11. Consider the following equilibrium: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Initial

Question

11. Consider the following equilibrium: Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Initial concentrations: [Fe3+] = 0.590; [SCN-] = 1.239; [FeSCN2+] = 0

The equilibrium concentration of [FeSCN2+]eq = 0.454 M.

What is the equilibrium concentration of Fe3+?

[Fe3+]eq =____ M

12. Consider the following equilibrium: Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Initial concentrations: [Fe3+] = 0.590; [SCN-] = 1.239; [FeSCN2+] = 0

The equilibrium concentration of [FeSCN2+]eq = 0.454 M.

What is the numerical value of KC for this equilibrium?

KC =

Explanation / Answer

Q11.

initially

[Fe3+] = 0.590

[SCN-] = 1.239

[FeSCN2+] = 0

in equilibrium

[Fe3+] = 0.590 - x

[SCN-] = 1.239 - x

[FeSCN2+] = 0 + x

and we know

[FeSCN2+] = 0 + x = 0.454 M

x = 0.454

so

[Fe3+] = 0.590 - x = 0.590-0.454 = 0.136 M

Q12.

Find numerical Kc

we need [SCN-] = 1.239 - x = 1.239-0.454 = 0.785 M

so

Kc = [FeSCN+2]/([Fe+3][SCn-])

Kc = 0.454 / (0.136 *0.785 ) = 4.252

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