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Question

(C) Barium hydroxide is a strong base and dissociates completely as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. Initial concentration of $Ba(OH)_2 = 1

Vedclass · Accepted Answer

$11.0$

Vedclass · Answer

(C) Barium hydroxide is a strong base and dissociates completely as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. Initial concentration of $Ba(OH)_2 = 1 	imes 10^{-3} \ M$. Initial concentration of $OH^- = 2 	imes 1 	imes 10^{-3} = 2 	imes 10^{-3} \ M$. Upon adding $50 \ mL$ of $H_2O$ to $50 \ mL$ of the solution, the total volume becomes $100 \ mL$. The new concentration of $OH^-$ is calculated using the dilution formula $M_1V_1 = M_2V_2$: $(2 	imes 10^{-3} \ M) 	imes (50 \ mL) = M_2 	imes (100 \ mL)$. $M_2 = \frac{2 	imes 10^{-3} 	imes 50}{100} = 1 	imes 10^{-3} \ M$. $pOH = -\log[OH^-] = -\log(1 	imes 10^{-3}) = 3$. $pH = 14 - pOH = 14 - 3 = 11.0$.

$50 \ mL$ 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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