In a titration of 25.0 mL of 0.100 M HCl with NaOH, what volume of 0.100 M NaOH is required to reach the equivalence point?

• A. 100.0 mL
• B. 50.0 mL
• C. 25.0 mL
• D. 12.5 mL

Answer: (C)

Verified step by step guidance

1. Identify the reaction between HCl and NaOH, which is a strong acid-strong base neutralization: \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}\).
2. Calculate the moles of HCl initially present using the formula \(\text{moles} = M \times V\), where \(M\) is molarity and \(V\) is volume in liters. So, \(\text{moles of HCl} = 0.100\,M \times 0.0250\,L\).
3. At the equivalence point, moles of NaOH added equal moles of HCl initially present, because they react in a 1:1 molar ratio.
4. Use the moles of NaOH needed and its molarity to find the volume required: \(V_{\mathrm{NaOH}} = \frac{\text{moles of NaOH}}{M_{\mathrm{NaOH}}}\).
5. Convert the volume of NaOH from liters to milliliters by multiplying by 1000 to get the final answer in mL.