DP Chemistry (first assessment 2025)

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Question 23M.2.SL.TZ2.5

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Date May 2023 Marks available [Maximum mark: 4] Reference code 23M.2.SL.TZ2.5
Level SL Paper 2 Time zone TZ2
Command term Calculate, Determine, Write Question number 5 Adapted from N/A
5.
[Maximum mark: 4]
23M.2.SL.TZ2.5

Methanoic acid is a monoprotic weak acid.

(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.

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+] = 1.00x10-140.200 = » 5.00 × 10−14 «mol dm−3» ✓
«pH = −log10(5.00 × 10−14)» = 13.301 ✓

 

Award [2] for correct final answer.

(b)

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.

(c)

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] = 0.200 mol dm-3×22.5×10-3dm325.0×10-3dm3» = 0.180 «mol dm−3» ✓