Concentration is an amount divided by a volume, and almost every mistake on this topic is a unit problem rather than a chemistry problem. Two relationships carry the page:
Volumes go in litres before they go into either formula. A millilitre left unconverted moves your answer by a factor of a thousand, which is large enough to be obvious and small enough to be missed if you are not checking magnitudes.
Molar masses to two decimal places; round once, at the end.
1. 5.85 g of sodium chloride is dissolved and made up to 250.0 mL of solution. Find the concentration in mol/L.
Answer 1
0.400 mol/L, to three significant figures.
Note what happened at the volume: 250.0 mL became 0.2500 L, and the trailing zeros were kept because the flask really does deliver four figures. Writing 0.25 L would have thrown away precision the volumetric flask gave you for free.
Note also the wording: made up to 250.0 mL of solution, not dissolved in 250.0 mL of water. Those are different volumes and only the first one is the one in the formula — which is the whole reason Preparing a Standard Solution insists on a volumetric flask.
2. What mass of glucose, , is needed to prepare 500.0 mL of a 0.200 mol/L solution?
Answer 2
Work backwards along the same chain: concentration and volume give moles, moles and molar mass give grams.
18.0 g of glucose, to three significant figures.
Sanity check on the size: a fifth of a mole of a compound with a molar mass near 180 should be somewhere around 36 g, and you are making half a litre rather than a full one, so about 18 g. If your answer had come out at 1.80 g or 180 g you converted a volume wrongly, and the estimate catches it before the calculator does.
3. What volume of 2.00 mol/L hydrochloric acid is needed to prepare 250.0 mL of 0.150 mol/L acid by dilution?
Answer 3
Dilution adds solvent and adds no solute, so the moles before equal the moles after. That single sentence is the whole of , and it is worth deriving rather than memorising.
18.8 mL, to three significant figures.
The volumes can stay in millilitres here because they appear on both sides and the units cancel — but only in this formula, and only when both volumes are in the same unit. Everywhere else on this page, convert to litres.
How you would actually do it. Measure 18.8 mL of the stock acid with a graduated pipette, run it into a 250.0 mL volumetric flask that already contains some distilled water, then make up to the mark and invert to mix.
The order in that sentence is a safety instruction, not a preference. Add acid to water, never water to acid. Diluting acid releases heat; with water already in the flask the heat spreads through a large volume, whereas acid poured into a few drops of water can boil and spit acid back out at your face.
4. A 250 mL water sample is found to contain 3.5 mg of chloride ion. Express this concentration in mg/L and in parts per million, and explain why those two numbers are the same.
Answer 4
14 mg/L, which is 14 ppm.
Why they coincide. Parts per million is a mass ratio: milligrams of solute per million milligrams of solution. One litre of a dilute aqueous solution has a density very close to that of water, about 1.00 g/mL, so it has a mass of about 1000 g, which is mg.
So 1 mg in 1 L is 1 mg in mg — one part per million, exactly. The two units are numerically interchangeable for dilute aqueous solutions and for nothing else. In a concentrated brine, or in a solvent that is not water, the density assumption fails and so does the equivalence.
Two significant figures throughout, because 3.5 mg had two. This is the unit that drinking water guidelines are written in, and you will use it in The Water Report.
5. What volume of 0.250 mol/L sodium hydroxide is needed to neutralise 25.0 mL of 0.200 mol/L sulfuric acid?
Answer 5
Balanced equation first, because the ratio here is not 1 to 1 and that is the entire question:
40.0 mL of sodium hydroxide.
Sulfuric acid supplies two acidic hydrogens per molecule, so it takes twice as much base per mole as hydrochloric acid would. Skip the equation and you get 20.0 mL, which is exactly half of the right answer and looks entirely plausible on the page.
Sanity check: the base is more concentrated than the acid but is needed in twice the amount, so the volume should be somewhere near the aliquot volume rather than wildly different. 40.0 mL against 25.0 mL is comfortable.
6. 50.0 mL of 0.100 mol/L barium chloride is added to an excess of sodium sulfate solution. What mass of barium sulfate precipitates?
