Question

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?

Answer 1

Answer: The value of equilibrium constant for the given reaction is 99.85

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(1)

• For
  `Fe^(3+)` ions:

Molarity of
`Fe^(3+)` solution =
`2.00* 10^(-2)M`

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

Putting values in equation 1, we get:

`2.00* 10^(-2)M=\frac{\text{Moles of }Fe^(3+)\text{ ions}}{0.007L}\\\\\text{Moles of }Fe^(3+)\text{ ions}=(2.00* 10^(-2)mol/L* 0.007L)=1.4* 10^(-4)mol`

• For
  `SCN^(-)` ions:

Molarity of
`SCN^(-)` solution =
`1.50* 10^(-3)M`

Volume of solution = 2.00 mL = 0.002 L

Putting values in equation 1, we get:

`1.50* 10^(-3)M=\frac{\text{Moles of }SCN^(-)\text{ ions}}{0.002L}\\\\\text{Moles of }SCN^-\text{ ions}=(1.50* 10^(-3)mol/L* 0.002L)=3* 10^(-6)mol`

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

`\text{Molarity of }Fe^(3+)\text{ ions}=(1.4* 10^(-4)mol)/(0.01)=1.4* 10^(-2)M`

`\text{Molarity of }SCN^(-)\text{ ions}=(3.0* 10^(-6)mol)/(0.01)=3.0* 10^(-4)M`

The chemical equation for the formation of
`[FeSCN^(2+)]`complex follows:

`Fe^(2+)+SCN^-\rightleftharpoons [FeSCN^(2+)]`

Initial: 0.014
`3.0* 10^(-4)`

At eqllm: 0.014-x
`3.0* 10^(-4)-x` x

We are given:

Equilibrium concentration of
`[FeSCN^(2+)]=1.74* 10^(-4)M=x`

Equilibrium concentration of
`[Fe^(2+)]\text{ ions}=(1.4* 10^(-2)-x)=(1.4-0.0174)* 10^(-3)=1.383* 10^(-2)M`

Equilibrium concentration of
`[SCN^(-)]\text{ ions}=(3.0* 10^(-4)-x)=(3.0-1.74)* 10^(-4)=1.26* 10^(-4)M`

The expression of
`K_(eq)` for above equation follows:

`K_(eq)=([FeSCN^(2+)])/([Fe^(3+)][SCN^-])`

Putting values in above equation, we get:

`K_(eq)=((1.74* 10^(-4)))/((1.383* 10^(-2))* (1.26* 10^(-4)))\\\\K_(eq)=99.85`

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