Curie Notes 09: Solutions and Concentration

On the bench in front of us there are four things: a small heap of blue crystals on a watch glass, a beaker of clean water, a tall glass flask with a thin neck and one line etched around it, and my notebook. In a few minutes the crystals will vanish into the water and the water will turn the colour of a summer sky. Nothing has been destroyed. The copper is still there, every atom of it, only spread out so evenly that no eye can find a grain. Most of the chemistry you will ever do happens in water like this. So today we learn three things: to say exactly how much of a substance is dissolved, to make a solution of a stated strength, and to make it weaker by known steps.

One of our joys was to go into our workroom at night; we then perceived on all sides the feebly luminous silhouettes of the bottles or capsules containing our products. It was really a lovely sight and one always new to us. The glowing tubes looked like faint, fairy lights.

Marie Curie, Autobiographical Notes, 1923

What you will learn

• Name the solute, the solvent and the solution in any mixture you dissolve.
• Explain, with a picture of the particles, why water dissolves salt and sugar but not oil.
• Calculate concentration as mass percent, as parts per million and as molarity.
• Work out the mass of solid you need to make a solution of a given molarity, and make it properly in a volumetric flask.
• Dilute a solution with the rule \(c_1V_1 = c_2V_2\), and plan a serial dilution.

What you need first: Lesson 1 (What Chemistry Studies: Matter, Mixtures and Pure Substances) for mixtures, Lesson 2 (Measurement, Units and Significant Figures) for units and significant figures, Lesson 5 (The Mole, or Counting Atoms by Weighing) and Lesson 6 (Chemical Formulas, Percent Composition and Empirical Formulas) for molar mass and hydrates, and Lesson 7 (Chemical Equations and How to Balance Them) for state symbols.

Solute, solvent and solution

When sugar disappears into tea, we have made a solution. A solution is a homogeneous mixture, the same in every part, in which one substance is spread through another as single molecules or ions. You met homogeneous mixtures in Lesson 1. A solution is the most perfect kind. You cannot see its parts even with a microscope, and it never settles on standing.

A solution has two roles in it. The solvent is the substance present in the larger amount, the one that does the dissolving. The solute is the substance dissolved in it. In sweet tea, water is the solvent and sugar is a solute. A solution can hold several solutes at once: the tea also holds the brown substances from the leaves.

When the solvent is water, we call the solution aqueous, from the Latin aqua. This is why the state symbol \(\ce{(aq)}\) appears in equations. When I write \(\ce{NaCl(aq)}\), I mean salt dissolved in water, which is a very different thing from \(\ce{NaCl(s)}\), the dry white crystals in a jar.

There is a limit to how much solute a given amount of solvent will take. A solution that holds as much solute as it can, at that temperature, is saturated. Add more crystals and they simply lie on the bottom. A solution with room for more is unsaturated. Most solids dissolve better in hot water than in cold, so a hot saturated solution that cools will often throw out crystals. We shall study this properly in Lesson 25.

An analogy from my Paris

Sit at the counter of any café on the boulevard Saint-Michel and watch a student sweeten his coffee. One lump of sugar in a small cup, and the coffee is sweet. The same lump in a large bowl of café au lait, and it is barely sweet at all. The amount of sugar did not change. What changed is how much sugar there is for each spoonful of drink. That is the whole idea of concentration: the amount of solute in a stated amount of solution. "How much sugar did you put in?" is a weak question. "How much sugar is there in each cup?" is the question a chemist asks.

In my laboratory

Almost every separation I made in the shed on rue Lhomond began by dissolving something. The crude barium and radium salts from the pitchblende residues were brought into solution, and I purified radium by crystallising barium chloride from water again and again. Radium chloride is a little less soluble than barium chloride. So each time crystals came out of a hot saturated solution, they carried a little more of the radium than the liquid kept. The whole method rested on knowing, at each step, how much salt that water could hold.

How water pulls a crystal apart

Why does salt vanish into water while a drop of lamp oil floats on top and never mixes? The answer lies in the shape of the water molecule and the way it shares its electrons.

