Question

Use an ICE table to calculate the equilibrium amounts of all 3 chemicals if the initial amounts are 0.20 M of each.

Fe3+(aq)+SCN-(aq)-----> (Fe(SCN)2+) Kc= 159

Answer 1

Fe3+(aq) + SCN-(aq) -----> (Fe(SCN)2+) Kc= 159

Initial 0.2 M 0.2 M 0.2 M

at equilibrium 0.2 -x 0.2 -x 0.2 +x

Kc =   [Fe(SCN)2+] / [ Fe3+(aq)] [SCN-(aq)]

159 =   (0.2 +x) / (0.2 -x) (0.2 -x)

159 [ x² -0.4 x + 0.04 ] = 0.2 +x

159 x²  - 63.6 x + 6.36 = 0.2 + x

159 x²  - 64.6 x + 6.16 = 0

On solving ,

x = 0.153 M

Therefore, equilibrium concentrations are

[Fe3+(aq)] = 0.2 -x = 0.2 - 0.153 = 0.047 M

[SCN-(aq)] = 0.2 -x = 0.2 - 0.153 = 0.047 M

[Fe(SCN)2+] = 0.2 + x = 0.2 + 0.153 = 0.353 M