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