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

Step-by-step 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 * 10^(-3)` M

`[SCN^(-)] = 8.60 * 10^(-3)` M

`[FeSCN^(2+)] = 6.25 * 10^(-2)` M

Now, first calculate the value of
`K_(eq)` as follows.

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

=
`(8.17 * 10^(-3) * 8.60 * 10^(-3))/(6.25 * 10^(-2))`

=
`11.24 * 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) = ([Fe^(3+)][SCN^(-)])/([FeSCN^(2+)])`

`11.24 * 10^(-4) = (8.12 * 10^(-3) * 7.84 * 10^(-3))/([FeSCN^(2+)])`

`[FeSCN^(2+)] = 5.66 * 10^(-2)` M

Thus, we can conclude that the concentration of
`[FeSCN^(2+)]` in the new equilibrium mixture is
`5.66 * 10^(-2)` M.