50 mL of water is added to 50 mL of 0.01 M | vedclass

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(A) $1$. Initial moles of $Ba(OH)_2 = 	ext{Molarity} 	imes 	ext{Volume (L)} = 0.01 \ M 	imes 0.050 \ L = 5 	imes 10^{-4} \ mol$. $2$. Since $Ba(O

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$12.0$

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(A) $1$. Initial moles of $Ba(OH)_2 = 	ext{Molarity} 	imes 	ext{Volume (L)} = 0.01 \ M 	imes 0.050 \ L = 5 	imes 10^{-4} \ mol$. $2$. Since $Ba(OH)_2$ is a strong base,it dissociates as $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. $3$. Moles of $OH^- = 2 	imes 	ext{moles of } Ba(OH)_2 = 2 	imes 5 	imes 10^{-4} = 10^{-3} \ mol$. $4$. Total volume of the solution after adding water = $50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. $5$. Concentration of $OH^- = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$. $6$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. $7$. $pH = 14 - pOH = 14 - 2 = 12.0$.

$50 \ mL$ of water is added to $50 \ mL$ of $0.01 \ M$ $Ba(OH)_2$ solution. Calculate the $pH$ of the resulting solution.

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