Question

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.

Answer 1

Explanation:

The given reaction equation will be as follows.

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

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

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

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

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

Now, first calculate the value of $K_{eq}$ as follows.

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

              = $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

              = $11.24 \times 10^{-4}$

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

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

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

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

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