Chemistry • Year 11 • Module 2 • Lesson 7
Standard Solutions & Dilutions
Build HSC Band 5–6 extended-response technique on preparing standard solutions, evaluating primary standard suitability, and designing dilution procedures in real-world analytical contexts.
1. Data + scenario: evaluating a standardisation experiment (Band 5–6)
9 marks Band 5–6
Scenario. A pharmaceutical analyst needs to prepare exactly 1.000 L of 0.1000 mol L−1 potassium iodate (KIO3) as a primary standard solution. The analyst weighed KIO3 on an analytical balance three times and recorded the following data.
| Trial | Mass of KIO3 weighed (g) | Volume of flask (mL) | Calculated concentration (mol L−1) |
|---|---|---|---|
| 1 | 21.40 | 1000.0 | |
| 2 | 21.37 | 1000.0 | |
| 3 | 21.43 | 1000.0 |
MM(KIO3) = K(39.098) + I(126.90) + 3×O(15.999) = 214.00 g mol−1. Target: 21.40 g to give exactly 0.1000 mol L−1. Illustrative data.
Q1. Analyse and evaluate the data and the analyst’s procedure. In your response you must:
- Calculate the exact concentration for all three trials using n = m ÷ MM and c = n ÷ V. Show working for Trial 1 in full; state results only for Trials 2 and 3.
- State which trial, if any, achieves the target concentration of exactly 0.1000 mol L−1, and explain why achieving the exact target is less important than knowing the exact value.
- Identify and explain all four properties that make KIO3 suitable as a primary standard, using specific data or reasoning from the scenario.
- Describe one procedural step the analyst must take to ensure all KIO3 is included in the final solution (quantitative transfer), and explain the consequence of omitting it.
- State one limitation of this experimental design and one way to improve it.
2. Experimental design — preparing a calibration series by serial dilution (Band 5–6)
8 marks Band 5–6
Research context. A food science technician needs five standard solutions of glucose at 0.100, 0.0500, 0.0250, 0.0125, and 0.00625 mol L−1 in 100 mL volumes to calibrate a colorimeter. A 0.200 mol L−1 glucose primary standard stock solution is available.
Constraint: You must minimise the number of stock solutions prepared directly from the solid glucose and use serial dilution wherever possible. You have access to 100 mL volumetric flasks, pipettes, a wash bottle, and distilled water.
Q2. Design the procedure to prepare all five standard solutions and present it in the format below.
- Calculate the volume of 0.200 mol L−1 stock needed to prepare the first solution (0.100 mol L−1, 100 mL) using c1V1 = c2V2. Show full working.
- For the remaining four solutions (0.0500 down to 0.00625), show the serial dilution volume used at each step in a calculation table.
- Describe the general procedure in at least four numbered steps that apply to each dilution, including how to ensure accuracy at each step.
- Identify the key source of error in a serial dilution series and explain how errors can accumulate.
- State two advantages of serial dilution over preparing each solution independently from the solid.
Q1 — Sample Band 6 response (9 marks), annotated
Calculations (Trial 1 in full):
MM(KIO3) = 39.098 + 126.90 + 3(15.999) = 214.00 g mol−1.
Trial 1: n = 21.40 ÷ 214.00 = 0.10000 mol; V = 1.0000 L; c = 0.10000 ÷ 1.0000 = 0.10000 mol L−1. [1 mark]
Trial 2: n = 21.37 ÷ 214.00 = 0.09986 mol; c = 0.09986 mol L−1. [1 mark for both Trial 2 and 3 stated correctly]
Trial 3: n = 21.43 ÷ 214.00 = 0.10014 mol; c = 0.10014 mol L−1.
