Here is the Chemical Equation for the following problem: Fe (aq) + HSCN (aq) <--

ID: 1015370 • Letter: H

Question

Here is the Chemical Equation for the following problem:

Fe (aq) + HSCN (aq) <-----> FeSCN (aq) + H

When 10.0 mL of 2.50 x 10-3 Fe(NO3)3 and 10.0 mL of 2.50 x 10-3 M HSCN (both in 0.5 M nitric acid) are mixed together with 5.00 mL of 0.5 M nitric acid and allowed to come to equilibrium, the concentration of the iron thiocynate complex, FeSCN2+ , was fund to be 1.33 x 10-4 M. Set up an equilibrium table and use it to organize this data. Calculate the initial concentrations of the Fe3+ and HSCN assuming no complex formation, then give the equilibrium concentration of the FeSCN2+ complex and calculate all the equilibrium concentrations. Finally, calculate the value of the equilibrium constant (Kc) for this reaction.

Explanation / Answer

Initial Moles of Fe3+=0.01L*2.5*10^-3 M=2.5*10^-5 moles

Initial Moles of SCN-=0.01L*2.5*10^-3 M=2.5*10^-5 moles

Initial moles of HNO3=0.005L*0.5M=2.5*10^-4 moles

Initial concentration of Fe3+=0.01L*2.5*10^-3 M=2.5*10^-5 moles/V

Initial conc of SCN-=0.01L*2.5*10^-3 M=2.5*10^-5 moles/V

Initial conc of HNO3=0.005L*0.5M=2.5*10^-4 moles /V

Total volume of solution=10ml+10ml+5ml=25 ml=0.025 L=V (same so gets cancelled in the expression of kc)

[Fe3+]

[SCN-]

[H+]

[FescN-]

Initial

2.5*10^-5 moles

2.5*10^-4 moles

Change

-x

equilibrium

2.5*10^-5 moles-x

2.5*10^-5 moles+x

Given [FeSCN]equilibrium=x=1.33*10^-4M

Kc=[FeSCN2+][H+]/[Fe3+][SCN-]=(1.33*10^-4 moles)(2.5*10^-4 moles +x)/(2.5*10^-5 -xmoles)(2.5*10^-5 moles-x)

Kc=1.33*10^-4 moles)(2.5*10^-4 moles +1.33*10^-4 moles)/(2.5*10^-5 -13.3*10^-5 moles)(2.5*10^-5 moles-13.3*10^-5 moles

Kc=(1.33*10^-4 mol)(3.83*10^-4 mol)/(10.8*10^-5)^2

Kc=0.0437*10^2)=4.37