600 M Fe³⁺