Question #c50da

1 Answer
Nov 24, 2016

Answer:

#5.57 * 10^(-8)"M"#

Explanation:

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

#"Ba"("OH")_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#

Notice that for every #1# mole of barium hydroxide that dissociates, you get #1# mole of barium cations and #color(red)(2)# moles of hydroxide anions.

By definition, the solubility product constant, #K_(sp)#, for this equilibrium looks like this

#K_(sp) = ["Ba"^(2+)] * ["OH"^(-)]^color(red)(2)#

Now, the equilibrium concentration of hydroxide anions in a saturated barium hydroxide solution will always by #color(red)(2)# times that of the barium cations, as given by the aforementioned #1:color(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

#["OH"^(-)] = color(red)(2) xx ["Ba"^(2+)] = color(red)(2)s#

If you plug these into the expression for the #K_(sp)#, you will end up with

#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

#K_(sp) = 4s^3#

This will get you

#s = root(3)(K_(sp)/4)#

Plug in your value to find

#s = root(3)((6.92 * 10^(-22))/4) = color(darkgreen)(ul(color(black)(5.57 * 10^(-8)"M")))#

The answer must be rounded to three sig figs.