Question

What is the concentration of \( \text{NaOH} \) in a solution if \( 25.0 \, \text{mL} \) of \( 0.100 \, \text{M} \) \( \text{HCl} \) is neutralized by \( 50.0 \, \text{mL} \) of \( \text{NaOH} \)?

• A. \( 0.05 \, \text{M} \)
• B. \( 0.10 \, \text{M} \)
• C. \( 0.20 \, \text{M} \)
• D. \( 0.25 \, \text{M} \)

Hint:

For neutralization reactions, the number of moles of acid equals the number of moles of base. Use \( M_1 V_1 = M_2 V_2 \) to calculate the concentration of the unknown solution.

Answer 1

The Correct Option is B

The neutralization reaction between \( \text{HCl} \) and \( \text{NaOH} \) is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] The number of moles of \( \text{HCl} \) is: \[ n_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} = 0.100 \, \text{M} \times 0.0250 \, \text{L} = 0.00250 \, \text{mol} \] Since the mole ratio between \( \text{HCl} \) and \( \text{NaOH} \) is 1:1, the moles of \( \text{NaOH} \) required for neutralization are also \( 0.00250 \, \text{mol} \). Now, calculate the molarity of \( \text{NaOH} \): \[ M_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} = \frac{0.00250 \, \text{mol}}{0.0500 \, \text{L}} = 0.0500 \, \text{M} \] Thus, the concentration of \( \text{NaOH} \) is \( 0.10 \, \text{M} \).