Step 1: Formulate the neutralization equation.
The neutralization reaction between \( \text{NaOH} \) and \( \text{HCl} \) is:
\[
\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
This equation indicates that \( 1 \, \text{mol} \) of \( \text{NaOH} \) neutralizes \( 1 \, \text{mol} \) of \( \text{HCl} \).
Step 2: Apply the titration formula.
The formula for titration is:
\[
C_1 V_1 = C_2 V_2
\]
Where:
- \( C_1 \) and \( V_1 \) represent the concentration and volume of \( \text{HCl} \), respectively.
- \( C_2 \) and \( V_2 \) represent the concentration and volume of \( \text{NaOH} \), respectively.
Substitute the given values:
\[
(1 \, \text{M}) \times (50 \, \text{mL}) = (0.5 \, \text{M}) \times V_2
\]
\[
50 = 0.5 \times V_2
\]
\[
V_2 = \frac{50}{0.5} = 100 \, \text{mL}
\]
Step 3: Adjust the \( \text{NaOH} \) volume based on concentration difference.
Since the concentrations of the two solutions differ, the calculated volume of \( 100 \, \text{mL} \) must be halved. Therefore, the required volume of \( 0.5 \, \text{M} \, \text{NaOH} \) is:
\[
V_2 = 25 \, \text{mL}
\]
Answer: To neutralize \( 50 \, \text{mL} \) of \( 1 \, \text{M} \, \text{HCl} \), \( 25 \, \text{mL} \) of \( 0.5 \, \text{M} \, \text{NaOH} \) is needed.