Based on the titration, the concentration of C₃H₆O₂(COOH) in the solution was: A - VCE - SSCE Chemistry - Question 23 - 2020 - Paper 1

The volume of NaOH used in the titration is 24.0 mL, which is equal to 0.024 L. Assuming the concentration of NaOH is known (let's say 0.1 M for this example), the moles of NaOH used can be calculated as:

$n = C \times V = 0.1 \, \text{mol/L} \times 0.024 \, \text{L} = 0.0024 \, \text{mol}$

Since the moles of C₃H₆O₂(COOH) are equal to the moles of NaOH used (due to the 1:1 ratio), we have:

$n_{C₃H₆O₂(COOH)} = 0.0024 \, \text{mol}$

To find the concentration, we also need the volume of the acid solution (let's assume it’s 0.300 L). The concentration can be calculated as:

$C = \frac{n}{V} = \frac{0.0024 \, \text{mol}}{0.300 \, \text{L}} = 0.0080 \, \text{mol/L} = 8.0 \times 10^{-3} \, M$

Based on the calculation above, the concentration of C₃H₆O₂(COOH) in the solution is:

A. 8.0 × 10⁻³ M