alex198 alex198

05-05-2016
Chemistry

Answered

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

Verified Answer :

EjerciciosFyQ EjerciciosFyQ

08-05-2016

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.

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

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