This checkpoint covers Lessons 6–10: concentration (mol L⁻¹), standard solutions and dilutions, concentration in context (ppm), gravimetric analysis, and volumetric analysis (titration). 20 multiple choice + 5 short answer questions.
Lesson Summaries, Quick Review
Concentration c = n ÷ V, with n in mol and V in litres (convert mL to L first). Rearrange to n = c × V and V = n ÷ c. To make a solution of known concentration, calculate the mass: n = c × V then m = n × MM.
A primary standard is pure, stable, of known composition and high molar mass, used to make a solution of accurately known concentration. Dilution conserves moles: c₁V₁ = c₂V₂. Diluting strong acids is technician/teacher work (always add acid to water, with PPE and ventilation).
For dilute aqueous solutions (density ≈ 1.00 kg L⁻¹), 1 ppm ≈ 1 mg L⁻¹ and 1 ppb ≈ 1 µg L⁻¹. Convert mol L⁻¹ to mg L⁻¹ with mass concentration = c × MM × 1000. The ppm = mg L⁻¹ equivalence is an approximation that relies on the density assumption.
Gravimetric analysis finds the amount of an ion by precipitating it as an insoluble solid, filtering, drying to constant mass, weighing, and applying stoichiometry. Excess precipitant ensures complete precipitation; the precipitate must be dry (residual water inflates the mass).
Titration delivers titrant from a burette into a measured aliquot of analyte until the equivalence point (stoichiometric amounts per the balanced equation), shown by an indicator end point. Rinse the burette with titrant and the pipette with analyte; never rinse the conical flask with analyte. Average concordant titres (within 0.10 mL).
Updates as you answer. Short answer marks are self-assessed.
1. What is the concentration of a solution containing 0.50 mol of solute in 2.0 L? L06
2. How many moles of NaOH are in 250 mL of 0.40 mol L⁻¹ solution? L06
3. What mass of NaCl is needed to prepare 500 mL of 0.200 mol L⁻¹ solution? (MM = 58.44) L06
4. A solution is diluted by adding water. What happens to the moles of solute and the concentration? L06
5. 25.0 mL of 1.0 mol L⁻¹ solution is diluted to 250 mL. What is the new concentration? (c₁V₁ = c₂V₂) L07
6. Which property is essential for a substance to be a primary standard? L07
7. Why is solid NaOH unsuitable as a primary standard? L07
8. What volume of 2.00 mol L⁻¹ stock is needed to prepare 500 mL of 0.150 mol L⁻¹ solution? L07
9. For a dilute aqueous solution, 1 ppm is approximately equal to: L08
10. A water sample contains 5.0 mg L⁻¹ of fluoride ion. Expressed in ppm (dilute aqueous), this is: L08
11. The approximation ppm = mg L⁻¹ relies on the assumption that: L08
12. Convert 0.0020 mol L⁻¹ of Ca²⁺ to a mass concentration in mg L⁻¹. (Ca = 40.08) L08
13. Why is an excess of the precipitating reagent added in gravimetric analysis? L09
14. 0.466 g of dry BaSO₄ is collected. How many moles of SO₄²⁻ does this represent? (MM BaSO₄ = 233.4) L09
15. Why must a gravimetric precipitate be dried to constant mass before weighing? L09
16. A precipitate of AgCl (MM = 143.3) has a mass of 1.433 g. What mass of Cl⁻ does it contain? (Cl = 35.45) L09
17. In an acid–base titration, the equivalence point is reached when: L10
18. 25.0 mL of NaOH is exactly neutralised by 20.0 mL of 0.100 mol L⁻¹ HCl (1:1). What is the concentration of the NaOH? L10
19. Why should the conical flask NOT be rinsed with the analyte solution before titrating? L10
20. Concordant titres are titres that: L10
Attempt all five questions before checking answers.
21. Calculate the mass of glucose (C₆H₁₂O₆, MM = 180.16) needed to prepare 250.0 mL of a 0.100 mol L⁻¹ solution. 3 MARKS
22. A technician must prepare 2.00 L of 0.500 mol L⁻¹ sulfuric acid from an 18.0 mol L⁻¹ stock. (a) Calculate the volume of stock required. (b) State the key safety rule for this dilution. 3 MARKS
23. A 500 mL water sample contains 1.2 mg of dissolved lead. (a) Calculate the concentration in mg L⁻¹. (b) Express this in ppm and state the assumption required. 3 MARKS
24. A 250 mL water sample is treated with excess BaCl₂; 0.583 g of dry BaSO₄ precipitate forms. (a) Explain why excess BaCl₂ is used. (b) Calculate the concentration of SO₄²⁻ in mol L⁻¹. (MM BaSO₄ = 233.4) 4 MARKS
25. 20.00 mL of vinegar (acetic acid, CH₃COOH) is titrated with 0.500 mol L⁻¹ NaOH; the average concordant titre is 16.70 mL. CH₃COOH + NaOH → CH₃COONa + H₂O. (a) Calculate the moles of NaOH used. (b) Calculate the concentration of acetic acid in the vinegar. 4 MARKS
