Question #c50da
1 Answer
Explanation:
Start by looking at the dissociation equilibrium for barium hydroxide,
#"Ba"("OH")_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#
Notice that for every
By definition, the solubility product constant,
#K_(sp) = ["Ba"^(2+)] * ["OH"^(-)]^color(red)(2)#
Now, the equilibrium concentration of hydroxide anions in a saturated barium hydroxide solution will always by
This means that if you take
#["OH"^(-)] = color(red)(2) xx ["Ba"^(2+)] = color(red)(2)s#
If you plug these into the expression for the
#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.