A 25.0 mL sample of a 0.100 mol L^-1 hydrochloric acid solution completely reacted with 23.4 mL of sodium hydroxide solution - HSC - SSCE Chemistry - Question 17 - 2013 - Paper 1

To determine the volume of sodium hydroxide (NaOH) required to react with the acetic acid (CH₃COOH), we start by understanding the neutralization reaction.

1. Calculate moles of hydrochloric acid (HCl):

   Using the concentration (0.100 mol L^-1) and volume (25.0 mL = 0.025 L):

   $ext{Moles of HCl} = 0.100 imes 0.025 = 0.0025 \text{ moles}$

2. Determine moles of sodium hydroxide (NaOH) used:

   Since HCl and NaOH react in a 1:1 molar ratio, the moles of NaOH used is also 0.0025 moles.

3. Find the concentration of NaOH solution:

   The volume of NaOH solution used is 23.4 mL = 0.0234 L, therefore, the concentration of NaOH is:

   $C_{NaOH} = \frac{0.0025}{0.0234} \approx 0.1064 \text{ mol L}^{-1}$

4. Calculate moles of acetic acid (CH₃COOH):

   For the acetic acid, using the same process, we have:

   $ext{Moles of CH₃COOH} = 0.100 imes 0.025 = 0.0025 \text{ moles}$

5. Determine the volume of NaOH required for acetic acid:

   Since acetic acid also reacts in a 1:1 ratio with NaOH, the moles of NaOH needed is also 0.0025 moles:

   $V_{NaOH} = \frac{0.0025 \text{ moles}}{0.1064 \text{ mol L}^{-1}} \approx 0.0235 \text{ L} = 23.5 \text{ mL}$

The final step indicates the volume of NaOH required to neutralize the acetic acid is approximately 23.5 mL, which is greater than the 23.4 mL of sodium hydroxide used for the HCl. Therefore, the correct answer is (C) More than 23.4 mL.