What is the pH of a solution in which 10.0 m | vedclass

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(C) $1$. Calculate the moles of $Sr(OH)_2$: $n(Sr(OH)_2) = M 	imes V = 0.010 \ M 	imes 0.010 \ L = 1.0 	imes 10^{-4} \ mol$. $2$. Since $Sr(OH)_2$

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$11.70$

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(C) $1$. Calculate the moles of $Sr(OH)_2$: $n(Sr(OH)_2) = M 	imes V = 0.010 \ M 	imes 0.010 \ L = 1.0 	imes 10^{-4} \ mol$. $2$. Since $Sr(OH)_2$ is a strong base,it dissociates as $Sr(OH)_2 ightarrow Sr^{2+} + 2OH^-$. $3$. Moles of $OH^-$ produced: $n(OH^-) = 2 	imes n(Sr(OH)_2) = 2 	imes 1.0 	imes 10^{-4} = 2.0 	imes 10^{-4} \ mol$. $4$. Calculate the moles of $HCl$: $n(HCl) = M 	imes V = 0.010 \ M 	imes 0.010 \ L = 1.0 	imes 10^{-4} \ mol$. $5$. Moles of $H^+$ produced: $n(H^+) = 1.0 	imes 10^{-4} \ mol$. $6$. Neutralization reaction: $H^+ + OH^- ightarrow H_2O$. $7$. Remaining moles of $OH^-$: $n(OH^-)_{rem} = 2.0 	imes 10^{-4} - 1.0 	imes 10^{-4} = 1.0 	imes 10^{-4} \ mol$. $8$. Total volume of the solution: $10.0 \ mL + 10.0 \ mL = 20.0 \ mL = 0.020 \ L$. $9$. Concentration of $OH^-$: $[OH^-] = \frac{1.0 	imes 10^{-4} \ mol}{0.020 \ L} = 5.0 	imes 10^{-3} \ M$. $10$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5.0 	imes 10^{-3}) = 3 - \log(5) = 3 - 0.699 = 2.301$. $11$. Calculate $pH$: $pH = 14 - pOH = 14 - 2.301 = 11.699 \approx 11.70$.

What is the $pH$ of a solution in which $10.0 \ mL$ of $0.010 \ M$ $Sr(OH)_2$ is added to $10.0 \ mL$ of $0.010 \ M$ $HCl$?

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