7.1 Oxalic acid, (COOH)₂, ionises in two steps as shown below - NSC Physical Sciences - Question 7 - 2016 - Paper 2

The temperature is given because the value of the equilibrium constant (Kₐ) is dependent on temperature. Variations in temperature can affect the degree of dissociation of the acid, thereby altering the Kₐ value.

The formula for a substance that acts as an amphiprotic species in the reactions is H₂O (water).

The net equation for the ionisation of oxalic acid can be represented as:

$(COOH)₂(s) + 2H₂O(l) ⇌ (COO)₂²⁻(aq) + 2H⁺(aq)$

To calculate the mass of sodium hydroxide (NaOH) required, use the formula:

$n = c imes V$ Where:

• $n$ is the number of moles,
• $c$ is the concentration (1 mol·dm⁻³), and
• $V$ is the volume (40 cm³ = 0.04 L).

Thus, $n = 1 imes 0.04 = 0.04 \text{ mol}$

Next, calculate the mass using: $m = n \times M$ Where M (molar mass of NaOH) = 40 g/mol, $m = 0.04 \times 40 = 1.6 \text{ g}$

To calculate the initial number of moles of sulphuric acid (H₂SO₄) in the flask:

$n = c \times V$ Where:

• $c$ is the concentration (0.06 mol·dm⁻³), and
• $V$ is the volume (250 cm³ = 0.25 L).

Thus, $n = 0.06 \times 0.25 = 0.015 \text{ mol}$

After the reaction, to find the pH of the solution, first determine the moles of NaOH and H₂SO₄ after the reaction occurs:

Assuming NaOH is in excess, calculate the remaining moles:

Initial moles of NaOH = 0.04 mol, Initial moles of H₂SO₄ = 0.015 mol.

The reaction ratio is 2:1, thus:

• Moles of NaOH that reacts = $2 \times n(H₂SO₄) = 2 \times 0.015 = 0.03 \text{ mol}$
• Remaining moles of NaOH = $0.04 - 0.03 = 0.01 \text{ mol}$$

At this point, calculate the concentration of ext{OH}⁻ ions:

• Total volume = 250 cm³ + 40 cm³ = 290 cm³ = 0.29 L,
• Concentration of ext{OH}⁻ = $\frac{0.01}{0.29} \approx 0.0345 \text{ mol·dm}^{-3}$

Using the relation:

The term for the underlined phrase is 'endpoint'.

Apparatus X is typically a burette.

The purpose of the white tile is to provide a contrasting background that helps in easily observing the change in color of the indicator during the titration.

When the learner added three additional drops of acid, the amount of acid in the solution surpassed that needed to neutralize the base present. As a result, there was an excess of H⁺ ions in the solution, thus pushing the pH above 7, indicating an acidic solution. The balanced equation would illustrate the neutralization reaction and the subsequent shift in balance toward the acidic side as H⁺ ions increase.