Question

Question #9e97b

Answer 1

Here's what I got.

Explanation:

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

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

`"CaSO"_4 ->` 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

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

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

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

Notice that the stoichiometric coefficients present in the dissociation equilibrium become exponents in the expression for `K_(sp)` and that the equilibrium concentration of the undissolved solid is not included.

Do the same for calcium sulfate.

`"CaSO"_ (4(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "SO"_ (4(aq))^(2+)`

This time, the solubility product constant will be

`K_(sp) = ["Ca"^(2+)] * ["SO"_4^(2-)]`

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