School Chemistry — Grades 1 to 12 · Grades 1–12

26Concentration and Dilution

A hospital drip bag hangs above a patient’s bed. Its label reads “0.9 % sodium chloride”: salt in water, nothing more. Yet the figure 0.9 is not a detail. Too little salt, and the patient’s blood cells would swell with water; too much, and they would shrivel. A solution is not described by its ingredients alone, but by how much solute each volume of it holds, its concentration. This chapter measures concentrations, prepares solutions of an exact concentration, and dilutes them.

You already know

A solution is made of a solvent and one or more dissolved solutes; the solubility is the largest mass of solute that a given quantity of solvent can dissolve (Chapter 9). The amount of a species is $n = m/M$ , in moles (Chapter 25).

A drip bag of salt solution: the concentration must be right.

26.1 Mass concentration

Definition 26.1 (Mass concentration)

The mass concentration $C_m$ of a solute in a solution is the mass of dissolved solute per volume of solution:

C_m = \frac{m_{\text{solute}}}{V_{\text{solution}}},

in grams per litre ( $\mathrm{g}/\mathrm{L}$ ).

Example 26.2 (Sugared water)

$20\,\mathrm{g}$ of sugar are dissolved in water, and the solution is made up to $0.50\,\mathrm{L}$ . Its mass concentration in sugar is $C_m = 20\,\mathrm{g} / 0.50\,\mathrm{L} = 40\,\mathrm{g}/\mathrm{L}$ . Any part of it has the same concentration: a glass of $0.20\,\mathrm{L}$ of it holds $40\,\mathrm{g}/\mathrm{L} \times 0.20\,\mathrm{L} = 8.0\,\mathrm{g}$ of sugar.

Remark 26.3 (The volume is that of the solution)

The volume in $C_m$ is the volume of the whole solution, not that of the water poured in: dissolving a solute changes the volume a little. That is why a solution is always made up to a known final volume, in a flask marked for it.

Remark 26.4 (Concentration is not solubility)

The solubility of a solute is the largest concentration it can reach; a solution may have any concentration from zero up to it. Sodium chloride dissolves up to about $360\,\mathrm{g}$ per kilogram of water at $25\,{}^{\circ}\mathrm{C}$ ; the drip bag holds only $9\,\mathrm{g}$ per litre, very far from saturation.

Remark 26.5 (Percentages on labels)

On medical and food labels, “0.9 %” for a solid dissolved in water usually means $0.9\,\mathrm{g}$ of solute per $100\,\mathrm{mL}$ of solution. In grams per litre it is ten times more: the drip bag holds $9.0\,\mathrm{g}/\mathrm{L}$ of sodium chloride.

26.2 Molar concentration

Definition 26.6 (Molar concentration)

The molar concentration $c$ of a solute in a solution is the amount of dissolved solute per volume of solution:

c = \frac{n_{\text{solute}}}{V_{\text{solution}}},

in moles per litre ( $\mathrm{mol}/\mathrm{L}$ ).

Proposition 26.7 (From one concentration to the other)

For a solute of molar mass $M$ ,

C_m = c \times M .

Proof. $C_m = m / V = (n \times M) / V = (n / V) \times M = c \times M$ . ∎

Example 26.8 (The drip bag in moles)

The drip bag holds $9.0\,\mathrm{g}/\mathrm{L}$ of sodium chloride, of molar mass $58.5\,\mathrm{g}/\mathrm{mol}$ . Its molar concentration is

c = \frac{C_m}{M} = \frac{9.0\,\mathrm{g}/\mathrm{L}}{58.5\,\mathrm{g}/\mathrm{mol}} = 0.154\,\mathrm{mol}/\mathrm{L}.

The manufacturer’s label gives the same figure the other way round: 154 thousandths of a mole of sodium ions per litre.

