Analogy
Imagine the scenario where a sachet of milo powder is dissolved in 100 cm3 of water.
Dilution – Another 100 cm3 of water is added to the solution.
- Is the amount of milo powder the same in 1 & 2?
- Is the sweetness of 1 & 2 the same?
Sampling – 100 cm3 of the milo solution is poured into a second cup.
- Is the amount of milo powder the same in 2 & 3?
- Is the sweetness of 2 & 3 the same?
From the above analogy, you should be able to distinguish between the consequences of dilution & sampling.
| Dilution | Sampling |
| No. of moles unchanged | No. of moles changed
ns = c x Vs ns: amount of substance in sample (mol) Vs: vol. of sample |
| Conc. changed
ci Vi = cf Vf ci Vi: conc. & vol. before dilution cf Vf:conc. & vol. after dilution |
Conc. unchanged |
It is important to highlight the presence of ‘dilution’ and/ or ‘sampling’ in a question. Often, students fail to account for them which results in incomplete answers.
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Example 10 cm3 of HCl was diluted to 250 cm3 with deionised water. A 25 cm3 aliquot of the diluted solution required 16.00 cm3 of 0.05 moldm–3 NaOH for complete neutralisation. Calculate the concentration of the undiluted HCl.
NaOH + HCl → NaCl + H2O
nNaOH = 16/1000 x 0.05 = 8.0 x 10–4 mol nHCl in 25 cm3 = nNaOH = 8.0 x 10–4 mol nHCl in 250 cm3 = 8.0 x 10–3 mol cHCl = 8.0 x 10–3 / (10/1000) = 0.8 moldm–3
Common Error #1 (Neglect sampling) nNaOH = 16/1000 x 0.05 = 8.0 x 10–4 mol nHCl = nNaOH = 8.0 x 10–4 mol cHCl = 8.0 x 10–4 / (10/1000) = 0.08 moldm–3 Common Error #2 (Neglect dilution) cHCl VHCl = cNaOH VNaOH cHCl x 25/1000 = 0.05 x 16/1000 cHCl = 0.032 moldm–3 |
