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

1. Moles of ${\text{H}}^{+}$ from $\text{HCl}=0.05\times 0.02=0.001\text{mol}$.2. Moles of ${\text{OH}}^{-}$ from ${\text{Ba(OH)}}_2=2\times 0.1\times 0.03=0.006\text{mol}$ (Note the factor of 2).3. Net moles of ${\text{OH}}^{-}$ after neutralization $=0.006-0.001=0.005\text{mol}$.4. Total volume $=20+30=50\text{mL}=0.05\text{L}$.5. Final concentration $[{\text{OH}}^{-}]=\frac{0.005}{0.05}=0.10\text{M}$.