1) A student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO3)3 with 5.00 mL of 2.00 x 10-3

Question

1) A student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO3)3 with 5.00 mL of 2.00 x 10-3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.40 x 10-4 M

a) What is the initial concentration in solution of the Fe3+ and SCN- ?

b) What is the equilibrium constant for the reaction?

2. Assume that the reaction studied is actually: Fe3+ (aq) + 2 SCN- (aq) Fe(SCN)2+ (aq)

a) What is the equation for determining the equilibrium constant?

b) Using the information from question 1 and assuming [Fe(SCN)2+] = 1.40 x 10-4 M, calculate the equilibrium concentration of Fe3+ and SCN- .

c) Determine the numerical value of K.

Explanation / Answer

1a)

This is a redox reaction, identify then the reaction

FeSCN+2 foramtion

calcualte concnetrations

initially

VT = V1+V2 = 5+5 = 10

[Fe3+] = MV/Vt = 5/10*(2*10^-3) = 10^-3 M

[KSCN] = MV/Vt = 5/10*2*10^-3) = 10^-3 M

1b)

initially

[Fe3+] = 10^-3 M

[KSCN] = 10^-3 M

[FeSCN+2] = 0

in equilibrium

[Fe3+] = 10^-3 - x

[SCN-] = 10^-3 -2x

[FeSCN+2] = 0 +x

and we know that

[SCN] = 10^-3 -2x = 1.4*10^-4

then x = 4.3*10^-4

then

[Fe3+] = 10^-3 -4.3*10^-4 = 0.00057

[SCN-] = 10^-3 -2*4.3*10^-4 = 0.00014

[FeSCN+2] = 0 +4.3*10^-4 = 0.00043

subsitutte in K

K = [Fe(SCN)+2] / [Fe3+][SCN+2]^2

K = (0.00043)/((0.00057)*(0.00014^2)) = 38489079.8425

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