Remark 26.9 (Ions in solution)

Sodium chloride dissolves as separate ions $\ce{Na+}$ and $\ce{Cl-}$ (Chapter 17): a solution of sodium chloride at $0.154\,\mathrm{mol}/\mathrm{L}$ holds $0.154\,\mathrm{mol}/\mathrm{L}$ of sodium ions and $0.154\,\mathrm{mol}/\mathrm{L}$ of chloride ions. How an ionic solid comes apart in water is the subject of a later chapter.

Squash from the bottle is diluted with water in a jug: the drink holds the same amount of fruit syrup, in a larger volume.

26.3 Preparing a solution by dissolving

Method 26.10 (Preparing a solution by dissolving a solid)

To prepare a volume $V$ of solution of molar concentration $c$ :

Compute the amount needed, $n = c \times V$ , and the mass to weigh, $m = n \times M$ .
Weigh this mass of solid in a small dish on a balance.
Pour the solid through a funnel into a volumetric flask of volume $V$ ; rinse the dish and the funnel with distilled water into the flask, so that no solid is lost.
Add distilled water up to about half of the flask and swirl until all the solid has dissolved.
Add distilled water up to the mark, the last drops with a dropper; stopper the flask and turn it over several times to mix.

Preparing a solution by dissolving a solid: weigh, transfer, dissolve, then make up to the mark of the volumetric flask.

In the lab — Filling to the mark

The mark of a volumetric flask is a ring around its neck. Water in a narrow glass tube does not end flat: its surface curves down in the middle, forming a meniscus. The flask is full when the bottom of the meniscus touches the mark, seen with the eye at the level of the mark, so that the front and the back of the ring look like a single line. Seen from above or from below, the level can look right when it is not.

The neck of a volumetric flask, enlarged and seen with the eye at the level of the meniscus. The blue line is the mark. Only the left-hand flask is filled correctly.

26.4 Diluting

Definition 26.11 (Dilution, stock solution, dilution factor)

A dilution lowers the concentration of a solution by adding solvent. The concentrated solution taken at the start is the stock solution. The dilution factor $F$ is the ratio of the concentration of the stock solution to that of the diluted solution:

F = \frac{c_{\text{stock}}}{c_{\text{diluted}}} .

Proposition 26.12 (The amount of solute is kept)

During a dilution, the amount of solute does not change: if a volume $V_0$ of stock solution of concentration $c_0$ is made up to a volume $V_1$ , the diluted solution has the concentration $c_1$ given by

c_0 \times V_0 = c_1 \times V_1, \qquad\text{so}\qquad F = \frac{c_0}{c_1} = \frac{V_1}{V_0} .

The same holds for mass concentrations.

Proof. Only solvent is added: the solute taken with the volume $V_0$ , an amount $c_0 V_0$ , is all in the final volume $V_1$ , where it makes up the amount $c_1 V_1$ . ∎

Method 26.13 (Preparing a solution by dilution)

To prepare a volume $V_1$ of solution of concentration $c_1$ from a stock solution of concentration $c_0$ :

Compute the volume of stock solution to take, $V_0 = c_1 V_1 / c_0$ (that is, $V_1 / F$ ).
Pour a little stock solution into a clean beaker (never pipette straight from the bottle) and take exactly $V_0$ with a volumetric pipette, filled to its mark.
Let the pipette empty into a volumetric flask of volume $V_1$ .
Add distilled water up to the mark, stopper and mix.

Example 26.14 (A tenfold dilution)

A stock solution of copper sulfate has $c_0 = 0.10\,\mathrm{mol}/\mathrm{L}$ ; $100.0\,\mathrm{mL}$ at $c_1 = 0.010\,\mathrm{mol}/\mathrm{L}$ are needed. The dilution factor is $F = 0.10 / 0.010 = 10$ , so $V_0 = 100.0\,\mathrm{mL} / 10 = 10.0\,\mathrm{mL}$ : a $10.0\,\mathrm{mL}$ pipette and a $100.0\,\mathrm{mL}$ flask.

