5.00 mL of a 0.200 M Fe(NO 3 ) 3 solution is mixed with 5.00 mL of a 0.00200 M N

Question

5.00 mL of a 0.200 M Fe(NO3)3 solution is mixed with 5.00 mL of a 0.00200 M NaSCN solution. The blood-red FeNCS2+ at equilibrium has a concentration of 6.86 x 10-4 M.

Complete the following table (note: the initial molar concentration is the concentration after mixing but before reaction). Be sure to show your work.

Fe3+(aq)

SCN-(aq)

FeNCS2+(aq)

Initial Molar Concentration ([i])

Change ( )

Initial + Change ( [i] + )

Equilibrium Molar Concentration([f])

Fe3+(aq)

SCN-(aq)

FeNCS2+(aq)

Initial Molar Concentration ([i])

Change ( )

Initial + Change ( [i] + )

Equilibrium Molar Concentration([f])

Explanation / Answer

Fe3+(aq):
Fe(NO3)3 = Fe3+ + 3NO3-
volume of Fe(NO3)3 = 5ml = 5e-3 L
concentration = 0.2M
moles of Fe3+, n1 = 5e-3*0.2 = 1e-3
Total volume of solution, V = 5+5 = 10ml = 1e-2 L

SCN-(aq):
5.00 mL of a 0.00200 M NaSCN
NaSCN = Na+ + SCN-
volume of SCN- = 5ml = 5e-3 L
concentration = 0.002M
moles of SCN-, n2 = 5e-3*0.002 = 1e-5

FeNCS2+(aq):
5.00 mL of a 0.200 M Fe(NO3)3 solution is mixed with 5.00 mL of a 0.00200 M NaSCN solution.
Reaction:
            Fe3+(aq)    +   SCN(aq) = [FeNCS]2+(aq)
initial:    i1                 i2            0
change:    -delta            -delta        +delta
Equilibrium: (i1-delta)    (i2-delta)      delta

Given equilibrium concentration, [FeNCS]2+(eq) = delta = 6.86E-4 M
So change = delta = 6.86E-4 M= 0.0007 M

Fe3+(aq):
(i) Initial Molar Concentration ([i]) = i1= n1/V = 1e-3/1e-2 = 0.1 M
(ii) Change () = -delta = - 6.86E-4 M
(iii) Initial + Change ([i] + ) = 0.1- 6.86E-4 M = 0.099314 M
(iv) Equilibrium Molar Concentration([f]) = i1 - delta = 0.1- 6.86E-4 M = 0.099314 M
  
SCN-(aq):
(i) Initial Molar Concentration ([i]) = i2 = n2/V = 1e-5/1e-2 = 0.001 M
(ii) Change () = -delta = - 6.86E-4 M
(iii) Initial + Change ([i] + ) = 0.001- 6.86E-4 M = 0.000314 M
(iv) Equilibrium Molar Concentration([f]) = i2 - delta = 0.1- 6.86E-4 M = 0.000314 M

FeNCS2+(aq):
(i) Initial Molar Concentration ([i]) = 0
(ii) Change () = 6.86E-4 M
(iii) Initial + Change ([i] + ) = 6.86E-4 M
(iv) Equilibrium Molar Concentration([f]) = 6.86E-4 M

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