A water molecule, \(\ce{H2O}\), is bent, like a shallow letter V, with the oxygen atom at the point. The oxygen atom pulls the shared electrons towards itself more strongly than the hydrogen atoms do. So the oxygen end carries a small negative charge, written \(\delta^{-}\) (read "delta minus"), and each hydrogen end carries a small positive charge, \(\delta^{+}\). A molecule with a positive end and a negative end is called polar. The charges are partial: smaller than the charge of an electron, but quite real.

Now drop a crystal of sodium chloride into the water. You will learn in Lesson 17 that this crystal is built of ions: atoms that have lost or gained electrons and so carry a whole charge. Here they are sodium ions, \(\ce{Na+}\), and chloride ions, \(\ce{Cl-}\), packed in a regular pattern. At the surface of the crystal, the water molecules turn to face them:

1. Around each positive \(\ce{Na+}\) ion, water molecules gather with their negative oxygen ends pointing inwards.
2. Around each negative \(\ce{Cl-}\) ion, water molecules gather with their positive hydrogen ends pointing inwards.
3. Each ion at the edge is tugged by many water molecules at once. One by one, the ions are pulled out of the crystal.
4. Once free, each ion travels wrapped in its own little cloak of water molecules. We say the ion is hydrated, and the cloak keeps it from finding its partner again.

We write the whole event as an equation, with water above the arrow to show the solvent:

\[ \ce{NaCl(s) ->[\text{water}] Na+(aq) + Cl-(aq)} \]

Sugar dissolves for a related reason. A sugar molecule has no whole charge, but it is covered with groups of oxygen and hydrogen atoms that are polar, like little pieces of water. Water molecules cling to these groups and draw the molecules away into the liquid. The sugar molecules stay whole: they do not break into ions. This difference, ions or whole molecules, will matter a great deal in Lesson 10.

Oil is made of molecules of carbon and hydrogen, which share their electrons almost equally. Oil molecules have no charged ends for the water to hold. The water molecules prefer to cling to each other, so they squeeze the oil out, and it rises in a layer. Chemists sum this up in an old rule: like dissolves like. Polar solvents dissolve polar and ionic solutes; non-polar solvents, such as petrol or turpentine, dissolve non-polar ones.

Margin

The idea that dissolved salt is present as separate charged ions came from the young Swede Svante Arrhenius, in his doctoral work of 1884 and his fuller theory of 1887. Many chemists laughed at it at first. Why should salt fall to pieces simply because it is wet? The evidence of electrical conductivity won the argument, and he received the Nobel Prize in Chemistry in 1903. We shall test his idea with our own lamp in the next lesson.

Measuring concentration: mass percent and parts per million

There are several ways to state a concentration. Each one answers the question "how much solute in how much solution?", with different units for "how much". We begin with the simplest, which needs only a balance.

The mass percent of a solute is the mass of solute in every hundred grams of solution. In words: divide the mass of solute by the mass of the whole solution, then multiply by one hundred.

\[ \text{mass percent} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100\% \]

The mass of the solution is the mass of solute plus the mass of solvent. Students forget this more often than any other step.

Worked example 1

I dissolve 12.0 g of sugar in 238 g of water. What is the mass percent of sugar?

1. Mass of solute: \(12.0\,\mathrm{g}\).
2. Mass of solution: \(12.0\,\mathrm{g} + 238\,\mathrm{g} = 250\,\mathrm{g}\).
3. Divide: \(12.0\,\mathrm{g} / 250\,\mathrm{g} = 0.0480\). The grams cancel, so the result is a pure number.
4. Multiply by 100: \(0.0480 \times 100\% = 4.80\%\).

Check: 12 g is about a twentieth of 250 g, and a twentieth is 5%. Our 4.80% is close to that.

An analogy from Warsaw

When I was a governess at Szczuki, the family I lived with ran a factory that made sugar from beets. Everything in that factory turned on one number: how many kilograms of sugar were in each hundred kilograms of beet juice. A cart of beets with a poor number was worth less, however heavy it was. That number is a mass percent. The factory chemist did not care how big the vat was; he cared what fraction of it was sugar, because the fraction tells you what you get from any amount you take.

