This question is about pH - AQA - A-Level Chemistry - Question 6 - 2021 - Paper 1

The value of Kw increases with temperature because the dissociation of water is an endothermic process. As temperature rises, the equilibrium shifts to the right, producing more H⁺ and OH⁻ ions.

The expression for pH is given by: pH = -\log[H^+]

At 50 °C, using the value of Kw from Table 5, we have: Kw = [H⁺][OH⁻] = 5.48 × 10⁻¹⁴mol² dm⁻⁶. Since [H⁺] = [OH⁻] in pure water, we can set [H⁺] = [OH⁻] = x:

$x^2 = 5.48 × 10^{-14}$

Thus, $x = \sqrt{5.48 × 10^{-14}} \approx 7.39 × 10^{-7} mol/dm³$. Now, calculating pH: pH = -\log(7.39 × 10^{-7}) \approx 6.13.

Water is considered neutral at 50 °C because at this temperature, [H⁺] and [OH⁻] are equal, meaning the solution maintains a pH of approximately 6.13, reflecting neither acidic nor basic conditions.

From Figure 3, a pH meter reading of 5.6 corresponds to a true pH value of approximately 5.5.

The pH probe is washed with distilled water to avoid contamination from previous solutions, which could affect the accuracy of the pH readings.

As the endpoint of the titration is approached, the solution's pH changes more rapidly with the addition of NaOH. Therefore, smaller volumes must be added to avoid overshooting the endpoint.

All three indicators change colour within the pH range of the titration, providing clear visual cues at the endpoint, which is crucial for accurate determination.

Total moles of HCl initially = concentration × volume = 0.150 mol/dm³ × 0.025 dm³ = 0.00375 mol. Total moles of NaOH added = 0.200 mol/dm³ × (36.25/1000 dm³) = 0.00725 mol.

Moles of NaOH remaining = 0.00725 - 0.00375 = 0.0035 mol. Final total volume = 25.00 cm³ + 36.25 cm³ = 61.25 cm³ = 0.06125 dm³.

Concentration of excess NaOH = $\frac{0.0035 mol}{0.06125 dm³} \approx 0.0571 mol/dm³$. Therefore, the pOH = -\log(0.0571) \approx 1.24, pH = 14 - pOH = 14 - 1.24 = 12.76.