Question #9e97b

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Question #9e97b

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Answer 1 · 2017-03-22T01:55:02

Here's what I got.

Explanation:

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

$Fe {(OH)}_{2} \to$ $"Fe"("OH")_2 ->$ iron(II) hydroxide

${CaSO}_{4} \to$ $"CaSO"_4 ->$ 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

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

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

$K_{s p} = [{Fe}^{2 +}] \cdot {[{OH}^{-}]}^{2}$ $K_(sp) = ["Fe"^(2+)] * ["OH"^(-)]^color(red)(2)$

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

Do the same for calcium sulfate.

${CaSO}_{4 (s)} ⇌ {Ca}_{(a q)}^{2 +} + {SO}_{4 (a q)}^{2 +}$ $"CaSO"_ (4(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "SO"_ (4(aq))^(2+)$

This time, the solubility product constant will be

$K_{s p} = [{Ca}^{2 +}] \cdot [{SO}_{4}^{2 -}]$ $K_(sp) = ["Ca"^(2+)] * ["SO"_4^(2-)]$

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

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Question #9e97b

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