For very small amounts of solute, percent gives awkward numbers like 0.0021%. Then we use parts per million, written ppm: the mass of solute in every million grams of solution. The formula has the same shape, with one million in place of one hundred.

\[ \text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \]

One part per million is very little. It is one second in about eleven and a half days, or one gram of salt in a tonne of water. A litre of a dilute solution in water has a mass very close to one kilogram. So one milligram of solute in one litre of solution is very nearly one part per million. Chemists who analyse drinking water use this all the time: they write "mg/L" and "ppm" almost as if they were the same.

Worked example 2

A 2.5 kg sample of hard water from a well contains 0.21 g of dissolved calcium. What is the calcium concentration in ppm?

1. Put both masses in the same unit: \(2.5\,\mathrm{kg} = 2500\,\mathrm{g}\).
2. Divide: \(0.21\,\mathrm{g} / 2500\,\mathrm{g} = 8.4 \times 10^{-5}\).
3. Multiply by one million: \(8.4 \times 10^{-5} \times 10^{6} = 84\,\mathrm{ppm}\).

Check: the sample is 2.5 litres, and \(210\,\mathrm{mg} / 2.5\,\mathrm{L} = 84\,\mathrm{mg/L}\), which agrees with the rule that mg/L is close to ppm.

In my laboratory

When I spoke to the students at Vassar College in 1921, I told them that even good ore holds less than one part of radium in a million parts. In the language of this lesson, radium in pitchblende is below one ppm. That is why we needed a tonne of residues to obtain a decigram of radium chloride. A concentration is a promise about what you will find in any amount you take; a low concentration is a promise that you will have to take a great deal.

Molarity: moles in each litre

Mass percent is honest, but it is not the most useful measure for a chemist. Reactions happen between particles, counted in moles, as you learned in Lesson 5. And in the laboratory we measure liquids by volume, with flasks, pipettes and burettes, far more easily than we weigh them. So we want a concentration that tells us how many moles of solute are in each litre of solution.

That measure is the molarity, also called the molar concentration. Its symbol is \(c\). In words: the molarity is the amount of solute in moles divided by the volume of the solution in litres.

\[ c = \dfrac{n}{V} \]

Here \(n\) is the amount of solute in moles and \(V\) is the volume of solution in litres. The unit is moles per litre, \(\mathrm{mol/L}\). Chemists often shorten it to a capital M and say "molar": a \(0.10\,\mathrm{M}\) solution is "point one molar", and holds 0.10 mol of solute in every litre.

Notice the word solution. The volume is the volume of the finished solution, never the volume of water you started with. The two are not the same, because the solute takes up room too.

The same formula can be turned round two ways, and you will need all three forms:

• moles from concentration and volume: \(n = c \times V\);
• volume from moles and concentration: \(V = n / c\).

Worked example 3

I dissolve 5.85 g of sodium chloride in water and make the solution up to 250.0 mL. What is its molarity?

1. Molar mass of \(\ce{NaCl}\): \(22.99 + 35.45 = 58.44\,\mathrm{g/mol}\).
2. Moles of solute: \(n = 5.85\,\mathrm{g} / 58.44\,\mathrm{g/mol} = 0.1001\,\mathrm{mol}\).
3. Volume in litres: \(250.0\,\mathrm{mL} = 0.2500\,\mathrm{L}\), since there are 1000 mL in a litre.
4. Molarity: \(c = 0.1001\,\mathrm{mol} / 0.2500\,\mathrm{L} = 0.400\,\mathrm{mol/L}\).

Check: a tenth of a mole in a quarter of a litre is four tenths of a mole in a whole litre. The units are mol/L, as they must be. We keep three significant figures, because the mass 5.85 g has three.

Worked example 4

I need 250.0 mL of 0.100 M copper sulfate solution, and the bottle holds the blue crystals, copper(II) sulfate pentahydrate. What mass do I weigh out?

