7.1 Ethanoic acid is a weak acid that reacts with water according to the following balanced equation: CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) 7.1.1 Define an acid in terms of the Lowry-Brønsted theory - NSC Physical Sciences - Question 7 - 2022 - Paper 2

Ethanoic acid is classified as a weak acid because it only partially ionizes in water. This means that not all of the acid molecules dissociate to release H⁺ ions, resulting in a lower concentration of hydronium ions in solution.

The two bases in the equation are:

1. CH₃COO⁻ (the acetate ion)
2. H₂O (water)

To calculate the number of moles of sodium hydroxide:

1. Use the formula for moles: $n = c imes V$ where:

   • $n$ is the number of moles,
   • $c$ is the concentration (0.167 mol·dm⁻³), and
   • $V$ is the volume in dm³ (300 cm³ = 0.300 dm³).

2. Calculation: $n = 0.167 imes 0.300 = 0.0501 ext{ mol}$

The pH of the mixture is given as 11.4. To find the concentration of hydroxide ions:

1. Calculate the pOH, using: $pOH = 14 - pH \Rightarrow pOH = 14 - 11.4 = 2.6$
2. Use the pOH to find the concentration of OH⁻: $[OH⁻] = 10^{-pOH} = 10^{-2.6} \approx 2.51 \times 10^{-3} ext{ mol·dm}⁻³$

Using the balanced equation and the information from previous calculations, we can find the initial concentration of ethanoic acid:

1. The total number of moles of NaOH reacted can be found: $n(NaOH)_{initial} = 0.00251 ext{ mol} \text{ (from above)}$ $n(NaOH)_{reacted} = n(NaOH)_{initial} - n(NaOH)_{remaining} = 0.00251 - 0.0002 = 0.00231 ext{ mol}$
2. Given that 500 cm³ of ethanoic acid was added, convert to dm³: $500 ext{ cm}³ = 0.500 ext{ dm}³$
3. Therefore, concentration of ethanoic acid: $c(CH₃COOH) = \frac{n(CH₃COOH)}{V} = \frac{0.00231}{0.500} = 0.096 ext{ mol·dm}⁻³$