A mixture is prepared by adding $50.0$ mL of $0.200$ M $\ce{NaOH}$ to $75.0$ mL of $0.100$ M $\ce{NaOH}$. What is the $\[[OH^-]$ in the mixture?

$ \textbf{(A) }\text{0.0600 M}\qquad\textbf{(B) }\text{0.0800 M}\qquad\textbf{(C) }\text{0.140 M}\qquad\textbf{(D) }\text{0.233 M}\qquad$
1. Calculate the moles of $\ce{OH^-}$ from the $50.0$ mL of $0.200$ M $\ce{NaOH}$:
   \[
   \text{Moles of } \ce{OH^-} = 0.200 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{mol}
   \]

2. Calculate the moles of $\ce{OH^-}$ from the $75.0$ mL of $0.100$ M $\ce{NaOH}$:
   \[
   \text{Moles of } \ce{OH^-} = 0.100 \, \text{M} \times 0.075 \, \text{L} = 0.0075 \, \text{mol}
   \]

3. Add the moles of $\ce{OH^-}$ from both solutions to get the total moles of $\ce{OH^-}$:
   \[
   \text{Total moles of } \ce{OH^-} = 0.010 \, \text{mol} + 0.0075 \, \text{mol} = 0.0175 \, \text{mol}
   \]

4. Calculate the total volume of the mixture:
   \[
   \text{Total volume} = 0.050 \, \text{L} + 0.075 \, \text{L} = 0.125 \, \text{L}
   \]

5. Calculate the concentration of $\ce{OH^-}$ in the mixture:
   \[
   \left[ \ce{OH^-} \right] = \frac{0.0175 \, \text{mol}}{0.125 \, \text{L}} = 0.140 \, \text{M}
   \]

Conclusion:
\[
\boxed{0.140 \, \text{M}}
\]