Answer 6
“An excess” tells you the barium chloride is limiting, so you never need the other solution’s numbers.
The ratio to the precipitate is 1 to 1, so mol.
1.17 g of barium sulfate.
This is a theoretical yield. What comes off the filter paper will be less, and by how much is the subject of Percentage Yield of a Precipitate.
7. Two conceptual questions. (a) A student has 0.0100 mol of solid and wants 0.100 mol/L. They add 100.0 mL of water to it in a beaker. What is wrong? (b) When you dilute a solution, what changes and what does not?
Answer 7
(a) Two separate problems, and both of them make the concentration wrong in the same direction.
The volume is of the wrong thing. Concentration is moles per litre of solution, not per litre of solvent. Dissolved solid takes up room, so 0.0100 mol of solute plus 100.0 mL of water gives slightly more than 100.0 mL of solution — and therefore a concentration slightly below 0.100 mol/L. This is why the technique is to dissolve in less water than you need and then make up to the mark.
The container is not calibrated. Beaker graduations are moulded into the glass during manufacture, not calibrated, and are commonly out by several percent. A volumetric flask is calibrated to a single line, to a stated tolerance, at a stated temperature, and that is the only reason its number can be trusted.
(b) The amount of solute in moles does not change — that is the entire content of , which is just the statement with both sides written as .
What changes: the volume goes up, the concentration goes down, and they do so in exact inverse proportion. Halve the concentration by doubling the volume.
A useful consequence: once you have added too much water, you cannot fix it by pouring some off. Removing solution removes solute along with it, so the concentration stays exactly where it was and you have less of it. The only repair is to start again.
8. A student writes: “I need 100.0 mL of 0.500 mol/L hydrochloric acid and I have a 2.00 mol/L stock. I measured 25.0 mL of the stock into a beaker and added 100.0 mL of water. Then I poured the water in first and added the acid to it — no, wait, the other way round, I poured the water into the acid. Anyway, diluting it means there are fewer moles of acid now, so it is safer.” Find every error.
Answer 8
Error 1 — the arithmetic gives the wrong concentration. The moles taken are right:
But adding 100.0 mL of water to 25.0 mL of stock gives a final volume of about 125.0 mL, not 100.0 mL. So
— not 0.500 mol/L. The student diluted to an added volume when the formula requires a final volume.
The correct method: run 25.0 mL of the stock into a 100.0 mL volumetric flask containing some distilled water, then make up to the mark so that the final volume is 100.0 mL. That gives mol/L exactly, and the flask does the measuring rather than the arithmetic.
Error 2 — the water went into the acid. This is the one that could hurt somebody. Add acid to water, never water to acid. Dilution releases heat. When the water is already in the flask, that heat is spread through a large volume and the temperature rise is small; when water is poured onto acid, the heat is released in the few drops at the surface, which can boil and throw acid out of the container at whoever is standing over it. The student noticed the rule, restated it backwards, and did the dangerous version.
Error 3 — diluting does not remove moles. The number of moles of hydrochloric acid is unchanged at 0.0500 mol before and after. What changed is how many of them are in each litre. All of the acid is still in the beaker and it will still neutralise exactly as much base as it would have before.
Dilute acid is less hazardous than concentrated acid, so the student’s conclusion is not wrong — but the reason given is, and a right answer resting on a wrong mechanism will fail the next question it meets.
Reference: Concentration and Water and Solutions. Making one for yourself: Preparing a Standard Solution. Measuring one you did not make: Titrating an Acid.
Curriculum connection
E2.1
use appropriate terminology related to aqueous solutions and solubility, including, but not limited to: concentration, solubility, precipitate, ionization, dissociation, pH, dilute, solute, and solvent [C]
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E2.2
solve problems related to the concentration of solutions by performing calculations involving moles, and express the results in various units (e.g., moles per litre, grams per 100 mL, parts per million or parts per billion, mass, volume per cent) [AI, C]
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E2.6
use stoichiometry to solve problems involving solutions and solubility [AI]
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