1. Moles needed: \(n = c \times V = 0.100\,\mathrm{mol/L} \times 0.2500\,\mathrm{L} = 0.0250\,\mathrm{mol}\).
2. The blue crystals are the hydrate, \(\ce{CuSO4.5H2O}\), so we use the molar mass of the whole hydrate, as in Lesson 6: \(63.546 + 32.06 + 4(15.999) + 5(18.015) = 249.68\,\mathrm{g/mol}\).
3. Mass: \(m = n \times M = 0.0250\,\mathrm{mol} \times 249.68\,\mathrm{g/mol} = 6.24\,\mathrm{g}\).

Check: a quarter of a litre at a tenth molar is a fortieth of a mole, and a fortieth of about 250 g is about 6 g. The answer is sensible.

A common mistake

A student uses the molar mass of plain copper sulfate, \(\ce{CuSO4}\), which is 159.60 g/mol, and weighs 3.99 g of the blue crystals. But 3.99 g of the hydrate is only \(3.99 / 249.68 = 0.0160\,\mathrm{mol}\), so the flask holds \(0.0640\,\mathrm{mol/L}\), well short of the 0.100 she wrote on the label. The water in the crystals has mass but brings no copper. Always ask what is really in the bottle.

Worked example 5

How many moles of silver nitrate are in 25.0 mL of a 0.150 M solution, and what mass of the solid is that?

1. Volume in litres: \(25.0\,\mathrm{mL} = 0.0250\,\mathrm{L}\).
2. Moles: \(n = c \times V = 0.150\,\mathrm{mol/L} \times 0.0250\,\mathrm{L} = 3.75 \times 10^{-3}\,\mathrm{mol}\).
3. Molar mass of \(\ce{AgNO3}\): \(107.87 + 14.007 + 3(15.999) = 169.87\,\mathrm{g/mol}\).
4. Mass: \(3.75 \times 10^{-3}\,\mathrm{mol} \times 169.87\,\mathrm{g/mol} = 0.637\,\mathrm{g}\).

Check: the litres cancel against the "per litre" of the molarity, leaving moles. The mass is less than a gram, which suits a small volume of a dilute solution.

The concentration of each ion

When an ionic solid dissolves, each formula unit gives several ions, and each kind of ion has its own concentration. Dissolve sodium sulfate, and every unit gives two sodium ions and one sulfate ion:

\[ \ce{Na2SO4(s) -> 2Na+(aq) + SO4^2-(aq)} \]

So in a \(0.250\,\mathrm{M}\) solution of \(\ce{Na2SO4}\), the sodium ions are at \(2 \times 0.250 = 0.500\,\mathrm{mol/L}\) and the sulfate ions at \(0.250\,\mathrm{mol/L}\). Read the small numbers in the formula; they tell you how many of each ion you get. This will be our daily bread in the next lesson.

Try it

On Plate 9.1 choose a salt, move the mass slider and pick a flask. The plate weighs the solid, counts the moles, and fills the flask to its mark. Watch the colour deepen as the molarity rises, and watch the magnifying glass: the more moles in each litre, the more crowded the particles. Then put a large mass of permanganate in a small flask and see what saturation looks like.

Preparing a solution in a volumetric flask

A molarity is only as good as the volume it is divided by. To make a solution of exactly known concentration, we use a volumetric flask. It is a round flask with a long thin neck and a single line etched around the neck, the calibration mark. When the liquid reaches the mark, the flask holds exactly the volume printed on it, such as 250.0 mL, at the temperature printed beside it.

The neck is thin for a good reason. A small error in height there means a very small error in volume, so you can fill it to the mark very precisely. A beaker, with its wide mouth and rough scale, could never do this.

Here is the method, step by step. Follow it every time.

