Question

Calculate the pOH of a 0.410 M Ba(OH)2 solution.

Answer 1

The dissociation of this basic compound is:
`Ba(OH)_2\ \to\ Ba^(2+) + 2OH^-`

There will be the double of
`OH^-` ions than molecules of
`Ba(OH)_2`, therefore, the concentration of
`OH^-` at the end will be 0.820 M.


`pOH = -log\ [OH^-] = - log (0.820) = \bf 0.09`