Question

20.00 mL of 3.00E-3 M Fe(NO3)3 is mixed with 8.00 mL of 2.48E-3 M KSCN and 12.00 mL of water. The equalibrium molarity of Fe(SCN)2+ is found to be 6.82E-5 M. If the reaction proceeds as shown below, what is the equilibrium molarity of Fe3+ and SCN-?    Fe3+(aq) + 2 SCN-(aq) Fe(SCN)2+(aq)

I have the answers (they are correct according to my online homework), but need to know how to get them. Thank you! The correct answeers are: Fe = 1.43E-3 SCN = 3.60E-4

Answer 1

Answer – Given, [Fe(NO₃)₃] = 3.00*10^-3 M , volume = 20.0 mL , [KSCN] =2.48*10^-3 M , volume = 8.00 mL , water volume = 12.0 mL

At equilibrium [Fe(SCN)₂²⁺] = 6.85*10^-5 M

Moles of

We know, moles of Fe(NO₃)₃] = 3.00*10^-3 M * 0.020 L = 6.0*10^-4 moles

Moles of KSCN = 2.48*10^-3 M * 0.008 L = 1.98*10^-4 moles

At equilibrium, total volume = 20+8.0+12.0 = 40 mL

Moles of Fe(SCN)₂²⁺ = 6.85*10^-5 M * 0.040 L = 2.73*10^-6 moles

    Fe³⁺    +      2SCN^- ------> Fe(SCN)₂²⁺

I   6.0*10^-4      1.98*10^-4             0

C    -x                -2x                    +x

E   6.0*10^-4-x    1.98*10^-4-2x   2.73*10^-6

So at equilibrium

Moles of Fe³⁺ = 6.0*10^-4-x

                        = 6.0*10^-4- 2.73*10^-6

                        = 5.73*10^-5 moles

Moles of SCN^- = 1.98*10^-4-2x  

                         = 1.98*10^-4-2 *2.73*10^-6

                         = 1.44*10^-5 moles

So, molarity of Fe³⁺ = 5.73*10^-5 moles / 0.040 L

                                 = 1.43*10^-3 M

Molarity of SCN^- = 1.44*10^-5 moles /0.040 L

                             = 3.60*10^-4 M