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(C) Barium hydroxide is a strong base and dissociates as follows: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.Initially,the volume is $50 \ mL$ and the con

Vedclass · Accepted Answer

$11.0$

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(C) Barium hydroxide is a strong base and dissociates as follows: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$.Initially,the volume is $50 \ mL$ and the concentration is $1 	imes 10^{-3} \ M$.Upon adding $50 \ mL$ of $H_2O$,the total volume becomes $100 \ mL$.Since the volume is doubled,the concentration of $Ba(OH)_2$ is halved: $M_2 = \frac{M_1 V_1}{V_2} = \frac{1 	imes 10^{-3} \ M 	imes 50 \ mL}{100 \ mL} = 0.5 	imes 10^{-3} \ M$.Since each mole of $Ba(OH)_2$ produces $2$ moles of $OH^-$,the concentration of $OH^-$ is: $[OH^-] = 2 	imes 0.5 	imes 10^{-3} \ M = 1 	imes 10^{-3} \ M$.$pOH = -\log[OH^-] = -\log(1 	imes 10^{-3}) = 3$.$pH = 14 - pOH = 14 - 3 = 11.0$.

$50 \ mL$ of $H_2O$ is added to $50 \ mL$ of $1 \times 10^{-3} \ M$ barium hydroxide solution. What is the $pH$ of the resulting solution?

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