# Question c50da

Nov 24, 2016

$5.57 \cdot {10}^{- 8} \text{M}$

#### Explanation:

Start by looking at the dissociation equilibrium for barium hydroxide, "Ba"("OH")_2#

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

Notice that for every $1$ mole of barium hydroxide that dissociates, you get $1$ mole of barium cations and $\textcolor{red}{2}$ moles of hydroxide anions.

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

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

Now, the equilibrium concentration of hydroxide anions in a saturated barium hydroxide solution will always by $\textcolor{red}{2}$ times that of the barium cations, as given by the aforementioned $1 : \textcolor{red}{2}$ mole ratio that exists between the two ions in solution.

This means that if you take $s$ to be the equilibrium concentration of the barium cations, you will have

$\left[{\text{OH"^(-)] = color(red)(2) xx ["Ba}}^{2 +}\right] = \textcolor{red}{2} s$

If you plug these into the expression for the ${K}_{s p}$, you will end up with

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

${K}_{s p} = 4 {s}^{3}$

This will get you

$s = \sqrt[3]{{K}_{s p} / 4}$

Plug in your value to find

$s = \sqrt[3]{\frac{6.92 \cdot {10}^{- 22}}{4}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.57 \cdot {10}^{- 8} \text{M}}}}$

The answer must be rounded to three sig figs.