# Question 9e97b

Mar 22, 2017

Here's what I got.

#### Explanation:

Start by writing out the chemical formulas for these two ionic compounds

"Fe"("OH")_2 -># iron(II) hydroxide

${\text{CaSO}}_{4} \to$ calcium sulfate

Next, you're dealing with insoluble ionic compounds, so you must write a dissociation equilibrium that is established when these salts are dissolved in water.

For iron(II) hydroxide, you will have

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

By definition, the solubility product constant for this equilibrium will be

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

Notice that the stoichiometric coefficients present in the dissociation equilibrium become exponents in the expression for ${K}_{s p}$ and that the equilibrium concentration of the undissolved solid is not included.

Do the same for calcium sulfate.

${\text{CaSO"_ (4(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "SO}}_{4 \left(a q\right)}^{2 +}$

This time, the solubility product constant will be

${K}_{s p} = \left[{\text{Ca"^(2+)] * ["SO}}_{4}^{2 -}\right]$

Once again, the stoichiometric coefficients become exponents and the equilibrium concentration of the undissolved solid is not included.