An aqueous solution was prepared at 21 oC by mixing 7.00 mL 2.00 x 10-2mol L-1Fe3+, 2.00 mL 1.50 x 10-3 mol L-1SCN−, and 1.00 mL water. At equilibrium, the concentration of the product complex, [Fe(SCN)2+]eq was determined to be 1.74 x 10-4mol L-1. What is the value of the equilibrium constant K for the reaction of interest at 21 oC?

The value of equilibrium constant for the given reaction is 99.85

Explanation:

To calculate the number of moles for given molarity, we use the equation:

   .....(1)

Molarity of solution =

Volume of solution = 7.00 mL = 0.007 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

Molarity of solution =

Volume of solution = 2.00 mL = 0.002 L

Putting values in equation 1, we get:

Volume of the container = [7 + 2 + 1] = 10 mL = 0.010 L

The chemical equation for the formation of complex follows:

Initial:                     0.014    

At eqllm:             0.014-x       x

We are given:

Equilibrium concentration of

Equilibrium concentration of

Equilibrium concentration of

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the value of equilibrium constant for the given reaction is 99.85

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