# Question 7445a

Apr 17, 2017

${K}_{s p} = 1.97 \cdot {10}^{- 3}$

#### Explanation:

You know that ${\text{AB}}_{2}$ is insoluble in water, which implies that an equilibrium is established when you dissolve this salt in aqueous solution

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

By definition, the solubility product constant, ${K}_{s p}$, for this equilibrium takes the form

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

Right from the start, you should be able to say that

${K}_{s p} < 1$

This means that the equilibrium lies to the left, which is why the salt is considered to be insoluble in water.

Now, the expression of the solubility product constant uses the equilibrium concentrations of the dissolved ions.

You know that at ${25}^{\circ} \text{C}$, only $0.0790$ moles of ${\text{AB}}_{2}$ dissolve in $\text{1.00 L}$ of water to produce ions. Notice that every mole of ${\text{AB}}_{2}$ that dissolves produces $1$ mole of ${\text{A}}^{2 +}$ cations and $2$ moles of ${\text{B}}^{-}$ anions.

This means that the solution will contain

$\text{0.0790 moles A"^(2+)" }$ and ${\text{ "2 xx "0.0790 moles B}}^{-}$

The molarities of the dissolved ions will be equal to

["A"^(2+)] = "0.0790 moles"/"1.00 L" = "0.0790 M"

["B"^(-)] = (2 xx "0.0790 moles")/"1.00 L" = "0.158 M"

This means that the solubility product constant will be equal to

K_(sp) = "0.0790 M" * ("0.158 M")^2#

${K}_{s p} = 1.97 \cdot {10}^{- 3}$ ${\text{M}}^{3}$

We usually express the solubility product constant without added units

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{s p} = 1.97 \cdot {10}^{- 3}}}}$

The answer is rounded to three sig figs.