Question

Question #7fcb7

Answer 1

`2.30 * 10^(-6)` `"mol L"^(-1)`

Explanation:

The problem wants you to determine the molar solubility of iron(II) hydroxide, which essentially means that you must find the number of moles of iron(II) hydroxide that will dissociate per `"1 L"` of solution to produce iron(II) cations and hydroxide anions.

Iron(II) hydroxide is considered insoluble in water, which implies that the position of its dissociation equilibrium lies far to the left.

`"Fe"("OH")_ (color(red)(2)(s)) rightleftharpoons "Fe"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)`

In other words, only a very, very small number of moles of iron(II) hydroxide will actually dissociate when dissolved in water.

If you take `s` to be the molar solubility of the salt, you can say that, at equilibrium, you will have

`["Fe"^(2+)] = s ->` every mole of iron(II) hydroxide that dissociates produces `1` mole of iron(II) cations in solution

`["OH"^(-)] = color(red)(2) * s ->` every mole of iron(II) hydroxide that dissociates produces `color(red)(2)` moles of hydroxide anions in solution

Now, the solubility product constant, `K_(sp)`, is equal to

`K_(sp) = ["Fe"^(2+)] * ["OH"^(-)]^color(red)(2)`

In your case, this will be equal to

`4.87 * 10^(-17) = s * (color(red)(2)s)^color(red)(2)`

`4.87 * 10^(-17) = 4s^3`

Rearrange to solve for `s`

`s = root(3)( (4.87 * 10^(-17))/4) = 2.30 * 10^(-6)`

This means that a saturated solution of iron(II) hydroxide will contain `2.30 * 10^(-6)` moles of dissociated salt for every `"1 L"` of solution, presumably at room temperature.

In other words, the molar solubility of the salt is equal to

`color(darkgreen)(ul(color(black)("molar solubility" = 2.30 * 10^(-6)color(white)(.)"mol L"^(-1))))`

The answer is rounded to three sig figs.