The number of moles of hydroxide (OH^{-}) ion | vedclass

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(B) The dissociation of $Ba(OH)_2$ is given by: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^{-}$.Since $1 \ mol$ of $Ba(OH)_2$ produces $2 \ mol$ of $OH^{-}$

Vedclass · Accepted Answer

$0.0030$

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(B) The dissociation of $Ba(OH)_2$ is given by: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^{-}$.Since $1 \ mol$ of $Ba(OH)_2$ produces $2 \ mol$ of $OH^{-}$ ions,the concentration of $OH^{-}$ ions is $[OH^{-}] = 2 	imes [Ba(OH)_2] = 2 	imes 0.005 \ M = 0.01 \ M$.The number of moles of $OH^{-}$ ions is calculated as: $	ext{Moles} = 	ext{Molarity} 	imes 	ext{Volume (in L)}$.$	ext{Moles of } OH^{-} = 0.01 \ M 	imes 0.3 \ L = 0.0030 \ mol$.

The number of moles of hydroxide $(OH^{-})$ ion in $0.3 \ L$ of $0.005 \ M$ solution of $Ba(OH)_2$ is

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