Exact target vs known exact value: Trial 1 achieves exactly 0.1000 mol L−1; Trials 2 and 3 do not. However, achieving the exact target is less important than knowing the exact concentration of the solution actually prepared. In all downstream titration or dilution calculations, the analyst will use the measured concentration (e.g. 0.09986 or 0.10014) rather than 0.1000. Using an approximate target value introduces systematic error; using the calculated actual concentration does not. This is why the exact mass weighed is recorded and used in calculations. [2 marks]
Four KIO3 properties:
- High purity: KIO3 is available in analytical-reagent grade (≥99.5% pure). Impurities would introduce additional moles of material, making the actual moles of KIO3 less than calculated from the total mass. [1 mark]
- Stable in air: KIO3 does not react with O2, CO2, or water vapour at room temperature. The mass weighed accurately reflects the moles of KIO3 to be dissolved; it will not change between weighing and dissolving. [1 mark]
- High molar mass (214 g mol−1): 0.1000 mol requires 21.40 g. A balance error of 0.001 g represents only 0.005% of 21.40 g — negligible. If the molar mass were 40 g mol−1, only 4.00 g would be needed, and the same error would be 0.025% of the mass — five times worse. [1 mark]
- Readily soluble in water: KIO3 dissolves completely in water, forming a homogeneous solution. All the KIO3 weighed is in solution, so the concentration can be accurately calculated. [1 mark]
Quantitative transfer: After dissolving in a beaker, the analyst must rinse the beaker and glass rod at least three times with small volumes of distilled water, adding each rinse to the volumetric flask. If this step is omitted, some KIO3 remains on the inner surfaces of the beaker. The moles of KIO3 in the flask are then less than calculated, and the actual concentration is lower than the value used in titrations, introducing systematic error in all downstream results. [1 mark: procedure. Consequence embedded in mark.] [Total 2 marks for this criterion if consequence is explained separately.]
Limitation and improvement: One limitation is that only one analyst performed the weighings; random errors in balance readings between trials can only be assessed within that person’s technique, not across independent observers [1]. Improvement: have a second analyst perform the same three weighings independently, compare results, and use the average concentration across all six trials to reduce the effect of random weighing error and improve reliability [1].
Marking criteria summary (9 marks): 1 = Trial 1 full working (n and c correctly calculated); 1 = Trials 2 and 3 stated correctly; 2 = discussion of why exact target matters less than knowing exact value (1 for stating it, 1 for explaining use in downstream calc); 4 = all four KIO3 properties each with explanation (1 each); 1 = quantitative transfer described with consequence of omitting; 1 = limitation plus improvement (0.5 each, rounded). Total markers may award 1 for limitation/improvement as a combined mark.
Q2 — Sample Band 6 response (8 marks), annotated
Step 1 calculation (0.200 → 0.100 mol L−1):
c1 = 0.200 mol L−1; c2 = 0.100 mol L−1; V2 = 100 mL. V1 = c2V2 ÷ c1 = (0.100 × 100) ÷ 0.200 = 50.0 mL of stock. [1 mark — full working]
Serial dilution table (each step: take 50.0 mL from previous, make to 100 mL):
| Step | Volume taken (mL) | Final volume (mL) | Concentration (mol L−1) |
|---|---|---|---|
| 1 | 50.0 from stock | 100 | 0.100 |
| 2 | 50.0 from Step 1 | 100 | 0.0500 |
| 3 | 50.0 from Step 2 | 100 | 0.0250 |
| 4 | 50.0 from Step 3 | 100 | 0.0125 |
| 5 | 50.0 from Step 4 | 100 | 0.00625 |
[1 mark for complete, correct table]
General four-step procedure: (1) Using a calibrated pipette, measure out exactly 50.0 mL of the source solution. (2) Transfer to a clean 100 mL volumetric flask. Rinse the pipette once with the source solution before use to ensure the correct concentration is transferred. (3) Add distilled water to just below the calibration mark. Use a dropper pipette to bring the meniscus to exactly the mark. (4) Stopper the flask and invert 10 times to mix thoroughly. Label the flask immediately. [1 mark — four steps present; 1 mark — accuracy measures (pipette rinse, dropper pipette, stopper-and-invert)] [2 marks total for procedure]
Key source of error — cumulative pipetting error: Each step introduces a small random error in the volume transferred (e.g. ±0.05 mL in a 50 mL pipette). Because each solution is made from the previous one, this error is carried forward and compounded: an error in Step 1 affects Steps 2–5; an error in Step 2 affects Steps 3–5. By Step 5 (0.00625 mol L−1), the accumulated percentage error can be considerably larger than in Step 1. [1 mark]
Two advantages of serial dilution: (1) It is faster and uses less material than separately weighing and dissolving glucose for each concentration, because only a single primary-standard solution is prepared from the solid and all subsequent solutions are made by dilution. (2) Consistent dilution steps reduce the risk of random errors arising from separate weighings; all solutions trace back to the same primary-standard preparation, so any systematic error in that preparation affects all solutions equally and can be corrected by calibration. [1 mark each, 2 marks total]
Marking criteria summary (8 marks): 1 = Step 1 full working with correct V1 = 50.0 mL; 1 = complete serial dilution table (all 5 concentrations correct); 2 = four-step procedure with accuracy measures (2 marks); 1 = cumulative error identified and explained; 2 = two valid advantages of serial dilution (1 each); 1 = precision in chemical terminology throughout (c1V1 = c2V2, quantitative transfer, meniscus, calibration mark, pipette).