This question is about pH - AQA - A-Level Chemistry - Question 6 - 2021 - Paper 1
Question 6
This question is about pH.
Pure water dissociates slightly.
H2O(l) ⇌ H+(aq) + OH−(aq)
ΔHf = +57 kJ mol⁻¹
The equilibrium constant, K_w = [H⁺][OH⁻] / [H2O]
T... show full transcript
Worked Solution & Example Answer:This question is about pH - AQA - A-Level Chemistry - Question 6 - 2021 - Paper 1
Step 1
Explain why [H2O] is not shown in the K_w expression.
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Answer
[H2O] is not included in the K_w expression because it is in a pure liquid state and its concentration remains essentially constant. In a dilute solution, the change in [H2O] is negligible, allowing it to be treated as a constant.
Step 2
Explain why the value of K_w increases as the temperature increases.
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The value of K_w increases with temperature because the dissociation of water is an endothermic reaction. As temperature rises, the system shifts to favor the production of H⁺ and OH⁻ ions to absorb the additional heat, thus increasing K_w.
Step 3
Give the expression for pH.
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The expression for pH is given by:
pH = -log[H⁺]
Step 4
Calculate the pH of pure water at 50 °C.
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At 50 °C, K_w is approximately 5.55 × 10⁻¹⁴. Since [H⁺] is equal to [OH⁻], we can use the formula:
K_w = [H⁺]²
Thus, [H⁺] = √(K_w) = √(5.55 × 10⁻¹⁴) ≈ 7.43 × 10⁻⁷ mol/dm³.
Using this to calculate pH:
pH = -log(7.43 × 10⁻⁷) ≈ 6.13.
Step 5
Explain why water is neutral at 50 °C.
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Water is neutral at 50 °C because in pure water at this temperature, the concentrations of H⁺ and OH⁻ ions are equal (both being 7.43 × 10⁻⁷ mol/dm³), hence pH = 7.00, indicating neutrality.
Step 6
Use Figure 3 to give the true pH value when the pH meter reading is 5.6.
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Referencing Figure 3, when the pH meter reading is 5.6, the true pH value corresponds to a reading of approximately 5.5.
Step 7
Suggest why the pH probe is washed with distilled water between each of the calibration measurements.
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The pH probe is washed with distilled water between calibrations to prevent cross-contamination of solutions which could lead to inaccurate pH readings.
Step 8
Explain why the volume of sodium hydroxide solution added between each pH measurement is smaller as the end point of the titration is approached.
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As the end point is approached, the pH of the solution changes more rapidly with smaller additions of sodium hydroxide. Therefore, to accurately detect the endpoint and prevent overshooting, smaller volumes are added.
Step 9
State why all three of the indicators in Table 6 are suitable for this titration.
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All three indicators are suitable because they each change color within the pH range expected during the titration of hydrochloric acid with sodium hydroxide, allowing clear visibility of the end point.
Step 10
Calculate the pH of the final solution at 25 °C.
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To find the final concentration of HCl after the addition of NaOH, calculate:
Moles of HCl = 0.150 mol/dm³ × 0.025 dm³ = 0.00375 mol.
Moles of NaOH = 0.200 mol/dm³ × 0.03625 dm³ = 0.00725 mol.
HCl reacts with NaOH in a 1:1 ratio, so after the reaction, the remaining moles of NaOH = 0.00725 - 0.00375 = 0.0035 mol.
Total volume = 25.00 cm³ + 36.25 cm³ = 61.25 cm³ = 0.06125 dm³.
The final concentration of NaOH = 0.0035 mol / 0.06125 dm³ ≈ 0.0571 mol/dm³.
The pH is calculated as:
pOH = -log(0.0571) ≈ 1.24
Thus, pH = 14 - pOH ≈ 12.76.
