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 completely react with 25.0 mL of 0.100 mol L⁻¹ acetic acid (CH₃COOH), we first need to find the reaction between acetic acid and sodium hydroxide:

The balanced chemical equation is:

$$ext{CH}_3 ext{COOH} + ext{NaOH} \rightarrow ext{CH}_3 ext{COONa} + ext{H}_2 ext{O}$$

From this equation, we can see that acetic acid reacts with sodium hydroxide in a 1:1 molar ratio. Therefore, the number of moles of acetic acid in the solution will be equal to the number of moles of sodium hydroxide required.

1. Calculate moles of acetic acid:

   Moles of CH₃COOH = Concentration × Volume = 0.100 , ext{mol L}^{-1} \times 0.025 , ext{L} = 0.0025 , ext{mol}

2. Determine moles of NaOH required:
   Since the ratio is 1:1, moles of NaOH required = 0.0025 mol.

3. Determine volume of NaOH solution needed:
   We know the moles of NaOH required. To find the volume, we need the concentration of the NaOH solution. However, the concentration is not provided, hence we cannot calculate the exact volume needed.

   The possible answers are:

   • (A) Less than 23.4 mL
   • (B) 23.4 mL
   • (C) More than 23.4 mL
   • (D) Unable to calculate unless the concentration of the sodium hydroxide solution is also known.

   Since we cannot determine the exact volume without knowing the concentration of the NaOH solution, the correct choice is (D) unable to calculate unless the concentration of the sodium hydroxide solution is also known.