AQA A-Level Chemistry Paper 3, 2023: Question 6
7 marks · Easy difficulty · State/Explain/Numerical
Calcium Hydroxide and pH Calculation This question involves verifying that calcium hydroxide is in excess when reacting with hydrochloric acid and calculating the pH of a saturated solution of calcium hydroxide using solubility and the ionic product of water Kw
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Question text
06 Calcium hydroxide is almost insoluble in water, but it reacts with
dilute hydrochloric acid.
Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
A student adds 100 cm3 of 0.100 mol dm–3 hydrochloric acid to
0.600 g of solid calcium hydroxide.
06.1 Show, by calculation, that the calcium hydroxide is in excess.
[2 marks]
06.2 The final mixture contains a saturated solution of Ca(OH)2 at 293 K
At 293 K
• the solubility of Ca(OH) in this solution is 0.400 g dm–3
• K = 6.80 × 10–15 mol2 dm–6
w
Calculate the pH of this solution.
Give your answer to two decimal places.
[5 marks]
pH
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Question Answers Additional comments/Guidelines Mark
6.1 Alternative:
M1 amount of HCl = 0.010 mol AND
0.6 M1 amount of HCl = 0.010 mol
amount Ca(OH)2 (= ) = 0.00810
74.1 amount of Ca(OH) needed = 0.0050 mol
NB There must be an indication that 0.005 is the
M2 0.00810 mol of Ca(OH)2 requires 0.0162 mol HCl OR amount of Ca(OH)2 needed and not just
0.01/2
0.01 mol of HCl requires 0.005 mol of Ca(OH)2
M2 mass of Ca(OH)2 needed = 0.0050 x 74.1 =
NB There must be an indication that 0.005 is the amount of
0.3705 g (so 0.6 g is excess)
Ca(OH)2 needed and not just 0.01/2
6.2 –3 –3 Correct answer to 2 dp = 5 marks
M1 0.400 g dm means 0.400 ÷ 74.1 (= 0.00540 mol dm )
ALLOW 12.21 for 5 marks
M2 [OH–] = M1 x 2 (= 0.0108 mol dm–3)
M3 [H+] = 6.80 x 10–15 ÷ M2 (= 6.30 x 10–13 mol dm–3) 5
M4 pH = –log[H+] M4 ALLOW if calculation shown containing a
number that has been calculated as [H+]
M5 = –log M3 with answer to 2dp (= 12.20)
How to answer it
Step-by-Step Guide: Calcium Hydroxide and pH Calculation
Question (a): Show Calcium Hydroxide is in Excess
Step 1: Calculate the Moles of HCl
Moles of HCl = concentration × volume
Calculation:
- Concentration of HCl = 0.100 mol dm-3
- Volume of HCl = 100 cm3 = 0.100 dm3
- Moles of HCl = 0.100 × 0.100 = 0.010 mol
Step 2: Calculate the Moles of Ca(OH)2
Moles of Ca(OH)2 = mass ÷ molar mass
Calculation:
- Mass of Ca(OH)2 = 0.600 g
- Molar mass of Ca(OH)2 = 74.1 g mol-1
- Moles of Ca(OH)2 = 0.600 ÷ 74.1 = 0.00810 mol
Step 3: Check Reaction Requirements
1 mole of Ca(OH)2 reacts with 2 moles of HCl.
Moles of Ca(OH)2 required = 0.010 ÷ 2 = 0.0050 mol
Available Ca(OH)2 = 0.00810 mol, so it is in excess .
Question (b): Calculate the pH of the Solution
Step 1: Determine the Moles of Dissolved Ca(OH)2
Solubility of Ca(OH)2 = 0.400 g dm-3.
Moles of Ca(OH)2 = solubility ÷ molar mass
Calculation:
- Moles of Ca(OH)2 = 0.400 ÷ 74.1 = 0.00540 mol dm-3
Step 2: Calculate [OH-]
1 mole of Ca(OH)2 produces 2 moles of OH-.
Calculation:
- [OH-] = 0.00540 × 2 = 0.0108 mol dm-3
Step 3: Calculate [H+]
Kw = [H+] × [OH-]
Calculation:
- Kw = 6.80 × 10-15
- [H+] = Kw ÷ [OH-]
- [H+] = 6.80 × 10-15 ÷ 0.0108 = 6.30 × 10-13 mol dm-3
Step 4: Calculate pH
pH = -log[H+]
Calculation:
- pH = -log(6.30 × 10-13)
- pH = 12.20
Topics
Physical Chemistry · 3.1.12 Acids and Bases · 3.1.2 Amount of Substance
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