Skip to content
MarkSchemeMarkScheme
Sign up — no card

9701 · 18.1

Carboxylic acids — practice questions

Practice and worked examples for 9701 Carboxylic acids. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 25.0 cm³ sample of a solution of ethanoic acid, CH3COOHCH_3COOH, was titrated with 0.150 mol dm⁻³ sodium hydroxide solution. 22.50 cm³ of the NaOH solution was required for complete neutralisation. Calculate the concentration of the ethanoic acid solution in mol dm⁻³.

Show solution outline
  1. Write the balanced equation: CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l)CH_3COOH(aq) + NaOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)
  2. Calculate the moles of NaOH used: Moles = Concentration × Volume Moles of NaOH = 0.150 mol dm−3×(22.50/1000) dm30.150 \text{ mol dm}^{-3} \times (22.50 / 1000) \text{ dm}^3 Moles of NaOH = 3.375×10−3 mol3.375 \times 10^{-3} \text{ mol}
  3. Determine the moles of CH₃COOH reacted: From the balanced equation, the mole ratio of CH3COOH:NaOHCH_3COOH : NaOH is 1:1. Therefore, moles of CH3COOHCH_3COOH = 3.375×10−3 mol3.375 \times 10^{-3} \text{ mol}
  4. Calculate the concentration of CH₃COOH: Concentration = Moles / Volume Concentration of CH3COOHCH_3COOH = (3.375×10−3 mol)/(25.0/1000) dm3(3.375 \times 10^{-3} \text{ mol}) / (25.0 / 1000) \text{ dm}^3 Concentration of CH3COOHCH_3COOH = 0.135 mol dm⁻³ (to 3 s.f.)

Worked example 2

A 1.85 g sample of a pure straight-chain carboxylic acid, Y, was reacted with excess sodium carbonate. The reaction produced 480 cm³ of carbon dioxide gas, measured at room temperature and pressure (RTP). Determine the molecular formula of the carboxylic acid Y. (Molar volume of a gas at RTP = 24.0 dm³ mol⁻¹).

Show solution outline
  1. Write a general balanced equation: Let the carboxylic acid be RCOOHRCOOH. The reaction is: 2RCOOH+Na2CO3→2RCOONa+H2O+CO22RCOOH + Na_2CO_3 \rightarrow 2RCOONa + H_2O + CO_2
  2. Calculate moles of CO₂ produced: Volume of CO2=480 cm3=0.480 dm3CO_2 = 480 \text{ cm}^3 = 0.480 \text{ dm}^3 Moles of CO2CO_2 = Volume / Molar Volume = 0.480/24.0=0.0200 mol0.480 / 24.0 = 0.0200 \text{ mol}
  3. Determine moles of carboxylic acid Y: From the stoichiometry, the mole ratio of RCOOH:CO2RCOOH : CO_2 is 2:1. Moles of Y = 2×2 \times Moles of CO2=2×0.0200=0.0400 molCO_2 = 2 \times 0.0200 = 0.0400 \text{ mol}
  4. Calculate the molar mass (Mr) of Y: Mr = Mass / Moles = 1.85 g/0.0400 mol=46.25 g mol−11.85 \text{ g} / 0.0400 \text{ mol} = 46.25 \text{ g mol}^{-1}. Let's round this to 46 as Mr must be an integer.
  5. Determine the molecular formula: The general formula is CnH2n+1COOHC_n H_{2n+1} COOH, which can be written as Cn+1H2n+2O2C_{n+1}H_{2n+2}O_2. Let's test small values of n. The simplest acid is methanoic acid (HCOOHHCOOH, n=0). Mr of HCOOH=1.0+12.0+16.0+16.0+1.0=46.0 g mol−1HCOOH = 1.0 + 12.0 + 16.0 + 16.0 + 1.0 = 46.0 \text{ g mol}^{-1}. This matches our calculated Mr. Therefore, the carboxylic acid Y is methanoic acid, HCOOH.
Mark a 9701 question on this topicFull lessonOpen in course studio