7.1 Ethanoic acid (CH₃COOH) is an ingredient of household vinegar - NSC Physical Sciences - Question 7 - 2020 - Paper 2

To calculate the concentration of hydronium ions, we first use the formula for pH:

$pH = - ext{log}[H₃O⁺]$

Rearranging provides:

$[H₃O⁺] = 10^{-pH}$

Substituting the pH value:

$[H₃O⁺] = 10^{-3.85} = 1.41 imes 10^{-4} ext{ mol·dm}^{-3}$

The pH of a sodium ethanoate solution will be GREATER THAN 7. This is because sodium ethanoate is the salt of a weak acid and a strong base (sodium hydroxide), leading to a basic solution. The hydrolysis of the acetate ion (CH₃COO⁻) generates hydroxide ions (OH⁻), which increase the pH above 7.

When sodium ethanoate (CH₃COONa) dissolves in water, it dissociates into sodium ions (Na⁺) and acetate ions (CH₃COO⁻). The acetate ions can react with water as follows:

$CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)$

This reaction produces hydroxide ions (OH⁻), which results in a pH greater than 7.

Given the percentage concentration of ethanoic acid in vinegar, we can calculate the mass in 25 cm³ (0.025 dm³):

ext{Mass of } CH₃COOH = 4.52 ext{%} imes 1 ext{ g/cm}³ imes 25 ext{ cm}³ = 1.13 ext{ g}

Using the molar mass of CH₃COOH (60 g/mol), the number of moles is:

n(CH₃COOH) = rac{1.13 ext{ g}}{60 ext{ g/mol}} = 0.01883 ext{ mol}

After reacting with NaOH, the moles of unreacted CH₃COOH is:

$n_{ ext{unreacted}} = n(CH₃COOH) - n(NaOH) = 0.01883 ext{ mol} - 0.0145 ext{ mol} = 0.0043 ext{ mol}$

First, calculate the moles of calcium carbonate that reacted using the balanced equation:

$CaCO₃ + 2CH₃COOH → Ca(CH₃COO)₂ + H₂O + CO₂$

Given the mass of CaCO₃ is 1.2 g, and using its molar mass (100.09 g/mol):

n(CaCO₃) = rac{1.2 ext{ g}}{100.09 ext{ g/mol}} = 0.0120 ext{ mol}

Now, we need to find the total mass of the impure sample:
The percentage of calcium carbonate in the impure sample is:

ext{Percentage} = rac{0.0120 ext{ mol} imes 100.09 ext{ g/mol}}{1.2 ext{ g}} imes 100 = 10 ext{%}