Diluting: a volume V_0 of stock solution, measured with a pipette, is made up to V_1 in a volumetric flask. The colour fades by the dilution factor V_1 / V_0. — Diluting: a volume $V_0$ of stock solution, measured with a pipette, is made up to $V_1$ in a volumetric flask. The colour fades by the dilution factor $V_1 / V_0$ .

26.5 A calibration scale

Definition 26.15 (Standard solution, calibration scale)

A standard solution is a solution whose concentration is accurately known. A calibration scale is a series of standard solutions of the same solute, at increasing concentrations, usually made by diluting one stock solution; comparing an unknown solution with it gives an estimate of the unknown’s concentration.

Example 26.16 (A blue scale)

From a stock solution of copper sulfate at $0.20\,\mathrm{mol}/\mathrm{L}$ , six standards are prepared at 0.02, 0.04, 0.06, 0.08, 0.10 and $0.12\,\mathrm{mol}/\mathrm{L}$ , each in identical tubes filled to the same height. A solution of unknown concentration, in the same kind of tube, looks darker than the $0.06\,\mathrm{mol}/\mathrm{L}$ tube and paler than the $0.08\,\mathrm{mol}/\mathrm{L}$ one: its concentration lies between the two, about $0.07\,\mathrm{mol}/\mathrm{L}$ .

A calibration scale of copper sulfate solutions and an unknown solution. The unknown lies between the third and fourth tubes.

Safety

Copper sulfate solutions: harmful if swallowed, irritating to the skin, damaging to the eyes, very toxic to aquatic life. Goggles and gloves; the used solutions are collected for treatment, never poured down the sink.

Remark 26.17 (The limits of the eye)

The eye can only say “between these two tubes”. To measure a concentration more finely from the colour of a solution, chemists measure how much light it absorbs; a later chapter does exactly that.

26.6 Exercises

Exercise 26.1 ★

$5.0\,\mathrm{g}$ of sugar are dissolved to make $250\,\mathrm{mL}$ of solution. Compute the mass concentration of sugar in grams per litre.

Solution

Solution of Exercise 26.1.

$C_m = 5.0\,\mathrm{g} / 0.250\,\mathrm{L} = 20\,\mathrm{g}/\mathrm{L}$ .

Exercise 26.2 ★

A solution of $200\,\mathrm{mL}$ contains $0.050\,\mathrm{mol}$ of glucose. Compute its molar concentration.

Solution

Solution of Exercise 26.2.

$c = 0.050\,\mathrm{mol} / 0.200\,\mathrm{L} = 0.25\,\mathrm{mol}/\mathrm{L}$ .

Exercise 26.3 ★

What mass of anhydrous copper sulfate $\ce{CuSO4}$ must be weighed to prepare $100.0\,\mathrm{mL}$ of solution at $0.10\,\mathrm{mol}/\mathrm{L}$ ?

Solution

Solution of Exercise 26.3.

$M(\ce{CuSO4}) = 63.5 + 32.1 + 4 \times 16.0 = 159.6\,\mathrm{g}/\mathrm{mol}$ ; $n = 0.10 \times 0.1000 = 0.0100\,\mathrm{mol}$ ; $m = 0.0100 \times 159.6 = 1.60\,\mathrm{g}$ .

Exercise 26.4 ★

A stock solution at $1.0\,\mathrm{mol}/\mathrm{L}$ is diluted to $0.050\,\mathrm{mol}/\mathrm{L}$ . What is the dilution factor? What volume of stock is needed to make $100.0\,\mathrm{mL}$ of diluted solution?

Solution

Solution of Exercise 26.4.

$F = 1.0 / 0.050 = 20$ ; $V_0 = 100.0\,\mathrm{mL} / 20 = 5.0\,\mathrm{mL}$ .

Exercise 26.5 ★

A solution of sodium chloride has a molar concentration of $0.10\,\mathrm{mol}/\mathrm{L}$ . What is its mass concentration?

