Chemistry • Year 11 • Module 2 • Lesson 7
Standard Solutions & Dilutions
Apply your understanding of the dilution formula, primary standard criteria, and standard solution preparation to real data, calculations, and scenario analysis.
1. Dilution calculations — all rearrangements of c1V1 = c2V2
Show all working for each calculation. 12 marks (3 each)
1.1 A student pipettes 20.0 mL of a 1.50 mol L−1 NaCl solution into a 250 mL volumetric flask and makes it up to the mark with distilled water. Calculate the concentration of the diluted solution.
1.2 What volume of 12.0 mol L−1 hydrochloric acid must be diluted to 500 mL to give a 0.300 mol L−1 solution?
1.3 A student adds 150 mL of water to 50.0 mL of a 0.800 mol L−1 glucose solution. Calculate the concentration of the resulting solution.
1.4 A student dilutes 5.00 mL of a stock solution to 250 mL and measures the final concentration as 0.0400 mol L−1. What was the concentration of the original stock solution?
2. Classify potential primary standards
The table below lists five substances considered for use as primary standards. Complete the table by ticking which criteria each substance meets, then write the overall verdict (suitable / not suitable) and one reason. 10 marks
| Substance | High purity available? | Stable in air? | High molar mass (>100)? | Readily soluble? | Verdict & reason |
|---|---|---|---|---|---|
| Na2CO3 MM = 106 g mol−1 |
Yes | Yes (dried) | Yes | ||
| NaOH MM = 40 g mol−1 |
Yes | No | Yes | ||
| KIO3 MM = 214 g mol−1 |
Yes | Yes | Yes | Yes | |
| HCl (conc. aq.) MM = 36.5 g mol−1 |
No | No | Yes | ||
| BaSO4 MM = 233 g mol−1 |
Yes | Yes | Yes |
3. Calculate the concentration of a standard solution
6 marks
Scenario. A student weighs out 4.763 g of anhydrous sodium carbonate (Na2CO3) on an analytical balance, dissolves it in approximately 50 mL of distilled water in a beaker, then transfers the solution quantitatively to a 250.0 mL volumetric flask and makes up to the mark with distilled water. (Na = 22.990, C = 12.011, O = 15.999)
3.1 Calculate the molar mass of Na2CO3. (1 mark)
3.2 Calculate the number of moles of Na2CO3 dissolved. (1 mark)
3.3 Calculate the exact concentration of the standard solution in mol L−1. (1 mark)
3.4 The student then takes 25.0 mL of this standard solution and dilutes it to 500 mL in a second volumetric flask. Using c1V1 = c2V2, calculate the concentration of the diluted solution. (2 marks)
3.5 State why the student uses the exact mass actually weighed (4.763 g) rather than the intended target mass in the calculation. (1 mark)
4. Interpret a serial dilution sequence
A student performs a 1:5 serial dilution of a 2.00 mol L−1 copper sulfate stock solution. At each step, 20.0 mL of the current solution is transferred to a 100 mL volumetric flask and made up to 100 mL with distilled water. The graph below shows concentration vs step number. 7 marks
Figure 4.1. Concentration of CuSO4(aq) after each step of a 1:5 serial dilution from a 2.00 mol L−1 stock. Illustrative data.
4.1 Use c1V1 = c2V2 to verify the concentration at Step 1. Show your working. (2 marks)
4.2 Describe the overall pattern shown in the graph and explain the mathematical relationship between successive concentrations. (2 marks)
4.3 A student claims that “each step removes some of the copper sulfate from solution.” Identify the error in this statement and write a correct explanation. (2 marks)
4.4 Calculate the number of moles of CuSO4 present in 50.0 mL of the Step 3 solution. (1 mark)
5. Predict and justify — an error in the procedure
5 marks
Scenario. A student intends to prepare 500 mL of exactly 0.200 mol L−1 NaOH by dissolving the required mass in a beaker of distilled water and transferring it to a 500 mL volumetric flask. Before doing so they read the following procedure: “Step 3: When most of the solid has dissolved, pour the hot solution directly into the volumetric flask and immediately fill to the mark.”
5.1 Identify two procedural errors in the student’s step 3 and explain the consequence of each. (4 marks)
5.2 In addition to these procedural errors, state why NaOH cannot be used as a primary standard in the first place. (1 mark)
Q1 — Dilution calculations
1.1 c1 = 1.50 mol L−1; V1 = 20.0 mL; V2 = 250 mL. c2 = (1.50 × 20.0) ÷ 250 = 0.120 mol L−1.
