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Questions on concentration, volumetric solutions and titration calculations
Concentration in mol dm⁻³, g dm⁻³ and parts per million, n = cV with the volume in dm³, dilution, primary standards and making a standard solution in a volumetric flask, deducing a reacting ratio from a titration, indicators and the end point, rough and concordant titres and the mean titre, calculating a concentration from a titre with the mole ratio, back titrations, the uncertainty in a titre, and AQA required practical 1 in full with its errors.
18 original questions · 60 marks · the concentration, volumetric solutions and titration calculations notes · Amount of substance · Flashcards for this unit
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Describe how to make a standard solution of sodium carbonate and use it in a titration to find the concentration of a solution of hydrochloric acid.
Mark scheme
Indicative content
- weigh the sodium carbonate by difference, dissolve it in distilled water in a beaker, and transfer it to a volumetric flask with the rinsings of the beaker, rod and funnel
- make up to the line with distilled water, adding the last drops so the bottom of the meniscus is on the line at eye level, then stopper and invert to mix
- rinse a pipette with the carbonate solution and transfer 25.0 cm³ to a conical flask, using a pipette filler, and add a few drops of methyl orange
- rinse the burette with the acid, fill it, remove the funnel, make sure the tip is full and record the initial reading at eye level
- carry out a rough titration, then accurate titrations adding the acid dropwise near the end point with swirling, until one drop turns the indicator from yellow to orange
- repeat until titres are concordant within 0.10 cm³ and use their mean, with the 1 : 2 ratio of Na₂CO₃ to HCl, to calculate the concentration of the acid
Levels
- Level 3 (5–6 marks) the standard solution is made with weighing by difference, rinsings transferred and the flask made up to the line, and the titration is carried out with the pipette, burette, indicator, dropwise end point and concordant titres all described, finishing with the 1 : 2 ratio in the calculation
- Level 2 (3–4 marks) the solution is made up in a volumetric flask and the titration done with an indicator and repeats; transferring the rinsings, reading the meniscus, the dropwise approach to the end point, or the concordance criterion is left out
- Level 1 (1–2 marks) sodium carbonate is dissolved and titrated with the acid; the calculation commonly uses a 1 : 1 ratio, and the steps that make the solution standard are missing
Calculate the percentage by mass of calcium carbonate in an indigestion tablet from the back-titration results in the table. The tablet is added to an excess of hydrochloric acid, and the acid left over is titrated with sodium hydroxide. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
Quantity Value Mass of tablet 0.600 g Volume of hydrochloric acid added 50.0 cm³ Concentration of hydrochloric acid 0.250 mol dm⁻³ Mean titre of sodium hydroxide 27.30 cm³ Concentration of sodium hydroxide 0.100 mol dm⁻³ Mr of CaCO₃ 100.1 Mark scheme
- M1 n(HCl) added = 0.250 × 0.0500 = 0.0125 mol
- M1 n(HCl) left = n(NaOH) = 0.100 × 0.02730 = 2.73 × 10⁻³ mol
- M1 n(HCl) used by the tablet = 0.0125 − 0.00273 = 9.77 × 10⁻³ mol, so n(CaCO₃) = 4.885 × 10⁻³ mol
- M1 mass of CaCO₃ = 4.885 × 10⁻³ × 100.1 = 0.4890 g
- A1 percentage = 0.4890 ÷ 0.600 × 100 = 81.5%, to three significant figures
Calculate the volume of sodium hydroxide needed to neutralise a portion of a diluted sulfuric acid, using the data in the table: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
Quantity Value Concentration of the stock sulfuric acid 1.50 mol dm⁻³ Volume of stock acid diluted 20.0 cm³ Volume after dilution with distilled water 500 cm³ Portion of diluted acid titrated 25.0 cm³ Concentration of sodium hydroxide 0.100 mol dm⁻³ Mark scheme
- M1 concentration of the diluted acid = 1.50 × 20.0 ÷ 500 = 0.0600 mol dm⁻³
- M1 n(H₂SO₄) in 25.0 cm³ = 0.0600 × 0.0250 = 1.50 × 10⁻³ mol
- M1 n(NaOH) = 2 × 1.50 × 10⁻³ = 3.00 × 10⁻³ mol, from the 1 : 2 ratio
- A1 V = 3.00 × 10⁻³ ÷ 0.100 = 0.0300 dm³, which is 30.0 cm³, to three significant figures; allow error carried forward from a wrong dilution
Calculate the concentration of a sulfuric acid solution, given that 20.50 cm³ of it neutralises 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
Mark scheme
- M1 n(NaOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol
- M1 the ratio H₂SO₄ : NaOH is 1 : 2, so n(H₂SO₄) = 1.25 × 10⁻³ mol
- M1 c = 1.25 × 10⁻³ ÷ (20.50 ÷ 1000)
- A1 c = 0.0610 mol dm⁻³, to three significant figures
Calculate the mass of anhydrous sodium carbonate, Na₂CO₃, needed to make 250 cm³ of a 0.100 mol dm⁻³ standard solution, and the concentration of that solution in g dm⁻³. Use Na 23.0, C 12.0, O 16.0.
