Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the following concentrations are found.
A) [Fe3+ ] = 8.12 ✕ 10−3 M,
B) [SCN − ] = 7.84 ✕ 10−3 M
What is the concentration of FeSCN2+ in the new equilibrium mixture?
Hint: Use your value of K to solve for the unknown concentration.

Respuesta :

Explanation:

The given reaction equation will be as follows.

          [tex][FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}][/tex]

Let is assume that at equilibrium the concentrations of given species are as follows.

        [tex][Fe^{3+}] = 8.17 \times 10^{-3}[/tex] M

        [tex][SCN^{-}] = 8.60 \times 10^{-3}[/tex] M

        [tex][FeSCN^{2+}] = 6.25 \times 10^{-2}[/tex] M

Now, first calculate the value of [tex]K_{eq}[/tex] as follows.

     [tex]K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}[/tex]

              = [tex]\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}[/tex]

              = [tex]11.24 \times 10^{-4}[/tex]

Now, according to the concentration values at the re-established equilibrium the value for [tex][FeSCN^{2+}][/tex] will be calculated as follows.

             [tex]K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}[/tex]

        [tex]11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}[/tex]

         [tex][FeSCN^{2+}] = 5.66 \times 10^{-2}[/tex] M

Thus, we can conclude that the concentration of [tex][FeSCN^{2+}][/tex] in the new equilibrium mixture is [tex]5.66 \times 10^{-2}[/tex] M.