What is the [ OH^- ]in the final solution prepared by mixing 20.0 ~mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba( OH)_2 ? -

$20 \mathrm{~mL}$20 mL of $0.50 \mathrm{M} \mathrm{HCl}=20 \times 0.050 \mathrm{~m} \mathrm{~mol}$0.50MHCl=20×0.050 m mol $=1.0 \mathrm{~m} \mathrm{~mol}=1.0$=1.0 m mol=1.0 meq. of $\mathrm{HCl}$HCl $30 \mathrm{~mL}$30 mL of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$0.10MBa(OH)2​ $$\begin{aligned} & =30 \times 0.1 \mathrm{~m} \mathrm{~mol} \\ & =3 \mathrm{~m} \mathrm{~mol}=3 \times 2 \mathrm{meq} \\ & =6 \mathrm{meq} \mathrm{Ba}(\mathrm{OH})_2 \end{aligned}$$​=30×0.1 m mol=3 m mol=3×2meq=6meqBa(OH)2​​ 1 meq of $\mathrm{HCl}$HCl will neutralize 1 meq of $\mathrm{Ba}(\mathrm{OH})_2$Ba(OH)2​ $$\mathrm{Ba}(\mathrm{OH})_2 \text { left }=5 \text { meq. }$$Ba(OH)2​ left =5 meq.  Tatal volume $=50 \mathrm{~mL}$=50 mL $\mathrm{Ba}(\mathrm{OH})_2$Ba(OH)2​ conc. in final solution $$=\frac{5}{50} \mathrm{~N}=0.1 \mathrm{~N}=0.05 \mathrm{M}$$=505​ N=0.1 N=0.05M $$\left[\mathrm{OH}^{-}\right]=2 \times 0.05 \mathrm{M}=0.10 \mathrm{M}$$[OH−]=2×0.05M=0.10M Alternatively, $$\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}$$Ba(OH)2​+2HCl→BaCl2​+2H2​O $2 \mathrm{~m} \mathrm{~mol}$2 m mol of $\mathrm{HCl}$HCl neutralize $1 \mathrm{~m}$1 m mole of $\mathrm{Ba}(\mathrm{OH})_2$Ba(OH)2​ $1 \mathrm{~m} \mathrm{~mol}$1 m mol of $\mathrm{HCl}$HCl neutralize $0.5 \mathrm{~m} \mathrm{~mol} \mathrm{of}$0.5 m molof $\mathrm{Ba}(\mathrm{OH})_2$Ba(OH)2​ $\mathrm{Ba}(\mathrm{OH})_2 1 \mathrm{eft}=3-0.5 \mathrm{~m} \mathrm{~mol}=2.5 \mathrm{mmol}$Ba(OH)2​1eft=3−0.5 m mol=2.5mmol $$\left[\mathrm{Ba}(\mathrm{OH})_2\right]=\frac{2.5}{50} \mathrm{M}=0.05 \mathrm{M}$$[Ba(OH)2​]=502.5​M=0.05M or $[\mathrm{OH}]^{-}=2 \times 0.05=0.1 \mathrm{M}$[OH]−=2×0.05=0.1M