Grade 11 Physical Sciences · Chemical Change

Calculate and interpret pH

Calculate pH with pH = −log[H₃O⁺], or use Kw to find a concentration (a [H₃O⁺] of 0,01 mol/dm³ gives pH 2).

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Question 1.8 · November 2023 · Paper 2 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10).

1.8 The apparatus in the diagram below is used for the titration between HCl(aq) and KOH(aq).

In a titration, the learner accidentally exceeds the endpoint. Which ONE of the following will be TRUE for the titration mixture?(2) Diagram for question 1.8

A $[H^{+}] > [OH^{-}]$ and pH $< 7$
B $[H^{+}] < [OH^{-}]$ and pH $< 7$
C $[H^{+}] < [OH^{-}]$ and pH $> 7$
D $[H^{+}] > [OH^{-}]$ and pH $> 7$

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

More past paper questions: calculate and interpret ph

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3. Question 7.2.1 · November 2025 · Paper 2 · National (DBE) · 4 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:
7.2.1 Final concentration of the hydroxide ions in the flask(4)
Show the memo (4 ticks)Memo
1. ✓
  (a) Formula $pH = - lo g [H_{3} O^{+}]$
  1 mark · Calculate and interpret pH
2. ✓
  (b) Substitute 12,62 $12, 62 = - lo g [H_{3} O^{+}]$
  $[H_{3} O^{+}] = 2, 4 \times 1 0^{- 13}$ mol·dm $^{- 3}$
  1 mark · Calculate and interpret pH
3. ✓
  (c) Substitute in Kw $[H_{3} O^{+}] [OH^{-}] = 1 0^{- 14}$
  $(2, 4 \times 1 0^{- 13}) [OH^{-}] = 1 \times 1 0^{- 14}$
  1 mark · Calculate and interpret pH
4. ✓
  (d) Final answer $[OH^{-}] = 0, 0417$ mol·dm $^{- 3}$ ( $0, 04$ )
  1 mark · Calculate and interpret pH
4. Question 7.2.1 · May/June 2022 · Paper 2 · National (DBE) · 3 marks
7.2 Learners add $150$ cm $^{3}$ of a sodium hydroxide solution, NaOH, of unknown concentration to $200$ cm $^{3}$ of a $0, 03$ mol·dm $^{- 3}$ hydrochloric acid solution, HC $ℓ$ , as illustrated below. They find that the pH of the final solution is 2.

Assume that the volumes are additive.

The balanced equation for the reaction is:

$HC ℓ (aq) + NaOH(aq) \to NaC ℓ (aq) + H_{2} O (ℓ)$

Calculate the:
7.2.1 Concentration of the $H_{3} O^{+}$ ions in the final solution(3)
Show the memo (3 ticks)Memo
1. ✓
  Formula $pH = - lo g [H_{3} O^{+}]$ OR $[H_{3} O^{+}] = 1 0^{- pH}$
  1 mark · Calculate and interpret pH
2. ✓
  Substitution $2 = - lo g [H_{3} O^{+}]$
  1 mark · Calculate and interpret pH
3. ✓
  Answer with unit $[H_{3} O^{+}] = 0, 01$ mol·dm $^{- 3}$ ( $1 \times 1 0^{- 2}$ mol·dm $^{- 3}$ )
  1 mark · Calculate and interpret pH
5. Question 7.2.2 · November 2022 · Paper 2 · National (DBE) · 4 marks
7.2 A flask contains $300$ cm $^{3}$ of dilute sodium hydroxide, NaOH(aq), of concentration $0, 167$ mol·dm $^{- 3}$ .

Answer the questions that follow.
Ethanoic acid of volume $500$ cm $^{3}$ and of unknown concentration, X, is now added to this flask to give a solution of volume $800$ cm $^{3}$ .

It is found that the pH of the mixture is $11, 4$ .

The balanced equation for the reaction is:

$NaOH(aq) + CH_{3} COOH(aq) \to CH_{3} COONa(aq) + H_{2} O (ℓ)$

Calculate the:
7.2.2 Concentration of the $OH^{-}$ (aq) in the mixture(4)
Show the memo (4 ticks)Memo
1. ✓
  Any formula $pH = - lo g [H_{3} O^{+}]$
  1 mark · Calculate and interpret pH
2. ✓
  Substitute 11,4 $11, 4 = - lo g [H_{3} O^{+}]$
  $[H_{3} O^{+}] = 1 0^{- 11, 4} = 3, 98 \times 1 0^{- 12}$ mol·dm $^{- 3}$
  1 mark · Calculate and interpret pH
3. ✓
  Substitute into Kw $[H_{3} O^{+}] [OH^{-}] = 1 \times 1 0^{- 14}$
  $(3, 98 \times 1 0^{- 12}) [OH^{-}] = 1 \times 1 0^{- 14}$
  1 mark · Calculate and interpret pH
4. ✓
  Final answer $[OH^{-}] = 2, 51 \times 1 0^{- 3}$ mol·dm $^{- 3}$ $(0, 003)$
  1 mark · Calculate and interpret pH
6. Question 7.2.2 · November 2023 · Paper 2 · National (DBE) · 5 marks
To identify metal M in an unknown metal carbonate, $MCO_{3}$ , the following procedure is carried out:

Step 1: $0, 198$ g of IMPURE $MCO_{3}$ is reacted with $25$ cm $^{3}$ of $0, 4$ mol·dm $^{- 3}$ nitric acid, $HNO_{3}$ (aq).

Step 2: The EXCESS $HNO_{3}$ (aq) is then neutralised with $20$ cm $^{3}$ of $0, 15$ mol·dm $^{- 3}$ barium hydroxide, $Ba(OH)_{2}$ (aq).

Assume that the volumes are additive.

The following reactions take place:

$2 HNO_{3} (aq) + MCO_{3} (s) \to M(NO_{3})_{2} (aq) + CO_{2} (g) + H_{2} O (ℓ)$

$2 HNO_{3} (aq) + Ba(OH)_{2} (aq) \to Ba(NO_{3})_{2} (aq) + 2 H_{2} O (ℓ)$
7.2 Calculate the:
7.2.2 pH of the solution after Step 1(5)
Show the memo (5 ticks)Memo
1. ✓
  Use ratio $n (HNO_{3})_{reacted} = 2 n (Ba(OH)_{2}) = 2 (0, 003) = 0, 006$ mol
  1 mark · Use the mole ratio
2. ✓
  Substitute n and 0,025 dm³ in c = n/V $n (H_{3} O^{+}) = n (HNO_{3}) = 0, 006$ mol
  $[H_{3} O^{+}] = \frac{n}{V} = \frac{0 , 006}{0 , 025} = 0, 24$ mol·dm $^{- 3}$
  1 mark · Calculate concentration of a solution
3. ✓
  FormulapH $= - lo g [H_{3} O^{+}]$
  1 mark · Calculate and interpret pH
4. ✓
  Substitution $= - lo g (0, 24)$
  1 mark · Calculate and interpret pH
5. ✓
  Final answerpH $= 0, 62$
  1 mark · Calculate and interpret pH

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