# Question #7fcb7

##### 1 Answer
Jul 6, 2017

$2.30 \cdot {10}^{- 6}$ ${\text{mol L}}^{- 1}$

#### 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 $\text{1 L}$ 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.

${\text{Fe"("OH")_ (color(red)(2)(s)) rightleftharpoons "Fe"_ ((aq))^(2+) + color(red)(2)"OH}}_{\left(a q\right)}^{-}$

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

$\left[{\text{Fe}}^{2 +}\right] = s \to$ every mole of iron(II) hydroxide that dissociates produces $1$ mole of iron(II) cations in solution

$\left[{\text{OH}}^{-}\right] = \textcolor{red}{2} \cdot s \to$ every mole of iron(II) hydroxide that dissociates produces $\textcolor{red}{2}$ moles of hydroxide anions in solution

Now, the solubility product constant, ${K}_{s p}$, is equal to

${K}_{s p} = {\left[{\text{Fe"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

In your case, this will be equal to

$4.87 \cdot {10}^{- 17} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

$4.87 \cdot {10}^{- 17} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt[3]{\frac{4.87 \cdot {10}^{- 17}}{4}} = 2.30 \cdot {10}^{- 6}$

This means that a saturated solution of iron(II) hydroxide will contain $2.30 \cdot {10}^{- 6}$ moles of dissociated salt for every $\text{1 L}$ of solution, presumably at room temperature.

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar solubility" = 2.30 * 10^(-6)color(white)(.)"mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.