# Question cb573

Jul 5, 2017

${K}_{s p} = 2.84 \cdot {10}^{- 4}$

#### Explanation:

Start by using the molar mass of the salt to convert its gram solubility to a molar solubility, i.e. the maximum number of moles of salt that can be dissolved per liter of water in order to produce a saturated solution.

8.90 color(white)(.)color(red)(cancel(color(black)("g")))/"L" * "1 mole AB"_2/(215color(red)(cancel(color(black)("g")))) = "0.041395 mol L"^(-1)

Now, the solubility product constant, ${K}_{s p}$ is essentially a measure of the degree of dissociation that a salt undergoes when dissolved in water.

The higher the value of the solubility product constant, the higher the molar solubility of the salt, i.e. the higher the concentrations of the dissolved ions in the solution.

In your case, you know that ${\text{AB}}_{2}$ dissociates according to the following dissociation equilibrium

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

By definition, the solubility product constant is equal to

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

Now, you know that the molar solubility of the salt is equal to ${\text{0.041395 mol L}}^{- 1}$ at ${25}^{\circ} \text{C}$.

As you can see, for every $1$ mole of salt that dissociates you get $1$ mole of ${\text{A}}^{2 +}$ cations and $\textcolor{red}{2}$ moles of ${\text{B}}^{-}$ anions.

This means that, at equilibrium, you will have

["A"^(2+)] = "0.041395 mol L"^(-1)

["B"^(-)] = color(red)(2) * "0.041395 mol L"^(-1)#

This means that the solubility product constant for this salt is equal to--I'll calculate it without added units

${K}_{s p} = 0.041395 \cdot {\left(\textcolor{red}{2} \cdot 0.041395\right)}^{\textcolor{red}{2}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.84 \cdot {10}^{- 4}}}}$

The answer is rounded to three sig figs.