Question

A mixture is made by combining 10.9 mL of 4.1×10−3 M KSCN, 6.1 mL of 4.5×10−2 M Fe(NO3)3, and 12.5 mL of 0.484 M HNO3. Calculate concentraion of Fe3+

Answer 1

The reaction here is

Fe³⁺(aq) + SCN^–(aq) FeSCN²⁺(aq)

[Fe³⁺]eq = [Fe³⁺]initial – [FeSCN²⁺]eq

[SCN^– ]eq = [SCN^– ]initial – [FeSCN²⁺]eq

10.9 mL of 4.1x10⁻³ M KSCN = (4.1x10⁻³ x 10.9)/1000 = 4.47 x 10^-5 moles

6.1 mL of 4.5x10−2 M Fe(NO₃)₃ =(4.5x10⁻² x 6.1)/1000 = 2.74 x 10^-4 moles

Since SCN is less than Fe the amount of Fe in solution will be the amount that does not react with SCN^-

which is 2.74 x 10^-4 moles - 4.47 x 10^-5 moles = 2.30 x 10^-4 moles

To get concentration we need to know the total volume which is 10.9 mL + 6.1 mL + 12.5 mL = 29.5 mL

2.30 x 10^-4 moles in  29.5 mL will be 7.79 x 10^-3 M