Assuming that Ba(OH)_{2} is completely ionise | vedclass

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(B) $Ba(OH)_{2}$ is a strong base and dissociates completely as: $Ba(OH)_{2} \rightarrow Ba^{2+} + 2OH^{-}$.Given concentration of $Ba(OH)_{2} = 0.005

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(B) $Ba(OH)_{2}$ is a strong base and dissociates completely as: $Ba(OH)_{2} ightarrow Ba^{2+} + 2OH^{-}$.Given concentration of $Ba(OH)_{2} = 0.005 \, M$.Therefore,$[OH^{-}] = 2 	imes 0.005 \, M = 0.01 \, M = 10^{-2} \, M$.At $298 \, K$,the ionic product of water is $K_{w} = [H_{3}O^{+}][OH^{-}] = 10^{-14}$.Substituting the value of $[OH^{-}]$:$[H_{3}O^{+}] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \, M$.Thus,the concentration is $1 	imes 10^{-12} \, mol \, L^{-1}$.

Assuming that $Ba(OH)_{2}$ is completely ionised in aqueous solution under the given conditions,the concentration of $H_{3}O^{+}$ ions in $0.005 \, M$ aqueous solution of $Ba(OH)_{2}$ at $298 \, K$ is $..... \times 10^{-12} \, mol \, L^{-1}$. (Nearest integer)

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