A solution is made by mixing 100mL Fe3+ solution of 1.00 x 10-3 M and 100 mL A-

Question

A solution is made by mixing 100mL Fe3+ solution of 1.00 x 10-3 M and 100 mL A- ion solution of 1.00 x 10-3 M. The mixture has a deep blue color, indicating the formation of a complex between Fe3+ and A-. By a spectrophotometric analysis, we find that 30.0% of Fe3+ is in the form of the complex and there are two probable products which are FeX2+ or FeX3

a. Balance the following equations and calculate the concentrations of the remaining species

Fe3+ + A- <--> FeA2+

Fe3+ + A- <--> FeA3

b. Assuming that the final state reached equilibrium, write the equilibrium expressions and calculate the equilibrium constants for both cases in (a)

Explanation / Answer

Formula used :

Molarity = moles/ volume in liters

So, no. of moles = molarity*volume in liters

Solution 1 : 100 mL ( 0.1 L ) of 0.001 M Fe3+ solution

So,initial moles of Fe3+ = 0.001*0.1 = 0.0001 moles Fe3+

Solution 2 : 100 mL ( 0.1 L ) of 0.001 M A- solution

So,initial moles of A- = 0.001*0.1 = 0.0001 moles A-

Solution 3 ( or the final solution ) : 100 mL Fe3+ + 100 mL A-

So, volume of final solution = 200 mL

Probable products : FeA2+ and FeA3 ( I had a confusion whether it was FeA2+ or FeA2+, so I have provided answer for FeA2+ in the end ! )

Balanced equations :

(i) Fe3+ + 2A- <--> FeA2+

(ii) Fe3+ + 3A- <--> FeA3

Given :

30% of Fe3+ gets consumed in forming the complex

So, remaining moles of Fe3+ = 70% of initial moles = 0.7*0.0001 = 0.00007 moles remaining Fe3+

and, moles of complex = 0.3*0.0001 = 0.00003 moles complex

Case (i) :

Fe3+ + 2A- <--> FeA2+

According to reaction stoichiometry, when 1 mole complex is formed, 2 moles A- are consumed

Applying unitary method, 0.00003 moles of complex are formed by consuming : 2*0.00003 = 0.00006 moles A-

So, moles of A- left = initial moles - moles consumed = 0.0001 - 0.00006 = 0.00004 moles A-

Concentration of A- = [A-] = moles of A- left / Volume of final solution in liters = 0.00004/0.2 = 0.0002 M [A-]

Concentration of Fe3+ = [Fe3+] = moles of Fe3+ left / Volume of final solution in liters = 0.00007/0.2 = 0.00035 M [Fe3+]

Concentration of complex FeA2+ = [FeA2+] = moles of FeA2+ left / Volume of final solution in liters = 0.00003/0.2 = 0.00015 M [FeA2+]

Equilibrium expression : Keq = [FeA2+] / ( [Fe3+]*[A-]2 )

Putting values, we get : Keq = 0.00015 / ( 0.00035*0.00022 ) = 1.07*107 M-2

Case (ii) :

Fe3+ + 3A- <--> FeA3

According to reaction stoichiometry, when 1 mole complex is formed, 3 moles A- are consumed

Applying unitary method, 0.00003 moles of complex are formed by consuming : 3*0.00003 = 0.00009 moles A-

So, moles of A- left = initial moles - moles consumed = 0.0001 - 0.00009 = 0.00001 moles A-

Concentration of A- = [A-] = moles of A- left / Volume of final solution in liters = 0.00001/0.2 = 0.00005 M [A-]

Concentration of Fe3+ = 0.00035 M [Fe3+]

Concentration of complex FeA3 = 0.00015 M [FeA3]

Equilibrium expression : Keq = [FeA3] / ( [Fe3+]*[A-]3 )

Putting values, we get : Keq = 0.00015 / ( 0.00035*0.000053 ) = 3.42*1012 M-3

Note : If the probable product was FeA2+, then the balanced equation is :

Fe3+ + A- <--> FeA2+

The equilibrium expression is : Keq = [FeA2+] / ( [Fe3+]*[A-] )

This time, for producing 0.00003 moles of complex, 0.00003 moles of A- are consumed.

So, [A-] = 0.00007 / 0.2 = 0.00035 [A-]

And, Keq = 0.00015 / ( 0.00035*0.00035) = 1.224*103 M-1

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