Grade 11 Physical Sciences · Chemical Change

Do titration calculations

Use titration results to find an unknown concentration or volume with the mole ratio, for example cₐVₐ/c_bV_b = nₐ/n_b.

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Question 7.5 · November 2024 · Paper 2 · National (DBE) · 5 marks

Hydrated potassium carbonate,

K_{2} CO_{3} \cdot x H_{2} O

, is a WEAK BASE. A solution is prepared by dissolving some of this solid in water.

7.5 Calculate the concentration of the

K_{2} CO_{3}

solution.(5)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: do titration calculations

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3. Question 7.2.1 · May/June 2023 · Paper 2 · National (DBE) · 3 marks
7.2 A hydrochloric acid solution, HCℓ(aq), is standardised by titrating it against $25$ cm $^{3}$ of a $0, 04$ mol·dm $^{- 3}$ sodium carbonate solution $Na_{2} CO_{3}$ (aq). At the endpoint, it is found that $19, 5$ cm $^{3}$ of HCℓ(aq) has reacted.

The balanced equation for the reaction is:

$Na_{2} CO_{3} (aq) + 2 HCl(aq) \to 2 NaCl(aq) + CO_{2} (g) + H_{2} O (ℓ)$

Answer the questions that follow.
7.2.1 Calculate the concentration of the HCℓ(aq).(3)
Show the memo (3 ticks)Memo
1. ✓
  Substitute 0,04 mol·dm⁻³, 25 cm³ and 19,5 cm³ $\frac{c _{a} V _{a}}{c _{b} V _{b}} = \frac{n _{a}}{n _{b}}$
  $\frac{c _{a} ( 19 , 5 )}{( 0 , 04 ) ( 25 )} = ...$
  1 mark · Do titration calculations
2. ✓
  Use mol ratio n(Na2CO3) : n(HCℓ) = 1 : 2 $\frac{c _{a} ( 19 , 5 )}{( 0 , 04 ) ( 25 )} = \frac{2}{1}$
  1 mark · Use the mole ratio
3. ✓
  Final answer $c_{a} = 0, 10$ mol·dm $^{- 3}$ ( $0, 103$ )
  1 mark · Do titration calculations
4. Question WE5 · Siyavula Grade 11 Physical Sciences, 8.1 Gases and solutions, Worked example 5 · 2 marks
WE5 Given the equation:
$NaOH (aq) + HCl (aq) \to NaCl (aq) + H_{2} O(ℓ)$
$25$ cm $^{3}$ of a $0, 2$ mol·dm $^{- 3}$ hydrochloric acid solution was pipetted into a conical flask and titrated with sodium hydroxide. It was found that $15$ cm $^{3}$ of the sodium hydroxide was needed to neutralise the acid. Calculate the concentration of the sodium hydroxide.(2)
Show the memo (2 ticks)Memo
1. ✓
  Titration formula and substitution $V_{NaOH} = \frac{15}{1 000} = 0, 015$ dm $^{3}$ ; $V_{HCl} = \frac{25}{1 000} = 0, 025$ dm $^{3}$
  $\frac{c _{A} V _{A}}{a} = \frac{c _{B} V _{B}}{b}$
  $\frac{( 0 , 2 ) ( 0 , 025 )}{1} = \frac{( c _{NaOH} ) ( 0 , 015 )}{1}$
  1 mark · Do titration calculations
2. ✓
  Answer with unit $c_{NaOH} = 0, 33$ mol·dm $^{- 3}$
  1 mark · Do titration calculations
5. Question WE6 · Siyavula Grade 11 Physical Sciences, 8.1 Gases and solutions, Worked example 6 · 6 marks
WE6 $4, 9$ g of sulfuric acid is dissolved in water and the final solution has a volume of $220$ cm $^{3}$ . Using an acid-base titration, it was found that $20$ cm $^{3}$ of this solution was able to completely neutralise $10$ cm $^{3}$ of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in mol·dm $^{- 3}$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Reactants and products $H_{2} SO_{4} (aq) + NaOH(aq) \to Na_{2} SO_{4} (aq) + H_{2} O(ℓ)$
  1 mark · Write acid-base reaction equations
2. ✓
  Balancing $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(ℓ)$
  1 mark · Balance a chemical equation
3. ✓
  Moles of sulfuric acid $n = \frac{m}{M} = \frac{4 , 9}{98 , 12} = 0, 0499...$ mol
  1 mark · Calculate number of moles
4. ✓
  Concentration of the acid $V = \frac{220}{1 000} = 0, 22$ dm $^{3}$
  $c = \frac{n}{V} = \frac{0 , 0499...}{0 , 22} = 0, 226...$ mol·dm $^{- 3}$
  1 mark · Calculate concentration of a solution
5. ✓
  Titration substitutionOnly $20$ cm $^{3}$ ( $0, 02$ dm $^{3}$ ) of the acid is used.
  $\frac{c _{a} V _{a}}{n _{a}} = \frac{c _{b} V _{b}}{n _{b}}$
  $\frac{( 0 , 226... ) ( 0 , 02 )}{1} = \frac{( c _{NaOH} ) ( 0 , 01 )}{2}$
  1 mark · Do titration calculations
6. ✓
  Answer with unit $c_{NaOH} = 0, 908$ mol·dm $^{- 3}$
  1 mark · Do titration calculations
6. Question 7.2.2 · November 2025 · Paper 2 · National (DBE) · 8 marks
7.2 Barium hydroxide, $Ba(OH)_{2}$ , dissolves in water according to the balanced equation:

