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

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This question is about pH. Pure water dissociates slightly. H2O(l) ⇌ H+(aq) + OH-(aq) ΔH = +57 kJ mol^-1 The equilibrium constant, Kc = [H+][OH-]/[H2O] The ioni... 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 Kc expression.

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Answer

[H2O] is considered to be almost constant in dilute solutions. The concentration of water does not change significantly during the dissociation process and therefore is not included in the equilibrium expression.

Step 2

Explain why the value of Kc increases as the temperature increases.

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Answer

Kc increases with temperature because the dissociation of water is an endothermic reaction. An increase in temperature supplies more energy, shifting the equilibrium to the right, thus producing more H+ and OH- ions.

Step 3

Give the expression for pH.

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Answer

pH is defined as: pH = -log[H+]

Step 4

Calculate the pH of pure water at 50 °C. Give your answer to 2 decimal places.

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Answer

At 50 °C, Kw = 5.48 × 10^-14 mol² dm^-6. Therefore, [H+] = ext{sqrt}(Kw) = ext{sqrt}(5.48 × 10^-14)

Which gives [H+] ≈ 7.40 × 10^-7 mol dm^-3. Thus, pH ≈ -log(7.40 × 10^-7) ≈ 6.13.

Step 5

Explain why water is neutral at 50 °C.

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Answer

Water is neutral because at equilibrium, the concentrations of H+ and OH- ions are equal. Thus, [H+] = [OH-] at 50 °C, confirming that pure water remains neutral.

Step 6

Use Figure 3 to give the true pH value when the pH meter reading is 5.6.

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Answer

From Figure 3, when the pH meter reading is 5.6, the true pH value is approximately 6.25.

Step 7

Suggest why the pH probe is washed with distilled water between each of the calibration measurements.

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Answer

Washing the pH probe with distilled water prevents contamination from the previous solution, ensuring that the subsequent measurements are accurate.

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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Answer

As the titration nears the endpoint, the solution's pH approaches neutrality. Hence, small additions of NaOH lead to significant changes in pH, requiring smaller volumes to accurately monitor these changes.

Step 9

State why all three of the indicators in Table 6 are suitable for this titration.

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Answer

All three indicators in Table 6 have pH ranges that include the expected pH change during the titration, allowing them to effectively signal the end point.

Step 10

Calculate the pH of the final solution at 25°C.

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Answer

To determine the final pH after titration:

  1. Calculate moles of HCl:
    Moles of HCl = Concentration × Volume = 0.150 mol dm^-3 × 0.025 dm³ = 0.00375 mol.

  2. Calculate moles of NaOH:
    Moles of NaOH = 0.200 mol dm^-3 × 0.03625 dm³ = 0.00725 mol.

  3. Since NaOH is in excess, moles of NaOH remaining = 0.00725 - 0.00375 = 0.0035 mol.

  4. Total volume after addition = 25 cm³ + 36.25 cm³ = 61.25 cm³ = 0.06125 dm³.

  5. Concentration of OH- = 0.0035 mol / 0.06125 dm³ = 0.0571 mol dm^-3.

  6. Calculate pOH:
    pOH = -log[OH-] ≈ -log(0.0571) ≈ 1.24.

  7. Finally, calculate pH:
    pH = 14 - pOH = 14 - 1.24 = 12.76.

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