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 find the volume of sodium hydroxide (NaOH) needed to react with the acetic acid (CH₃COOH), we first need to analyze the neutralization reaction. The balanced chemical equation for the reaction between acetic acid and sodium hydroxide is as follows:

$\text{CH}_3\text{COOH}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{CH}_3\text{COONa}_{(aq)} + \text{H}_2\text{O}_{(l)}$

This shows that acetic acid reacts with sodium hydroxide in a 1:1 molar ratio. Therefore, to calculate the moles of acetic acid in 25.0 mL of a 0.100 mol L$^{-1}$ solution:

$\text{Moles of CH}_3\text{COOH} = 0.100 \text{ mol L}^{-1} \times 0.025 \text{ L} = 0.0025 \text{ mol}$

Since one mole of acetic acid reacts with one mole of sodium hydroxide, the same number of moles of NaOH will be required:

$\text{Moles of NaOH required} = 0.0025 \text{ mol}$

To find the volume of NaOH needed, we need the molarity of the sodium hydroxide solution. However, from the information given, we know that 23.4 mL of NaOH completely reacted with 25.0 mL of HCl at the same concentration. Thus, the NaOH also has a concentration of 0.100 mol L$^{-1}$.

Now we can calculate the volume of NaOH required:

$\text{Volume of NaOH} = \frac{\text{moles of NaOH}}{\text{Concentration of NaOH}} = \frac{0.0025 \text{ mol}}{0.100 \text{ mol L}^{-1}} = 0.025 \text{ L} = 25.0 \text{ mL}$

Since 25.0 mL > 23.4 mL, the answer to the question is (C) More than 23.4 mL.