DP Chemistry (last assessment 2024)

Question 23M.2.HL.TZ2.5a

Date	May 2023	Marks available	[Maximum mark: 8]	Reference code	23M.2.HL.TZ2.5a
Level	HL	Paper	2	Time zone	TZ2
Command term	Calculate, Determine, Identify, Write	Question number	a	Adapted from	N/A

[Maximum mark: 8]

23M.2.HL.TZ2.5a

(a)

The concentration of methanoic acid was found by titration with a 0.200 mol dm⁻³ standard solution of sodium hydroxide, NaOH (aq), using an indicator to determine the end point.

(a.i)

Calculate the pH of the sodium hydroxide solution.

[2]

Markscheme

«[OH⁻] = 0.200 mol dm⁻³»

ALTERNATIVE 1:
«pOH = −log₁₀(0.200) =» 0.699 ✓
«pH = 14.000 − 0.699 =» 13.301 ✓

ALTERNATIVE 2:
«[H⁺] = $\frac{1.00 \times 10^{- 14}}{0.200}$ = » 5.00 × 10⁻¹⁴ «mol dm⁻³» ✓
«pH = −log₁₀(5.00 × 10⁻¹⁴)» = 13.301 ✓

Award [2] for correct final answer.

(a.ii)

Write an equation for the reaction of methanoic acid with sodium hydroxide.

[1]

Markscheme

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H₂O(l) ✓

Accept ionic equation or net ionic equation.

(a.iii)

22.5 cm³ of NaOH(aq) neutralized 25.0 cm³ of methanoic acid. Determine the concentration of the methanoic acid.

[1]

Markscheme

«n(acid) = n(OH⁻)»
«[acid] = $\frac{0.200 m o l d m^{- 3} \times 22.5 \times 10^{- 3} d m^{3}}{25.0 \times 10^{- 3} d m^{3}}$ » = 0.180 «mol dm⁻³» ✓

(a.iv)

Calculate the pH of the original solution of methanoic acid. Use your answer to (a)(iii) and section 21 of the data booklet. If you did not get an answer to (a)(iii) use 0.300 mol dm⁻³, but this is not the correct answer.

[2]

Markscheme

ALTERNATIVE 1:
«K_a = 10^−3.75 = 1.78 × 10⁻⁴ »
[H⁺] = $\sqrt{K_{a} \times [HCOOH]}$ / $\sqrt{1.78 \times 10^{- 4} \times 0.180}$ / 5.66 × 10⁻³ «mol dm⁻³» ✓
pH «= −log₁₀ (5.66 × 10⁻³)» = 2.247 ✓