Mark scheme
- M1 n = 0.100 × 0.250 = 0.0250 mol
- M1 M(Na₂CO₃) = 2 × 23.0 + 12.0 + 3 × 16.0 = 106.0 g mol⁻¹, and m = 0.0250 × 106.0
- A1 m = 2.65 g, to three significant figures
- A1 mass concentration = 0.100 × 106.0 = 10.6 g dm⁻³, or 2.65 ÷ 0.250; allow error carried forward from the mass
Explain why, in a titration of sodium carbonate solution in a conical flask with hydrochloric acid from a burette, the burette is rinsed with the acid but the flask is rinsed only with distilled water.
Mark scheme
- B1 water left in the burette would dilute the acid, lowering its concentration
- B1 a larger titre would then be needed, so the calculated concentration of the acid would come out too low
- B1 distilled water in the flask adds no carbonate, so the amount delivered by the pipette is unchanged; ignore 'it dilutes the solution'
- B1 rinsing the flask with carbonate solution would leave extra carbonate in it, so the titre would be too large
Evaluate a student's mean titre of 23.89 cm³, calculated from all four titres in the table of burette readings given.
Reading Rough Run 1 Run 2 Run 3 Final reading / cm³ 24.30 24.10 25.05 23.85 Initial reading / cm³ 0.00 0.40 1.10 0.25 Titre / cm³ 24.30 23.70 23.95 23.60 Mark scheme
- B1 the rough titre should be left out, because it is run quickly and usually overshoots the end point
- B1 run 2 differs from runs 1 and 3 by more than 0.10 cm³, so it is not concordant and should also be left out
- B1 the correct mean is (23.70 + 23.60) ÷ 2 = 23.65 cm³; reject a mean that includes run 2
- B1 the student's mean is 0.24 cm³, about 1%, too high, which is larger than the ±0.10 cm³ uncertainty of a titre, so a concentration calculated from it would be wrong
Calculate the concentration of a sodium hydroxide solution, given that 25.0 cm³ of it needs a mean titre of 21.85 cm³ of 0.200 mol dm⁻³ hydrochloric acid.
Mark scheme
- M1 n(HCl) = 0.200 × 21.85 ÷ 1000 = 4.37 × 10⁻³ mol
- M1 NaOH and HCl react 1 : 1, so n(NaOH) = 4.37 × 10⁻³ mol, and c = 4.37 × 10⁻³ ÷ 0.0250
- A1 c = 0.175 mol dm⁻³, to three significant figures
Calculate the percentage uncertainty in a titre of 18.20 cm³, given that each burette reading has an uncertainty of ±0.05 cm³.
Mark scheme
- M1 the titre is the difference of two readings, so its uncertainty is 2 × 0.05 = ±0.10 cm³
- M1 percentage uncertainty = 0.10 ÷ 18.20 × 100
- A1 0.55%, to two significant figures
Explain why, when making a standard solution, the beaker and funnel are rinsed into the volumetric flask and the flask is inverted after making up to the line.