1. Calculate the mass of solid needed, as in Worked example 4.
2. Weigh the solid accurately on a balance, on a clean watch glass or weighing boat.
3. Tip it into a small beaker, rinse the watch glass into the beaker with distilled water, and stir with a glass rod until every crystal has dissolved. Use less water than the final volume: about half.
4. Pour the solution through a small funnel into the volumetric flask. Rinse the beaker, the rod and the funnel two or three times with a little distilled water, and pour every rinsing into the flask. The rinsings carry the last of the solute.
5. Add distilled water until the liquid is a little below the mark. Then add the last drops with a dropping pipette until the bottom of the curved surface just touches the line, with your eye level with it.
6. Stopper the flask and turn it upside down and back ten or fifteen times, so the solution is the same throughout.
7. Label it at once: the substance, the concentration, the date and your name.

The curved surface of the liquid in a narrow tube is called the meniscus. Water wets glass, so its meniscus dips in the middle. We always read the bottom of the dip, and always with the eye level with the mark. Look from above and you will stop too low; look from below and you will overfill.

A common mistake

Never put the solid into the flask and then pour in 250 mL of water. The solid takes up room, so the level would end above the mark. You would have more than 250 mL of solution and a lower concentration than you think. The flask is filled to the mark, never with a measured volume of water. The same mistake in another form: overshooting the mark and pouring a little out. The liquid you pour out carries solute with it. If you overshoot, start again.

An analogy from my Paris

In 1889 a bar of platinum and iridium was laid up at Sèvres, near Paris, as the international metre. A cylinder of the same metal became the kilogram. Every ruler and every weight in the world could be traced back to them. A volumetric flask is a little piece of the same idea on your bench. Its maker filled it, at a fixed temperature, with a mass of water whose volume was known, and etched the line exactly there. When you fill to the mark you borrow that care. Warm the flask in your hands, or fill it with hot solution, and you have broken the agreement: glass and water both expand, and the volume is no longer the one written on the flask.

In my laboratory

In August 1911 I prepared the international radium standard: 21.99 milligrams of very pure radium chloride, sealed in a thin glass tube a few centimetres long. That tube was kept as a solid. For the smallest quantities I devised a way of measuring radium through its emanation, the radioactive gas it gives off. That gas is collected most easily from a solution of the salt. Let us use the standard's mass for an exercise.

Worked example 6

Suppose 21.99 mg of radium chloride, \(\ce{RaCl2}\), is dissolved and made up to 100.0 mL in a volumetric flask. What is the molarity? Take the molar mass of radium as 226 g/mol.

1. Mass in grams: \(21.99\,\mathrm{mg} = 0.02199\,\mathrm{g}\).
2. Molar mass of \(\ce{RaCl2}\): \(226 + 2(35.45) = 296.9\,\mathrm{g/mol}\).
3. Moles: \(n = 0.02199\,\mathrm{g} / 296.9\,\mathrm{g/mol} = 7.407 \times 10^{-5}\,\mathrm{mol}\).
4. Volume: \(100.0\,\mathrm{mL} = 0.1000\,\mathrm{L}\).
5. Molarity: \(c = 7.407 \times 10^{-5}\,\mathrm{mol} / 0.1000\,\mathrm{L} = 7.41 \times 10^{-4}\,\mathrm{mol/L}\).

Check: dividing by a tenth of a litre multiplies by ten, so the moles and the molarity differ by exactly one power of ten. Such a solution looks like plain water, yet it would make an electrometer race.

Dilution: same solute, more water

We rarely make every solution from the solid. It is quicker and more accurate to keep a strong stock solution and make weaker ones from it by adding water. This is dilution.

The key idea is simple. When you add water, you add no solute. The number of moles of solute is the same before and after. Only the volume grows, so the concentration falls. Let us build the rule from that sentence.

1. Before: you take a volume \(V_1\) of stock with concentration \(c_1\). It holds \(n = c_1V_1\) moles of solute.
2. After: the same moles are spread through the final volume \(V_2\), at the new concentration \(c_2\). So \(n = c_2V_2\).
3. The moles are equal, so the two products are equal:

\[ c_1V_1 = c_2V_2 \]

Any three of these numbers give you the fourth. The volumes can be in millilitres, as long as both are in the same unit, because the unit cancels from the two sides.

