# I was given this problem without any explanation. Please help!!! The formula is Fe3+ + SCN- yields FeSCN 2+Then you have Kf= {FeSCN 2+} / {Fe3+} {SCN-}When 5.0ml of 2.0x 10^-4 M KSCN and 5.0ml of...

I was given this problem without any explanation. Please help!!!

The formula is Fe3+ + SCN- yields FeSCN 2+

Then you have Kf= {FeSCN 2+} / {Fe3+} {SCN-}

When 5.0ml of 2.0x 10^-4 M KSCN and 5.0ml of 2.0 x 10^-4 are combined, the resulting FeSCN2- solution has an absourbance of 0.15. A standard solution of 1.0x 10^-4 M FeSCN2+ has an absourabance of 0.40. Calculate Kf.

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The reaction is Fe3+ + SCN- ---> [Fe(SCN)] 2+

c(1-α) c(1-α) cα (Applying degree of dissociation concept, c is the initial concentration of both the reactants)

Kf = cα/ c^2 (1-α) ^2 = α/ c (1-α) ^2

When 5.0ml of 2.0x 10^-4 M KSCN and 5.0ml of 2.0 x 10^-4 M Fe3+ solutions are combined, the concentrations of both these ions will be halved, i.e. both these ions will have initial concentration, c= 1.0x 10^-4 M. Assuming that the spectrophotometer is set at the absorption maximum of Fe(SCN)] 2+, and Lambert-Beer’s law requires that A = Ɛ.c’.l (where c’ is the eqm. concentration of [Fe(SCN)] 2+ )

For a particular cell, while observing the same species, Ɛ & l remains constant, hence A varies as c’, or A1/A2 = c’1/c’2. Putting values for different concentrations of [Fe(SCN)] 2+ giving absorbance values of 0.15 and 0.40 we can write 0.15/0.40 = c’/1.0x 10^-4

Or, c’ = 3×10^-4/8

Now from the original reaction, c’ = cα = 1.0 x 10^-4×α

So, 1.0 x 10^-4 ×α = 3×10^-4/8, or α = 3/8 and 1-α = 5/8

Hence Kf = α/ c (1-α) ^2 = 3/8/(1.0 x 10^-4×5/8×5/8) = 24/25 × 10^4.

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