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 $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$ ? -
$0.12 \mathrm{M}$
$0.10 \mathrm{M}$
$0.40 \mathrm{M}$
$0.0050 \mathrm{M}$
Solution
$20 \mathrm{~mL}$ of $0.50 \mathrm{M} \mathrm{HCl}=20 \times 0.050 \mathrm{~m} \mathrm{~mol}$ $=1.0 \mathrm{~m} \mathrm{~mol}=1.0$ meq. of $\mathrm{HCl}$
$30 \mathrm{~mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$
$\begin{aligned}
& =30 \times 0.1 \mathrm{~m} \mathrm{~mol} \\
& =3 \mathrm{~m} \mathrm{~mol}=3 \times 2 \mathrm{meq} \\
& =6 \mathrm{meq} \mathrm{Ba}(\mathrm{OH})_2
\end{aligned}$
1 meq of $\mathrm{HCl}$ will neutralize 1 meq of $\mathrm{Ba}(\mathrm{OH})_2$
$\mathrm{Ba}(\mathrm{OH})_2 \text { left }=5 \text { meq. }$
Total volume $=50 \mathrm{~mL}$ $\mathrm{Ba}(\mathrm{OH})_2$ conc. in final solution
$=\frac{5}{50} \mathrm{~N}=0.1 \mathrm{~N}=0.05 \mathrm{M}$
$\left[\mathrm{OH}^{-}\right]=2 \times 0.05 \mathrm{M}=0.10 \mathrm{M}$
Alternatively,
$\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}$
$2 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $1 \mathrm{~m}$ mole of $\mathrm{Ba}(\mathrm{OH})_2$
$1 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $0.5 \mathrm{~m} \mathrm{~mol} \mathrm{of}$ $\mathrm{Ba}(\mathrm{OH})_2$
$\mathrm{Ba}(\mathrm{OH})_2 1 \mathrm{eft}=3-0.5 \mathrm{~m} \mathrm{~mol}=2.5 \mathrm{mmol}$
$\left[\mathrm{Ba}(\mathrm{OH})_2\right]=\frac{2.5}{50} \mathrm{M}=0.05 \mathrm{M}$
or $[\mathrm{OH}]^{-}=2 \times 0.05=0.1 \mathrm{M}$