An analogy from my Paris

In a Paris café a child asks for a sirop, and the waiter brings a glass with a finger of red grenadine syrup in the bottom and a jug of water beside it. Fill the glass and the drink turns pale pink. All the syrup that was in the finger is still in the glass, not a drop more or less; it is only spread through more water. If the waiter had poured two fingers of syrup, you would need twice the water for the same pinkness. That is \(c_1V_1 = c_2V_2\) in a glass.

Worked example 7

I want 500.0 mL of 0.100 M hydrochloric acid, and the stock bottle is 2.00 M. What volume of stock do I take?

1. Write what we know: \(c_1 = 2.00\,\mathrm{M}\), \(c_2 = 0.100\,\mathrm{M}\), \(V_2 = 500.0\,\mathrm{mL}\). We want \(V_1\).
2. Start from \(c_1V_1 = c_2V_2\) and divide both sides by \(c_1\): \(V_1 = c_2V_2 / c_1\).
3. Substitute: \(V_1 = (0.100\,\mathrm{M} \times 500.0\,\mathrm{mL}) / 2.00\,\mathrm{M}\).
4. Calculate: \(V_1 = 50.0 / 2.00 = 25.0\,\mathrm{mL}\). The units M cancel, leaving mL.
5. Method: measure 25.0 mL of stock with a pipette into a 500 mL volumetric flask half full of water, then fill to the mark and invert.

Check: the new solution is twenty times weaker (2.00 / 0.100 = 20), so the stock must be one twentieth of the final volume, and \(500 / 20 = 25\).

A common mistake

A student takes 25.0 mL of stock and adds 500.0 mL of water. Now the volume is 525 mL and the concentration is \(2.00 \times 25.0 / 525 = 0.0952\,\mathrm{M}\), not 0.100 M. In \(c_1V_1 = c_2V_2\), the volume \(V_2\) is always the final volume of the solution, never the water added. And a word of safety that I learned before any formula: with strong acids, put the acid into the water, never water into the acid. Water poured on concentrated acid can boil where it lands and spit.

Worked example 8

10.0 mL of 0.500 M copper sulfate is diluted to 250.0 mL. What is the new concentration?

1. We know \(c_1 = 0.500\,\mathrm{M}\), \(V_1 = 10.0\,\mathrm{mL}\) and \(V_2 = 250.0\,\mathrm{mL}\). We want \(c_2\).
2. Divide both sides of \(c_1V_1 = c_2V_2\) by \(V_2\): \(c_2 = c_1V_1 / V_2\).
3. Substitute: \(c_2 = (0.500\,\mathrm{M} \times 10.0\,\mathrm{mL}) / 250.0\,\mathrm{mL}\).
4. Calculate: \(c_2 = 5.00 / 250.0 = 0.0200\,\mathrm{M}\).

Check: the volume grew 25 times (250 / 10), so the concentration fell 25 times: \(0.500 / 25 = 0.0200\). The ratio \(V_2/V_1\) is called the dilution factor; here it is 25.

Try it

Plate 9.2 has two modes. In One dilution, set the stock concentration, the volume you pipette and the flask you fill. The dots in the pipette are the same dots that end up in the flask. Count them if you like: only the space around them grows. In Serial dilution, choose the factor and the number of steps, and watch the concentration fall tube by tube.

Serial dilutions

Suppose you need a solution a million times weaker than your stock. In one step you would need to put 1 mL of stock into a flask of a thousand litres. No laboratory has such a flask. Instead we dilute in a chain of small steps, each by the same factor. This is a serial dilution.

The usual step is tenfold: take 1.00 mL of the solution, add 9.00 mL of water, and mix, so that the 1.00 mL becomes 10.00 mL. Each step divides the concentration by 10. After \(k\) steps the concentration is the starting one divided by \(10^{k}\):

\[ c_k = \dfrac{c_0}{10^{k}} \]

The same formula works with any factor \(f\) in place of 10: \(c_k = c_0 / f^{k}\).

Worked example 9

A 0.100 M stock is diluted tenfold four times in a row (1.00 mL into 9.00 mL of water each time). What is the concentration in the fourth tube?

