Assuming that $B a {(O H)}_{2}$ is completely ionized in aqueous solution under the given conditions the concentration of $H_{3} O^{+}$ ions in 0.005M aqueous solution of $B a {(O H)}_{2}$ at 298K is $\times 10^{- 12} m o l L^{- 1}$

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Detailed Solution

$B a {(O H)}_{2} \to B a^{2 +} + 2 O H^{-} W e k n o w [H^{+}] [O H^{-}] = 10^{- 14} [H_{3} O^{+}] [O H^{-}] = 10^{- 14} [H_{3} O^{+}] = \frac{10^{- 14}}{10^{- 2}} = 10^{- 12}$

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