# Question cc92d

May 28, 2017

$\text{0.17 g}$

#### Explanation:

The idea here is that you can use the molar solubility of calcium sulfate to calculate the number of grams of salt that can be dissolved in $\text{250 mL}$ of water to make a saturated solution, presumably at room temperature.

So, you know that calcium sulfate is considered an insoluble salt, which implies that when you dissolve calcium sulfate in water, an equilibrium will be established between the undissolved solid and the dissolved ions.

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

By definition, the solubility product constant of calcium sulfate is equal to

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

Now, notice that every mole of calcium sulfate that dissociates produces $1$ mole of calcium cations and $1$ mole of sulfate anions.

This means that if you take $s$ to be the molar solubility of the salt, i.e. the concentration of calcium sulfate that dissociates, you can say that, at equilibrium, you will have

${K}_{s p} = s \cdot s$

which, in your case, is equivalent to

$2.4 \cdot {10}^{- 5} = {s}^{2}$

Solve for $s$ to find

$s = \sqrt{2.4 \cdot {10}^{- 5}} = 4.9 \cdot {10}^{- 3}$

So, you know that you have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{molar solubility CaSO"_4 = 4.9 * 10^(-3)color(white)(.)"mol L}}^{- 1}}}}$

This means that you can only hope to dissolve $4.9 \cdot {10}^{- 3}$ moles of calcium sulfate for every $\text{1 L}$ of solution.

Convert the number of moles to grams by using the compound's molar mass

4.9 * 10^(-3) color(red)(cancel(color(black)("moles CaSO"_4))) * "136.14 g"/(1color(red)(cancel(color(black)("mole CaSO"_4)))) = "0.667 g"#

Since this is how much calcium sulfate can be dissolved for every

$\text{1 L} = {10}^{3}$ $\text{mL}$

of solution, you can say that $\text{250 mL}$ of solution, which, for all intended purposes, is equivalent to $\text{250 mL}$ of water, will dissolve

$250 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mL solution"))) * "0.667 g CaSO"_4/(10^3color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.17 g CaSO}}_{4}}}}$