A solution of 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300 M KSCN, and 3.00

Question

A solution of 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300 M KSCN, and 3.00 mL of 1.0 M HNO3 is mixed together and allowed to reach equilibrium. The concentration of Fe(SCN)^2+ is found to be 2.72 x 10^-4 M at equilibrium. Calculate the equilibrium constant for this reaction. A solution of 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300 M KSCN, and 3.00 mL of 1.0 M HNO3 is mixed together and allowed to reach equilibrium. The concentration of Fe(SCN)^2+ is found to be 2.72 x 10^-4 M at equilibrium. Calculate the equilibrium constant for this reaction.

Explanation / Answer

equilibrium constant for this reaction:--

Fe^3+ (aq) + SCN- (aq) ---> Fe(SCN)^2+ (aq)

moles Fe(NO3)3 = 0.00300 M x 5.00 x 10^-3 L=1.50 x 10^-5
moles KSCN = 0.00300 M x 4.00 x 10^-3 L=1.2 x 10^-5
total volume = 5 + 4 + 3 = 12 mL = 0.012 L
[SCN-]= 1.2 x 10^-5 / 0.012 =0.00100 M
[Fe3+]= 1.50 x 10^-5 / 0.012 L=0.00125

at equilibrium
[Fe3+]= 0.00125 - 2.72 x 10^-4
[SCN-]= 0.00100 - 2.72 x 10^-4

so
[Fe3+]= 0.000978 M
[SCN-] = 0.000728 M

K = 2.72 x 10^-4 / ( 0.000978)( 0.000728)= 382

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