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10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3 M KSCN and...

10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3 M KSCN and 6.00 mL of water. The equalibrium molarity of FeSCN2+ is found to be 8.70E-5 M. If the reaction proceeds as shown below, what is the equilibrium molarity of Fe3+ and SCN-? Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Equilibrium Concentration Fe3+ ____________

Equilibrium Concentration SCN- ____________

Expert Solution

initial molarity of Fe+3 = 10 x 3.48 x 10^-3 / (10 +4 + 6)

= 1.74 x 10^-3 M

initial molarity of SCN- = 4 x 2.84 x 10^-3 / 20

= 5.68 x 10^-4 M

Fe+3 + SCN- ----------------------------> FeSCN2+

1.74 x 10^-3 5.68 x 10^-4 0 ----------------> initial

1.74 x 10^-3-x 5.68 x 10^-4 -x x -------------------> equilibrium

x = 8.70 x 10^-5 M

equilibrium concentration of Fe+3 = 1.74 x 10^-3-x

= 1.74 x 10^-3 - 8.70 x 10^-5

= 1.65 x 10^-3 M

equilibrium concentration of SCN- = 5.68 x 10^-4 -x

=5.68 x 10^-4 - 8.70 x 10^-5

= 4.81 x 10^-4 M

queen_honey_blossom answered 1 year ago

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