Mark scheme
- B1 some solution remains on the beaker, rod and funnel
- B1 rinsing carries all of the weighed solute into the flask, so the concentration matches the mass used
- B1 inverting mixes the solution so the concentration is the same throughout and every portion taken is identical
Calculate the concentration of lead ions in a 500 cm³ sample of drinking water that contains 0.0250 mg of lead ions, in ppm and in mol dm⁻³. Use Pb 207.2.
Mark scheme
- M1 mass per dm³ = 0.0250 ÷ 0.500 = 0.0500 mg dm⁻³, and 1 dm³ of dilute solution has a mass of about 1 kg, 10⁶ mg
- A1 0.0500 ppm, to three significant figures
- A1 0.0500 × 10⁻³ g dm⁻³ ÷ 207.2 = 2.41 × 10⁻⁷ mol dm⁻³, to three significant figures
Explain why sulfamic acid, H₂NSO₃H, is suitable as a primary standard for standardising a sodium hydroxide solution.
Mark scheme
- B1 it is a solid that can be obtained very pure and is stable in air
- B1 it does not absorb water from the air, so the mass weighed is all sulfamic acid
- B1 it has one acidic hydrogen and reacts with sodium hydroxide in a known 1 : 1 ratio, so the moles weighed give the moles of alkali
Calculate the number of moles of sodium hydroxide that react with one mole of an acid, from the titration results in the table.
Solution Volume / cm³ Concentration / mol dm⁻³ Acid, pipetted 25.0 0.0400 Sodium hydroxide, mean titre 20.00 0.100 Mark scheme
- M1 n(acid) = 0.0400 × 0.0250 = 1.00 × 10⁻³ mol
- M1 n(NaOH) = 0.100 × 0.02000 = 2.00 × 10⁻³ mol
- A1 ratio NaOH : acid = 2.00 × 10⁻³ ÷ 1.00 × 10⁻³ = 2, so two moles; the acid has two acidic hydrogens
Show that 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by exactly 12.50 cm³ of 0.100 mol dm⁻³ sulfuric acid: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
Mark scheme
- M1 n(NaOH) = c × V = 0.100 × 25.0 ÷ 1000 = 2.50 × 10⁻³ mol
- M1 the equation gives 2 mol of NaOH for 1 mol of H₂SO₄, so n(H₂SO₄) = 2.50 × 10⁻³ ÷ 2 = 1.25 × 10⁻³ mol
- A1 V = n ÷ c = 1.25 × 10⁻³ ÷ 0.100 = 0.0125 dm³ = 12.50 cm³, every step shown because the answer is given in the question
State what is meant by concordant titres, and state which titres are used to calculate the mean titre.
Mark scheme
- B1 concordant titres agree with each other to within 0.10 cm³
- B1 the mean is calculated from the concordant titres only, leaving out the rough titre and any that do not agree
Explain why a rough titration is carried out before the accurate titrations.
Mark scheme
- B1 it shows roughly where the end point lies
- B1 in the accurate runs the solution can be added quickly to just short of that volume and then dropwise, so the end point is not overshot
Suggest why a back titration is used to find the calcium carbonate content of an indigestion tablet, rather than titrating the tablet directly with acid.
Mark scheme
- B1 calcium carbonate is insoluble and reacts slowly, so the end point would be reached before all of it had reacted
- B1 the carbon dioxide given off makes the end point hard to see, whereas an excess of acid reacts with all of it and the leftover acid can be titrated cleanly
Suggest why solid sodium hydroxide is not used to make a standard solution by weighing.
Mark scheme
- B1 sodium hydroxide absorbs water and carbon dioxide from the air while it is being weighed
- B1 the mass weighed is then not all sodium hydroxide, so the concentration calculated from it would be wrong
Practise concentration, volumetric solutions and titration calculations one question at a time