1. A c = n ÷ V = 0.50 ÷ 2.0 = 0.25 mol L⁻¹.
2. A n = c × V = 0.40 × 0.250 = 0.10 mol.
3. A n = 0.200 × 0.500 = 0.100 mol; m = 0.100 × 58.44 = 5.84 g.
4. B Adding water does not change the moles of solute; spreading them through more volume lowers the concentration.
5. A c₂ = c₁V₁ ÷ V₂ = (1.0 × 25.0) ÷ 250 = 0.10 mol L⁻¹.
6. A A primary standard must be pure, stable and of accurately known composition.
7. C NaOH is hygroscopic and absorbs CO₂, so a weighed sample is not pure NaOH and its concentration is uncertain.
8. A V₁ = c₂V₂ ÷ c₁ = (0.150 × 500) ÷ 2.00 = 37.5 mL.
9. A For dilute aqueous solutions (density ≈ 1.00 kg L⁻¹), 1 ppm ≈ 1 mg L⁻¹.
10. A 5.0 mg L⁻¹ ≈ 5.0 ppm for a dilute aqueous solution.
11. B The equivalence holds only when the density is close to 1.00 kg L⁻¹ (dilute aqueous solution).
12. A mass conc = c × MM × 1000 = 0.0020 × 40.08 × 1000 = 80.2 mg L⁻¹.
13. B Excess precipitant drives the precipitation to completion so all the analyte is captured.
14. A n(BaSO₄) = 0.466 ÷ 233.4 = 2.00 × 10⁻³ mol; n(SO₄²⁻) = 2.00 × 10⁻³ mol (1:1).
15. B Any retained water adds to the weighed mass, making the result too high; dry to constant mass.
16. A n(AgCl) = 1.433 ÷ 143.3 = 0.01000 mol; n(Cl⁻) = 0.01000 mol; m = 0.01000 × 35.45 = 0.355 g.
17. B Equivalence point = stoichiometric amounts per the balanced equation (the end point is the indicator colour change).
18. A n(HCl) = 0.100 × 0.0200 = 2.00 × 10⁻³ mol; n(NaOH) = 2.00 × 10⁻³ mol; c = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 mol L⁻¹.
19. B Rinsing the flask with analyte adds extra moles of analyte, increasing the titre and overestimating the concentration.
20. B Concordant titres agree within 0.10 mL of each other; only these are averaged.
Q21 (3 marks): n = c × V = 0.100 × 0.2500 = 0.0250 mol [1]; m = n × MM = 0.0250 × 180.16 = 4.50 g [2].
Q22 (3 marks): (a) V₁ = c₂V₂ ÷ c₁ = (0.500 × 2.00) ÷ 18.0 = 0.0556 L = 55.6 mL [2]. (b) Always add the acid slowly to water (never water to acid), in a fume cupboard wearing eye protection and gloves; this dilution is done by the technician/teacher [1].
Q23 (3 marks): (a) mass concentration = 1.2 mg ÷ 0.500 L = 2.4 mg L⁻¹ [1]. (b) 2.4 ppm [1], assuming the solution density is ≈ 1.00 kg L⁻¹ (dilute aqueous), so 1 mg L⁻¹ ≈ 1 ppm [1].
Q24 (4 marks): (a) Excess BaCl₂ ensures all the SO₄²⁻ is precipitated as BaSO₄ (complete precipitation), so the mass reflects all the sulfate present [1]. (b) n(BaSO₄) = 0.583 ÷ 233.4 = 2.498 × 10⁻³ mol [1]; n(SO₄²⁻) = 2.498 × 10⁻³ mol (1:1) [1]; c = 2.498 × 10⁻³ ÷ 0.250 = 0.0100 mol L⁻¹ [1].
Q25 (4 marks): (a) n(NaOH) = c × V = 0.500 × 0.01670 = 8.35 × 10⁻³ mol [2]. (b) 1:1 ratio so n(CH₃COOH) = 8.35 × 10⁻³ mol; c = 8.35 × 10⁻³ ÷ 0.02000 = 0.418 mol L⁻¹ [2].
Tick when you've checked all answers and noted areas to review.