$Ba(OH)_{2} (s) \to Ba^{2 +} (aq) + 2 OH^{-} (aq)$

A $100$ cm $^{3}$ solution is prepared by dissolving an unknown amount of $Ba(OH)_{2}$ at $25$ °C.

$25$ cm $^{3}$ of this $Ba(OH)_{2}$ solution is reacted with $15$ cm $^{3}$ of a $0, 2$ mol·dm $^{- 3}$ HC $ℓ$ solution in a flask, according to the balanced equation:

$Ba(OH)_{2} (aq) + 2 HC ℓ (aq) \to BaC ℓ_{2} (aq) + 2 H_{2} O (ℓ)$

The final pH of the solution is $12, 62$ at $25$ °C.

Calculate the:
7.2.2 Number of moles of $Ba(OH)_{2}$ used to prepare the $100$ cm $^{3}$ solution(8)
Show the memo (7 ticks)Memo
1. ✓
  (a) Substitute 0,2 and 0,015 in n = cV $n (HC ℓ) = c V = (0, 2) (0, 015) = 3 \times 1 0^{- 3}$ mol
  1 mark · Calculate concentration of a solution
2. ✓
  (b) Ratio OH- : HCl = 1 : 1 $n (OH^{-}) = n (HC ℓ) = 3 \times 1 0^{- 3}$ mol (reacted)
  1 mark · Use the mole ratio
3. ✓
  (c) Substitute c(OH-) and 0,04 in n = cV $n (OH^{-})_{final} = c V = (0, 0417) (0, 04) = 1, 67 \times 1 0^{- 3}$ mol
  1 mark · Calculate concentration of a solution
4. ✓
  (d) Initial moles of OH- = reacted + left over $n (OH^{-})_{ini} = 3 \times 1 0^{- 3} + 1, 67 \times 1 0^{- 3} = 4, 67 \times 1 0^{- 3}$ mol
  2 marks · Identify the limiting reagent
5. ✓
  (e) Ratio Ba(OH)2 : OH- = 1 : 2 $n (Ba(OH)_{2}) = \frac{1}{2} n (OH^{-})_{ini} = \frac{1}{2} (4, 67 \times 1 0^{- 3}) = 2, 33 \times 1 0^{- 3}$ mol in $25$ cm $^{3}$
  1 mark · Use the mole ratio
6. ✓
  (f) Scale up to 100 cm3In $100$ cm $^{3}$ : $(4) (2, 33 \times 1 0^{- 3})$
  1 mark · Do titration calculations
7. ✓
  (g) Final answer $= 9, 34 \times 1 0^{- 3}$ mol
  1 mark · Do titration calculations

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