50 mL water is mixed with 50 mL 0.01 M Ba( | vedclass

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(A) Step $1$: Calculate the total volume of the solution. $V_{total} = 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. Step $2$: Calculate the moles of $Ba(O

Vedclass · Accepted Answer

$12.0$

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(A) Step $1$: Calculate the total volume of the solution. $V_{total} = 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. Step $2$: Calculate the moles of $Ba(OH)_2$. $n = M 	imes V = 0.01 \ mol/L 	imes 0.05 \ L = 5 	imes 10^{-4} \ mol$. Step $3$: Calculate the concentration of $OH^-$ ions. Since $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$,the moles of $OH^-$ are $2 	imes (5 	imes 10^{-4}) = 10^{-3} \ mol$. $[OH^-] = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$. Step $4$: Calculate $pOH$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. Step $5$: Calculate $pH$. $pH = 14 - pOH = 14 - 2 = 12.0$.

$50 \ mL$ water is mixed with $50 \ mL$ $0.01 \ M$ $Ba(OH)_2$. Calculate its $pH$.

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