What is the $\left[\mathrm{OH}^{-}\right]$in the final solution prepared by mixing $20.0 \mathrm{~mL}$ of $0…
What is the $\left[\mathrm{OH}^{-}\right]$in the final solution prepared by mixing $20.0 \mathrm{~mL}$ of $0.050 \mathrm{M} \mathrm{HCl}$ with $30.0 \mathrm{~mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$ ?
$0.10 \mathrm{M}$
$0.40 \mathrm{M}$
$0.0050 \mathrm{M}$
$0.12 \mathrm{M}$
Solution
Number of milliequivalents of $\mathrm{HCl}$
$=20 \times 0.050 \times 1=1$
Number of milliequivalents of $\mathrm{Ba}(\mathrm{OH})_2$
$=2 \times 30 \times 0.10=6$
$\left[\mathrm{OH}^{-}\right]$of final solution
$\begin{aligned}
& \text { milliequivalents of } \mathrm{Ba}(\mathrm{OH})_2 \\
& =\frac{- \text { milliequivalents of } \mathrm{HCl}}{\text { total volume }} \\
& =\frac{6-1}{50}=0.1 \mathrm{M} \\
&
\end{aligned}$