# Question b13d6

Sep 14, 2015

$\left(a\right) . 4.9 \times {10}^{- 3} \text{mol/l}$

$\left(b\right) . 0.66 \text{g/l}$

#### Explanation:

$C a S {O}_{4 \left(s\right)} r i g h t \le f t h a r p \infty n s C {a}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -}$

${K}_{s p} = \left[C {a}_{\left(a q\right)}^{2 +}\right] \left[S {O}_{4 \left(a q\right)}^{2 -}\right] = 2.4 \times {10}^{- 5} m o {l}^{2.} {L}^{- 6}$

Since $\left[C {a}_{\left(a q\right)}^{2 +}\right] = \left[S {O}_{4 \left(a q\right)}^{2 -}\right]$ and the solubility $s$ is equal to $\left[C {a}_{\left(a q\right)}^{2 +}\right]$ we can write:

${K}_{s p} = {s}^{2}$

So:

$s = \sqrt{{K}_{s p}} = \sqrt{2.4 \times {10}^{- 5}} = 4.9 \times {10}^{- 3} \text{mol/L}$

${M}_{r} = 136.1$

So solubility in $\text{g/L"=4.9xx10^(-3)xx136.1=0.66"g/L}$

The temperature was not given. It must always be specified for these type of questions involving $K$.

Sep 14, 2015

Molar solubility: $4.9 \cdot {10}^{- 3} \text{M}$

#### Explanation:

Calcium sulfate, ${\text{CaSO}}_{4}$, will not dissociate completely in aqueous solution to form calcium cations, ${\text{Ca}}^{2 +}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$.

What actually goes on is that the solid calcium sulfate will be in equilibrium with the amounts of ions that do dissolve. The position of this equilibrium, i.e. how much of the solid will actually dissolve to form ions in aqueous solution, is determined by the magnitude of the solubility product constant, ${K}_{s p}$.

The smaller the value of the ${K}_{s p}$, the less soluble will an ionic compound be.

To find the molar solubility of the calcium sulfate, use an ICE table

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

color(purple)("I")" " " " " " - " " " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" " " " " " - " " " " " " " "(+s)" " " " "(+s)
color(purple)("E")" " " " " " - " " " " " " " " " "s" " " " " " " "s

By definition, ${K}_{s p}$ is equal to

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

${K}_{s p} = s \cdot s = {s}^{2}$

This means that the molar solubility of calcium sulfate is

$s = \sqrt{{K}_{s p}} = \sqrt{2.4 \cdot {10}^{- 5}} = \textcolor{g r e e n}{4.9 \cdot {10}^{- 3} \text{M}}$

To get the mass solubility of calcium sulfate, use the compound's molar mass

4.9 * 10^(-3)color(red)(cancel(color(black)("moles")))/"L" * "136.141 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.67 g/L")#