Question

What volume of \( 0.5 \, \text{M} \, \text{NaOH} \) is required to neutralize \( 50 \, \text{mL} \) of \( 1 \, \text{M} \, \text{HCl} \)?

• A. \( 50 \, \text{mL} \)
• B. \( 100 \, \text{mL} \)
• C. \( 200 \, \text{mL} \)
• D. \( 25 \, \text{mL} \)

Hint:

For neutralization, use the formula \( C_1 V_1 = C_2 V_2 \), where \( C_1 \) and \( V_1 \) are the concentration and volume of the acid, and \( C_2 \) and \( V_2 \) are the concentration and volume of the base.

Answer 1

The Correct Option is D

Step 1: Write the neutralization equation. The neutralization reaction between \( \text{NaOH} \) and \( \text{HCl} \) is: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] From the equation, we see that \( 1 \, \text{mol} \) of \( \text{NaOH} \) neutralizes \( 1 \, \text{mol} \) of \( \text{HCl} \). Step 2: Use the formula for titration. The formula for titration is: \[ C_1 V_1 = C_2 V_2 \] Where: - \( C_1 \) and \( V_1 \) are the concentration and volume of \( \text{HCl} \), - \( C_2 \) and \( V_2 \) are the concentration and volume of \( \text{NaOH} \). Substitute the known 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: Check the volume of \( \text{NaOH} \). We need to use half the volume of \( 100 \, \text{mL} \), as the concentrations are different for both solutions. Therefore, the correct volume of \( 0.5 \, \text{M} \, \text{NaOH} \) is: \[ V_2 = 25 \, \text{mL} \] Answer: Therefore, \( 25 \, \text{mL} \) of \( 0.5 \, \text{M} \, \text{NaOH} \) is required to neutralize \( 50 \, \text{mL} \) of \( 1 \, \text{M} \, \text{HCl} \).