Solution

Solution of Exercise 26.5.

$C_m = c \times M = 0.10 \times 58.5 = 5.85\,\mathrm{g}/\mathrm{L}$ , about $5.9\,\mathrm{g}/\mathrm{L}$ .

Exercise 26.6 ★★

What volume of a stock solution at $0.50\,\mathrm{mol}/\mathrm{L}$ must be taken to prepare $250.0\,\mathrm{mL}$ of solution at $0.020\,\mathrm{mol}/\mathrm{L}$ ? Name the two pieces of glassware used.

Solution

Solution of Exercise 26.6.

$V_0 = c_1 V_1 / c_0 = 0.020 \times 250.0 / 0.50 = 10.0\,\mathrm{mL}$ (dilution factor 25). A $10.0\,\mathrm{mL}$ volumetric pipette and a $250.0\,\mathrm{mL}$ volumetric flask.

Exercise 26.7 ★★

Why is a solution of exact concentration made up in a volumetric flask rather than in a beaker with graduations? Why is a volume of stock measured with a volumetric pipette rather than with a graduated cylinder?

Solution

Solution of Exercise 26.7.

The volumetric flask has a single mark, made for one volume and accurate to a small fraction of a percent; the graduations of a beaker are only a rough guide. Likewise a volumetric pipette delivers one volume very accurately, while a graduated cylinder is wider and read less precisely.

Exercise 26.8 ★★

Look at the calibration scale of copper sulfate. A second unknown is paler than the $0.04\,\mathrm{mol}/\mathrm{L}$ tube but darker than the $0.02\,\mathrm{mol}/\mathrm{L}$ one. Estimate its concentration. How could the scale be changed to estimate it more precisely?

Solution

Solution of Exercise 26.8.

Between $0.02\,\mathrm{mol}/\mathrm{L}$ and $0.04\,\mathrm{mol}/\mathrm{L}$ : about $0.03\,\mathrm{mol}/\mathrm{L}$ . For more precision, add standards between these two, for example at 0.025, 0.030 and $0.035\,\mathrm{mol}/\mathrm{L}$ : a finer scale where it is needed.

Exercise 26.9 ★★

A glucose solution for infusion contains $50\,\mathrm{g}/\mathrm{L}$ of glucose $\ce{C6H12O6}$ . Compute its molar concentration.

Solution

Solution of Exercise 26.9.

$M(\ce{C6H12O6}) = 180.0\,\mathrm{g}/\mathrm{mol}$ ; $c = 50 / 180.0 = 0.28\,\mathrm{mol}/\mathrm{L}$ .

Exercise 26.10 ★★

A syrup is made by dissolving $342\,\mathrm{g}$ of sucrose $\ce{C12H22O11}$ to make $1.00\,\mathrm{L}$ of solution. Compute its molar concentration. It is then diluted ten times. What mass of sucrose does a $20\,\mathrm{mL}$ spoonful of the diluted syrup contain?

Solution

Solution of Exercise 26.10.

$M = 342.0\,\mathrm{g}/\mathrm{mol}$ , so $n = 1.00\,\mathrm{mol}$ and $c = 1.00\,\mathrm{mol}/\mathrm{L}$ . Diluted ten times: $c = 0.100\,\mathrm{mol}/\mathrm{L}$ , that is $C_m = 0.100 \times 342.0 = 34.2\,\mathrm{g}/\mathrm{L}$ . A spoonful of $0.020\,\mathrm{L}$ holds $34.2 \times 0.020 = 0.68\,\mathrm{g}$ of sucrose.

Exercise 26.11 ★★

A solution has a mass concentration of $12\,\mathrm{g}/\mathrm{L}$ . What is its concentration (a) after water is added to double its volume; (b) after half of its water has been evaporated, all the solute remaining dissolved?

Solution

Solution of Exercise 26.11.

