Question

Question #e8dce

Answer 1

`["Fe"^(3+)] = 1.0 * 10^(-9)` `"M"`

Explanation:

You know that iron(II) oxide-hydroxide is insoluble in water, which implies that a dissociation equilibrium is established in aqueous solution between the undissolved solid and the solvated ions.

`"Fe"("OH")_ (3(s)) rightleftharpoons "Fe"_ ((aq))^(3+) + 3"OH"_( (aq))^(-)`

By definition, the solubility product constant, `K_(sp`, for this solubility equilibrium looks like this

`K_(sp) = ["Fe"^(3+)] * ["OH"^(-)]^3`

Now, in order for precipitation to occur, you need to have

`K_(sp) <= ["Fe"^(3+)] * ["OH"^(-)]^3" "" "color(darkorange)(("*"))`

As you know, an aqueous solution at room temperature has

`color(blue)(ul(color(black)("pOH" = 14 - "pH")))`

Since you know that

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

you can say that

`["OH"^(-)] = 10^(-"pOH")`

which is equivalent to

`["OH"^(-)] = 10^(-(14 - "pH"))`

Plug in the value you have for the `"pH"` of the solution to get

`["OH"^(-)] = 10^(-(14 - 3.48))`

`["OH"^(-)] = 10^(-10.52)`

Next, rearrange equation `color(darkorange)(("*"))` to solve for the minimum concentration of iron(III) cations that will precipitate the solid.

`["Fe"^(3+)] >= K_(sp)/(["OH"^(-)]^3)`

Plug in your values to find

`["Fe"^(3+)] >= (2.79 * 10^(-39))/(10^(-10.52))^3`

`["Fe"^(3+)] >= (2.79 * 10^(-39))/( (10^(-10))^3 * (10^(-0.52))^3)`

`["Fe"^(3+)] >= 2.79 * 10^(1.56) * 10^(-39) * 10^(30)`

Therefore, you can say that you need

`color(darkgreen)(ul(color(black)(["Fe"^(3+)] >= 1.0 * 10^(-7)color(white)(.)"M")))`

The answer is rounded to two sig figs, the number of decimal places you have for the `"pH"` of the solution.