1. Each step: take 1.00 mL, make it up to 10.00 mL, so each step has a dilution factor of \(10.00 / 1.00 = 10\).
2. Tube 1: \(0.100 / 10 = 1.00 \times 10^{-2}\,\mathrm{M}\).
3. Tube 2: \(1.00 \times 10^{-2} / 10 = 1.00 \times 10^{-3}\,\mathrm{M}\).
4. Tube 3: \(1.00 \times 10^{-4}\,\mathrm{M}\).
5. Tube 4: \(1.00 \times 10^{-5}\,\mathrm{M}\).
6. In one line: \(c_4 = 0.100 / 10^{4} = 1.00 \times 10^{-5}\,\mathrm{M}\).

Check: four steps of ten is a total factor of 10,000, and 0.1 divided by 10,000 is 0.00001. Four pipettings of 1 mL and 36 mL of water did the work. A single step would need a whole litre flask and a pipette of a tenth of a millilitre.

A common mistake

Some students add the factors instead of multiplying them: "four tenfold steps make a fortyfold dilution." No. Each step works on what the step before it left. Ten times ten times ten times ten is ten thousand. And an error in any step is carried down the whole chain, so each pipetting must be done with care.

From the shed

My fractional crystallisations were a serial process run backwards. Each crystallisation enriched the radium in the crystals by a modest factor compared with the liquid. One step alone was almost nothing. But the factors multiplied: the crystals of one step were the starting material of the next, and after many steps the modest factor had become an enormous one. I learned in that shed what you learn from a serial dilution: a small factor repeated is a great power.

Try it

Exercise 9.3 brings everything together: mass percent, parts per million, molarity, the mass to weigh, the concentration of an ion, dilution and serial dilution, and the order of steps at the flask. Press "New numbers" for fresh values once you have the method.

What to remember

• A solution is a homogeneous mixture: the solute dissolves in the solvent; in water it is aqueous, \(\ce{(aq)}\).
• Water is polar. It pulls ions out of a crystal and wraps each one in a cloak of water molecules. Like dissolves like.
• Mass percent is the mass of solute over the mass of solution, times 100%; the solution's mass includes the solute.
• Parts per million is the same fraction times a million, \(10^{6}\); in dilute water, 1 mg/L is about 1 ppm.
• Molarity \(c = n/V\), in mol/L (M), with \(V\) the volume of solution in litres; also \(n = cV\).
• Each ion has its own concentration: 0.250 M \(\ce{Na2SO4}\) is 0.500 M in \(\ce{Na+}\).
• Make solutions in a volumetric flask: dissolve first, transfer with rinsings, fill to the mark at eye level, invert.
• Dilution keeps the moles: \(c_1V_1 = c_2V_2\), with \(V_2\) the final volume.
• Serial dilution multiplies the factors: \(c_k = c_0 / f^{k}\).

Problems for your notebook

Problem 1. I dissolve 15.0 g of salt in 135 g of water. What is the mass percent of salt in the solution?

Answer. Mass of solution: \(15.0 + 135 = 150\,\mathrm{g}\). Mass percent: \(15.0 / 150 \times 100\% = 10.0\%\).

Problem 2. 4.00 g of sodium hydroxide, \(\ce{NaOH}\), is dissolved and made up to 500.0 mL. What is the molarity?

Answer.

1. Molar mass: \(22.99 + 15.999 + 1.008 = 40.00\,\mathrm{g/mol}\).
2. Moles: \(4.00 / 40.00 = 0.100\,\mathrm{mol}\).
3. Molarity: \(0.100\,\mathrm{mol} / 0.5000\,\mathrm{L} = 0.200\,\mathrm{M}\).

Problem 3. What mass of potassium permanganate, \(\ce{KMnO4}\), do you need for 100.0 mL of a 0.0200 M solution?

Answer.

