Question

# 5.00 mL of 2.70E-3 M Fe(NO3)3 is mixed with 1.00 mL of 2.79E-3 M KSCN and...

5.00 mL of 2.70E-3 M Fe(NO3)3 is mixed with 1.00 mL of 2.79E-3 M KSCN and 4.00 mL of water. The equalibrium molarity of Fe(SCN)2+ is found to be 6.00E-5 M. If the reaction proceeds as shown below, what is the equilibrium molarity of Fe3+ and SCN-?

Start by calculating the initial concentrations of Fe3+ and SCN- in the mixture using M1V1 = M2V2
Fe: 2.70X10^-3M (5 mL) = M2 X 10 mL
M2 = 1.35X10^-3 M

SCN-: 2.79X10^-3 M (1 mL) = M2(10 mL)
M2 = 2.79X10^-4 M

Now, since [FeSCN2+] = 6X10^-5 M, the concentration of Fe3+ and SCN- at equilibrium must be reduced by this amount. So at equilibrium:

[Fe3+] = 1.35X10^-3 M - 6X10^-5 M = 1.29X10^-3 M
[SCN-] = 2.79X10^-4 M - 6X0^-5 M = 2.19X10^-4 M

The expression for Kc is:

Kc= [FeSCN2+]/[Fe3+][SCN-]
Kc = 6.00X10^-5 / (1.29X10^-3)(2.19X10^-4) = 212

By this equation euillibrium molarity can also be calculated.

Hope this helped.

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