1.2 c1 = 12.0; V2 = 500 mL; c2 = 0.300. V1 = c2V2 ÷ c1 = (0.300 × 500) ÷ 12.0 = 12.5 mL.
1.3 V2 = 50.0 + 150 = 200 mL (total final volume). c2 = (0.800 × 50.0) ÷ 200 = 0.200 mol L−1. Common error: using V2 = 150 mL (volume of water added, not total volume).
1.4 c1 = c2V2 ÷ V1 = (0.0400 × 250) ÷ 5.00 = 2.00 mol L−1.
Marking criteria (3 marks each): 1 mark — correct identification of variables; 1 mark — correct rearrangement and substitution; 1 mark — correct answer with units.
Q2 — Primary standard classification
Na2CO3: High molar mass = Yes (106 > 100). Verdict: Suitable — meets all four criteria when dried before use (anhydrous Na2CO3 is stable in dry air after drying).
NaOH: Stable in air = No (absorbs CO2 and water vapour). Verdict: Not suitable — fails stability criterion; composition changes continuously in air, making accurate weighing impossible.
KIO3: All four = Yes. Verdict: Suitable — widely used HSC primary standard; high MM (214 g mol−1), stable, pure, and soluble.
HCl (conc. aq.): High purity = No (variable concentration, volatile). Verdict: Not suitable — fails purity (variable concentration), stability (volatile, loses HCl gas), and high molar mass criteria.
BaSO4: Readily soluble = No (Ksp ≈ 1.1 × 10−10). Verdict: Not suitable — fails solubility criterion; a homogeneous solution of known concentration cannot be prepared.
Q3 — Standard solution concentration
3.1 MM(Na2CO3) = 2(22.990) + 12.011 + 3(15.999) = 45.980 + 12.011 + 47.997 = 105.988 g mol−1.
3.2 n = m ÷ MM = 4.763 ÷ 105.988 = 0.04494 mol.
3.3 V = 0.2500 L. c = n ÷ V = 0.04494 ÷ 0.2500 = 0.1798 mol L−1.
3.4 c1 = 0.1798 mol L−1; V1 = 25.0 mL; V2 = 500 mL. c2 = (0.1798 × 25.0) ÷ 500 = 8.99 × 10−3 mol L−1.
3.5 In practice, it is impossible to weigh out the exact target mass. The actual mass weighed is used in calculations to ensure the concentration calculated reflects the actual number of moles dissolved, not the intended number of moles.
Q4 — Serial dilution interpretation
4.1 c1 = 2.00 mol L−1; V1 = 20.0 mL; V2 = 100 mL. c2 = (2.00 × 20.0) ÷ 100 = 0.400 mol L−1. Matches Step 1 in the graph [verified].
4.2 The concentration decreases by a factor of 5 at each step (i.e. is multiplied by 0.2 each time): 2.00 → 0.400 → 0.0800 → 0.0160 → 0.00320. This is a geometric sequence (exponential decay pattern) because the same dilution factor is applied repeatedly.
4.3 The student’s statement is incorrect. No solute is removed; the 20.0 mL aliquot transferred to each new flask contains the same moles as in that 20.0 mL portion. Those moles are then spread through a larger (100 mL) volume, reducing the concentration, not the total moles in the original solution.
4.4 n = c × V = 0.0160 × 0.0500 = 8.00 × 10−4 mol (or 0.000800 mol).
Q5 — Procedural errors
5.1 Error 1: Pouring a hot solution into the volumetric flask. Volumetric flasks are calibrated at a specific temperature (usually 20 °C). A hot solution has a lower density (greater volume per mole) than a cool solution; when made up to the mark while hot, the solution will contract on cooling to below the calibration mark, so the final volume will be less than 500 mL and the concentration will be higher than intended. The solution must be cooled to room temperature before transfer. [2 marks]
Error 2: Dissolving only “most” of the solid, not all of it. If undissolved solid remains in the beaker and is not transferred, the moles of NaOH actually in the flask are less than calculated from the mass weighed. The concentration will be lower than intended, and the transfer is not quantitative. [2 marks]
5.2 NaOH is not a primary standard because it is not stable in air — it absorbs CO2 (forming Na2CO3) and water vapour from the atmosphere, continuously changing its mass and purity. The moles of NaOH in a weighed sample are therefore unknown. [1 mark]