The reaction between a sulphuric acid (H₂SO₄) solution and a sodium hydroxide (NaOH) solution is investigated using the apparatus illustrated below - NSC Physical Sciences - Question 7 - 2018 - Paper 2

The burette is used to measure the exact volume of NaOH solution needed to reach the endpoint of the titration, ensuring precise control over the addition of the base.

According to the Arrhenius theory, an acid is a substance that produces hydrogen ions (H⁺) or hydronium ions (H₃O⁺) when dissolved in water.

Sulphuric acid is considered a strong acid because it ionizes completely in aqueous solution, significantly increasing the concentration of H⁺ ions in the solution, leading to a very low pH.

The colour change will be from 'blue to yellow.'

To determine the volume of H₂SO₄ needed:

Using the equation: $c_a imes V_a = c_b imes V_b$ Where:

• $c_a$ = concentration of H₂SO₄ = 0.1 mol·dm⁻³
• $V_b$ = volume of NaOH = 25 cm³ = 0.025 dm³
• $c_b$ = concentration of NaOH = 0.1 mol·dm⁻³

Substituting into the formula: $0.1 imes V_a = 0.1 imes 0.025$ Thus, V_a = rac{0.1 imes 0.025}{0.1} = 0.025 ext{ dm}^3 = 25 ext{ cm}^3. Therefore, 25 cm³ of H₂SO₄ is needed for neutralisation.

After reaching the endpoint, if 5 cm³ of H₂SO₄ is added, the total volume in the flask becomes: $25 ext{ cm}^3 + 5 ext{ cm}^3 = 30 ext{ cm}^3$.

The amount of excess H₂SO₄ can be calculated using:

• $c_a = 0.1 ext{ mol/dm}^3$,
• $V_a = 0.005 ext{ dm}^3$,
• thus, total moles of H₂SO₄ added = $0.1 imes 0.005 = 0.0005$ moles.

The concentration of H⁺ ions in the final solution after the excess is: [H^+] = rac{0.0005}{0.03} ext{ mol/dm}^3 = 0.01667 ext{ mol/dm}^3.

Then, the pH is calculated as: $ext{pH} = - ext{log}[H^+] = - ext{log}(0.01667) \\ ext{which results in approximately} = 1.78.$ Therefore, the pH of the solution is approximately 1.78.