To find the final $[OH^-]$ concentration, we first need to calculate the milli-moles of $H^+$ and $OH^-$ ions. Milli-moles of $H^+$ from HCl = Molarity $\times$ Volume = $0.050 \text{ M} \times 20.0 \text{ mL} = 1.0 \text{ mmol}$. Since each $Ba(OH)_2$ provides two $OH^-$ ions, milli-moles of $OH^-$ from $Ba(OH)_2$ = $2 \times \text{Molarity} \times \text{Volume}$ = $2 \times 0.10 \text{ M} \times 30.0 \text{ mL} = 6.0 \text{ mmol}$. When mixed, $H^+$ will neutralize $OH^-$. The net milli-moles of $OH^-$ remaining = $6.0 \text{ mmol} - 1.0 \text{ mmol} = 5.0 \text{ mmol}$. The total volume of the mixture = $20.0 \text{ mL} + 30.0 \text{ mL} = 50.0 \text{ mL}$. Therefore, the final concentration of $[OH^-]$ = $\frac{\text{Net milli-moles of } OH^-}{\text{Total Volume}} = \frac{5.0 \text{ mmol}}{50.0 \text{ mL}} = 0.10 \text{ M}$.