Question

A mixture is made by combining 12.9 mL of 5.4×10−3 M KSCN, 8.3 mL of 2.2×10−2 M Fe(NO3)3, and 7.7 mL of 0.34 M HNO3. Calculate the concentraion of Fe3+ in the solution.

Answer 1

By adding HNO₃, we reduce the concentration of OH^- ions and thus the formation of colored Fe³⁺ and OH^- complex ions.

Moles of KSCN added (SCN^-) = 12.9 * 5.4 x 10^-3 = 0.0697 mmoles

Moles of Fe(NO₃)₃ added (Fe³⁺)= 8.3 * 2.2 x 10^-2 = 0.183 mmoles

Fe³⁺ + SCN^- ---> [Fe(SCN)]²⁺

The concentration of Fe³⁺ ions is much greater than the concentration of SCN^- ions. This shifts the equilibrium to the right so far that ALL the SCN^- ions will react to become FeSCN²⁺ ions. In case the concentration of Fe³⁺ ions and SCN^- ions is similar, the final concentrations are a function of the equilibrium constant.
Here the concentration of the FeSCN²⁺ ions will be equal to the initial concentration of the SCN^- ions.

Moles of Fe³⁺ left in the solution = 0.183 - 0.0697 = 0.113 mmol

Concentration of Fe³⁺ in the solution = 0.113 / (12.9 + 8.3 + 7.7) = 3.9 x 10^-3 M