Ethanedioic acid is a white crystalline solid. If excess aqueous potassium hydroxide, KOH(aq), is added to dilute ethanedioic acid, H₂C₂O₄(aq), full neutralisation occurs and potassium ethanedioate, K₂C₂O₄(aq), forms. H₂C₂O₄(aq) + 2KOH(aq) → K₂C₂O₄(aq) + 2H₂O(l) If a small amount of potassium hydroxide is added, partial neutralisation takes place and not all H⁺ ions in the acid are replaced by K⁺ ions. Instead an acid salt forms, which crystallises to form a solid with the formula KₐHᵦ(C₂O₄)c•dH₂O. The letters a, b and c represent a ratio of the numbers of species present in the compound and may not necessarily be whole numbers. The relative number of water molecules associated with one formula of the compound is represented by d. A student attempted to determine the values of a, b, c and d in a sample of an acid salt, KₐHᵦ(C₂O₄)c•dH₂O.
- (a)The student wants to make a 250.0 cm³ aqueous solution of KₐHᵦ(C₂O₄)c•dH₂O, solution A. The student adds 1.89g of KₐHᵦ(C₂O₄)c•dH₂O into a 100cm³ beaker. Describe the next steps the student should take to make solution A, containing exactly 1.89 g of KₐHᵦ(C₂O₄)c•dH₂O. Give the name and capacity of the apparatus which should be used and describe how the student should ensure the volume is exactly 250.0 cm³. Write your answer using a series of numbered steps.[4]
- (b)(i)When emptying the pipette in step 2, the student touches the surface of the solution in the flask with the tip of the pipette. Suggest why the student does this.[1]
- (b)(ii)Suggest the most appropriate piece of apparatus to measure H₂SO₄(aq) in step 3.[1]
- (b)(iii)Suggest why the student starts each titration with an initial burette reading of 0.10 cm³ rather than the usual 0.00 cm³.[1]
- (b)(iv)What is meant by the term concordant readings?[1]
- (b)(v)State the change of colour seen in the mixture in the conical flask at the end-point. from .................................................... to ....................................................[1]
- (b)(vi)Determine the number of moles of C₂O₄²⁻ ions in the 250.0 cm³ of solution A, KₐHᵦ(C₂O₄)c•dH₂O. Give your answer to three significant figures. moles of C₂O₄²⁻ ions in 250.0 cm³ of solution A = ................................... mol[3]
- (c)(i)Suggest the name of a suitable reagent to use in this titration.[1]
- (c)(ii)The student finds the concentration of H⁺(aq) in solution A is 6.10 × 10⁻² mol dm⁻³. Use this value and your answer to (b)(vi) to determine the ratio of c to b to two decimal places. Then deduce the value of a in KₐHᵦ(C₂O₄)c•dH₂O to two decimal places. ratio c: b = 1: .................................................... value of a = ....................................................[3]
- (c)(iii)Use your answer to (b)(vi), (c)(ii) and other information given in the question to determine the mass of 1 mol of KₐHᵦ(C₂O₄)c•dH₂O and hence determine the value of d to the nearest whole number. [Aᵣ: K, 39.1; H, 1.0; C, 12.0; O, 16.0] If you were unable to calculate an answer to (c)(ii), then you may use a = 0.30 and ratio c:b = 1:1.60. These are not the correct values. mass of 1 mol of KₐHᵦ(C₂O₄)c•dH₂O = ................................... g value of d = ....................................................[2]
- (d)A second student uses another method to determine d. Crystals of the sample, with known values of a, b and c, are heated in a crucible to remove the water molecules. Construct a results table to show the readings that would need to be taken during this experiment.[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.
- [1] M1: (250 cm³) volumetric flask
- [1] M2: dissolve the solid / acid salt (1.89 g) (in the beaker) (using distilled) water
- [1] M3: transfer / add to a (250 cm³ volumetric) flask AND rinse (with distilled water)
- [1] M4: top up to mark (with distilled water)
- [1] to ensure that (exactly) 25.0 cm³ of solution has been delivered into the flask
- [1] measuring cylinder
- [1] difficult to see the 0.00 cm³ line
- [1] (two) titres / readings within 0.1(0) (cm³) of each other
- [1] colourless to pale purple
- [1] M1: mol MnO₄⁻(aq) = 0.0200 × 24.40 / 1000 = 4.88 × 10⁻⁴ (mol)
- [1] M2: mol of C₂O₄²⁻(aq) (in 25.0 cm³) = M1 × 5/2 = 1.22 × 10⁻³ (mol)
- [1] M3: mol of C₂O₄²⁻(aq) (in solution A) = M2 × 250 / 25 = 1.22 × 10⁻² (mol)
- [1] (aqueous) sodium hydroxide
- [1] M1: mol H⁺(aq) in 250 cm³ = 6.10 × 10⁻²/4 = 1.525 × 10⁻² (mol)
- [1] M2: ratio C₂O₄²⁻(aq) : H⁺(aq) (1.22 × 10⁻²/1.22 × 10⁻²) : 1.525 × 10⁻²/1.22 × 10⁻² = 1 : 1.25 and b = 1.25
- [1] M3: 2 - 1.25 a = 0.75
- [1] M1: Mᵣ of hydrated K₀.₇₅H₁.₂₅C₂O₄•dH₂O = 1.89 / 1.22 × 10⁻² = 154.9 OR Mᵣ of anhydrous K₀.₇₅H₁.₂₅C₂O₄ = [(0.75 × 39.1) + (1.25 × 1.0) + (2 × 12.0) + (4 × 16.0)] = 118.575
- [1] M2: mass of water = hydrated mass – anhydrous mass = 154.9 – 118.575 = 36.325 AND 36.325 / 18 = 2
- [1] M1: initial readings mass of (empty) crucible / g mass of crucible + crystals before heating / g mass of crucible + crystals after heating / g
- [1] M2: second reading of mass of crucible + crystals after re-heating
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →