Consider the titration of 100.0 mL of 0.40 M HCl with 0.40 M NaOH. If sodium hydroxide is the titrant, which volume would place it in excess? 

• A. 70.0 mL
• B. 25.0 mL
• C. 100.0 mL
• D. 110.0 mL
• E. 9.0 mL

Answer: (D)

Verified step by step guidance

1. Identify the reaction: HCl (acid) reacts with NaOH (base) in a neutralization reaction: \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}\).
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. For HCl: \(\text{moles}_{\mathrm{HCl}} = 0.40\,M \times 0.100\,L\).
3. Determine the volume of NaOH needed to exactly neutralize the HCl by using the stoichiometry of the reaction (1:1 mole ratio). Use \(\text{moles}_{\mathrm{NaOH}} = \text{moles}_{\mathrm{HCl}}\) and calculate the volume of NaOH: \(V_{\mathrm{NaOH}} = \frac{\text{moles}_{\mathrm{HCl}}}{0.40\,M}\).
4. To have NaOH in excess, the volume of NaOH added must be greater than the volume calculated for neutralization. So, compare the given volumes to this neutralization volume to find which volume exceeds it.
5. Select the volume from the options that is greater than the neutralization volume, indicating that NaOH is in excess.