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...

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.

Homework Answers

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

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