Question

A student mixes 5.00 mL of 2.00 x 10‐3 M Fe(NO3)3 with 5.00 mL 2.00 x 10‐3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN+2 is 1.40 x 10‐4 M.

What is the initial concentration in solution of the Fe+3 and SCN‐?

What is the equilibrium constant for the reaction?

What happened to the K+ and the NO3‐ ions in this solution?

Answer 1

number of Fe +3 = M*V = 2*10^-3 M * 0.005 L = 1*10^-5 M
number of SCN-= M*V = 2*10^-3 M * 0.005 L = 1*10^-5 M
Total volume = 5 +5 = 10 mL = 0.01 L
Initial concentration of Fe+3,
[Fe+3] = number of moles / total volume
                =1*10^-5 / 0.01
                 =1*10^-3 M
Initial concentration of SCN-,
[SCN-] = number of moles / total volume
                =1*10^-5 / 0.01
                 =1*10^-3 M
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Fe+3      +      SCN - --->FeSCN+2
10^-3              10^-3              0              (initial)
10^-3-x         10^-3-x           x               (at equilibrium)
here x = 1.4*10^-4 M
At equilibrium:
[FeSCN+2] = x = 1.4*10^-4 M
[Fe+3] = 10^-3-x = 10^-3-1.4*10^-4 =8.6*10^-4 M
[SCN-]= 10^-3-x = 10^-3-1.4*10^-4 =8.6*10^-4 M

Kc = [FeSCN+2] / {[Fe+3]*[SCN-]}
Kc = (1.4*10^-4) / {(8.6*10^-4 )*(8.6*10^-4 )}
    =189.3
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K+ and NO3- are present in solution as K+ and NO3-