# Question 9f499

Nov 13, 2016

${K}_{\text{sp}} = 1.5 \cdot {10}^{- 12}$

#### Explanation:

Your starting point here is the pH of the solution. More specifically, you need to use the given pH to determine the concentration of hydroxide anions, ${\text{OH}}^{-}$, present in the saturated solution.

As you know, an aqueous solution kept at room temperature has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH " + " pOH} = 14}}}$

This means that your solution has

$\text{pOH} = 14 - 10.16 = 3.84$

Now, the pOH of the solution gives you its concentration of hydroxide anions

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))

To find the concentration of hydroxide anions, rearrange the equation as

log(["OH"^(-))] = - "pOH"

10^log(["OH"^(-)]) = 10^(-"pOH")

$\left[\text{OH"^(-)] = 10^(-"pOH}\right)$

Plug in your value to find

["OH"^(-)] = 10^(-3.84) = 1.445 * 10^(-4)"M"

Now, your unknown salt only paritally dissolves i nwater, meaning that an equilibrium is established between the dissolves ions and the undissolved solid

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

Notice that every mole of salt that dissociates produces $1$ mole of ${\text{M}}^{2 +}$ cations and $\textcolor{red}{2}$ moles of ${\text{OH}}^{-}$ anions.

This means that the equilibrium concentration of hydroxide anions will be twice as high as that of the metal cations. Therefore, you can say that

["M"^(2+)] = (["OH"^(-)])/color(red)(2)

["M"^(2+)] = (1.445 * 10^(-4)"M")/color(red)(2) = 7.225 * 10^(-5)"M"

Finally, the solubility product constant, ${K}_{s p}$, is equal to

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

Plug in your values to find

K_(sp) = 7.225 * 10^(-5)"M" * (1.445 * 10^(-4)"M")^color(red)(2)#

${K}_{s p} = 1.509 \cdot {10}^{- 12} {\text{M}}^{3}$

Rounded to two sig figs, the number of decimal places you have for the pH of the solution, and express without added units, the ${K}_{S p}$ will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{\text{sp}} = 1.5 \cdot {10}^{- 12}}}}$