Ethanoic acid (CH₃COOH) is a weak acid in water - VCE - SSCE Chemistry - Question 7 - 2005 - Paper 1

To calculate the hydrogen ion concentration, we can use the formula relating pH to hydrogen ion concentration:

$[H^+] = 10^{-pH}$

Substituting the given pH:

$[H^+] = 10^{-2.88} \approx 1.32 \times 10^{-3} \text{ M}$

The acidity constant (Kₐ) is calculated using the expression:

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Substituting the known values:

• Letting ([CH_3COOH] \approx 0.100 ext{ M} - [H^+] \approx 0.100 - 1.32 \times 10^{-3} \approx 0.100)

Kₐ calculation:

$K_a = \frac{(1.32 \times 10^{-3})(1.32 \times 10^{-3})}{0.100} = \frac{(1.74 \times 10^{-6})}{0.100} = 1.74 \times 10^{-5}$

Since methanoic acid (HCOOH) has a higher acidity constant than ethanoic acid (CH₃COOH), it dissociates more in solution. Therefore, the 0.10 M solution of ethanoic acid will have a higher pH compared to the 0.10 M solution of methanoic acid.

The solution of ethanoic acid would require a greater volume of NaOH to neutralise because it has a lower acidity constant, meaning that it produces fewer hydrogen ions compared to the stronger methanoic acid. Thus, more moles of NaOH are needed to neutralise the weaker acid.