1. Moles: \(n = 0.0200\,\mathrm{mol/L} \times 0.1000\,\mathrm{L} = 2.00 \times 10^{-3}\,\mathrm{mol}\).
2. Molar mass: \(39.098 + 54.938 + 4(15.999) = 158.03\,\mathrm{g/mol}\).
3. Mass: \(2.00 \times 10^{-3} \times 158.03 = 0.316\,\mathrm{g}\).

Problem 4. How much 6.0 M hydrochloric acid do you need to make 250 mL of 1.5 M acid? Describe how you would mix it.

Answer. \(V_1 = c_2V_2 / c_1 = 1.5 \times 250 / 6.0 = 62.5\,\mathrm{mL}\), which is 63 mL to two significant figures. Half fill a 250 mL volumetric flask with water, add the acid to the water, let it cool to room temperature, then fill to the mark and invert.

Problem 5. I dissolve 0.500 g of copper(II) sulfate pentahydrate and make it up to 100.0 mL (solution A). I then dilute 10.00 mL of A to 250.0 mL (solution B). What is the molarity of B, and how many milligrams of copper does each litre of B contain?

Answer.

1. Moles of hydrate: \(0.500 / 249.68 = 2.003 \times 10^{-3}\,\mathrm{mol}\).
2. Solution A: \(2.003 \times 10^{-3}\,\mathrm{mol} / 0.1000\,\mathrm{L} = 2.003 \times 10^{-2}\,\mathrm{M}\).
3. Solution B: \(c_2 = 2.003 \times 10^{-2} \times 10.00 / 250.0 = 8.01 \times 10^{-4}\,\mathrm{M}\).
4. Each mole of hydrate holds one mole of copper, so B has \(8.01 \times 10^{-4}\,\mathrm{mol}\) of copper per litre.
5. Mass of copper per litre: \(8.01 \times 10^{-4} \times 63.546 = 0.0509\,\mathrm{g} = 50.9\,\mathrm{mg}\), which is about 51 ppm.

After my time

In my day the mole was a chemist's habit, a convenient way of counting by weighing. In 1971 the international system of units made the mole one of its seven base units. In 2019 the mole was redefined by fixing the Avogadro constant at exactly \(6.02214076 \times 10^{23}\,\mathrm{mol^{-1}}\). Your molarities now rest on that fixed number, as your metres rest on the speed of light instead of a bar at Sèvres.

Modern chemists also write the litre as the cubic decimetre, \(\mathrm{dm^3}\), so you will see molarity printed as \(\mathrm{mol\,dm^{-3}}\). It is the same thing. And the analysis of water has gone far below parts per million. Instruments today measure parts per billion and even per trillion, and the laws on drinking water are written in those units. The method of thinking is the one in this lesson: how much solute, in how much solution.

Come and ask me about anything here that is still unclear. You might begin with this: come and ask me. In the next lesson, Reactions in Water: Precipitates and Net Ionic Equations, we pour two of these solutions together and watch a solid fall out of clear liquid.

Sources and liberties

The Curie Notes are a modern chemistry course written in Marie Curie's voice by Eternal Gardens. The chemistry is standard modern chemistry; dates, people, apparatus and quotations come from the sources below; the teaching, the analogies and the private remarks are our reconstruction.

• Marie Curie, Pierre Curie, with her Autobiographical Notes, translated by Charlotte and Vernon Kellogg (1923), Project Gutenberg.
• Marie Curie, The Discovery of Radium, address at Vassar College, 14 May 1921, Project Gutenberg.
• OpenStax, Chemistry 2e, chapter 3 (solution concentration and dilution) and chapter 11 (solutions), openstax.org.
• American Institute of Physics, Marie Curie and the Science of Radioactivity, history.aip.org.
• NIST, the international system of units and the 2019 redefinition of the mole, nist.gov.
• IUPAC, Compendium of Chemical Terminology (the Gold Book), entries for concentration and amount concentration, goldbook.iupac.org.
• The Nobel Prize in Chemistry 1903 (Svante Arrhenius), nobelprize.org.

The worked example dissolving 21.99 mg of radium chloride is an exercise: the international standard itself was kept as a sealed solid, and we do not claim it was ever dissolved.