(a) The same mass of solute in twice the volume: $6.0\,\mathrm{g}/\mathrm{L}$ . (b) The volume is about halved, the mass of solute unchanged: about $24\,\mathrm{g}/\mathrm{L}$ .

Exercise 26.12 ★★★

A solution at $0.10\,\mathrm{mol}/\mathrm{L}$ is diluted ten times, the result diluted ten times again, and once more.

What is the final concentration?
Why is this done in three steps rather than in a single dilution by 1000 (think of the volumes of glassware needed)?

Solution

Solution of Exercise 26.12.

Three dilutions by 10 make a dilution by $10^3$ : $0.10\,\mathrm{mol}/\mathrm{L} / 1000 = 1.0 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ .
A single dilution by 1000 would need either a tiny volume of stock ( $0.1\,\mathrm{mL}$ into $100\,\mathrm{mL}$ ), impossible to measure accurately with a pipette, or a very large flask ( $1\,\mathrm{mL}$ into $1\,\mathrm{L}$ ). Three tenfold dilutions use ordinary pipettes and flasks.

Exercise 26.13 ★★★

$11.1\,\mathrm{g}$ of calcium chloride $\ce{CaCl2}$ are dissolved to make $500.0\,\mathrm{mL}$ of solution. In water the solid separates into calcium ions $\ce{Ca^2+}$ and chloride ions $\ce{Cl-}$ .

Compute the molar concentration of calcium chloride dissolved.
Deduce the molar concentrations of the calcium ions and of the chloride ions in the solution.

Solution

Solution of Exercise 26.13.

$M(\ce{CaCl2}) = 40.1 + 2 \times 35.5 = 111.1\,\mathrm{g}/\mathrm{mol}$ ; $n = 11.1 / 111.1 = 0.100\,\mathrm{mol}$ ; $c = 0.100 / 0.5000 = 0.200\,\mathrm{mol}/\mathrm{L}$ .
Each $\ce{CaCl2}$ gives one $\ce{Ca^2+}$ and two $\ce{Cl-}$ : $[\ce{Ca^2+}] = 0.200\,\mathrm{mol}/\mathrm{L}$ and $[\ce{Cl-}] = 0.400\,\mathrm{mol}/\mathrm{L}$ .

Exercise 26.14 ★★★

A $100.0\,\mathrm{mL}$ volumetric flask has a neck of inner diameter $1.2\,\mathrm{cm}$ . A student fills it with the bottom of the meniscus $2\,\mathrm{mm}$ above the mark.

What extra volume of water has been added (volume of a cylinder: $\pi r^2 h$ )?
By what percentage is the concentration of the solution too low?

Solution

Solution of Exercise 26.14.

$r = 0.60\,\mathrm{cm}$ , $h = 0.20\,\mathrm{cm}$ : $V = \pi \times 0.60^2 \times 0.20 = 0.23\,\mathrm{cm}^{3}$ , that is $0.23\,\mathrm{mL}$ too much water.
The same solute is in $100.23\,\mathrm{mL}$ instead of $100.00\,\mathrm{mL}$ : the concentration is too low by $0.23 / 100.23 \approx 0.23\,\%$ .

Exercise 26.15 ★★★

$100\,\mathrm{mL}$ of sodium chloride solution at $0.20\,\mathrm{mol}/\mathrm{L}$ are mixed with $300\,\mathrm{mL}$ of sodium chloride solution at $0.10\,\mathrm{mol}/\mathrm{L}$ . Assuming the volumes add, compute the concentration of the mixture. Is it the average of the two concentrations? Explain.

Solution

Solution of Exercise 26.15.

$n = 0.20 \times 0.100 + 0.10 \times 0.300 = 0.020 + 0.030 = 0.050\,\mathrm{mol}$ in $0.400\,\mathrm{L}$ : $c = 0.125\,\mathrm{mol}/\mathrm{L}$ . It is not the plain average, $0.15\,\mathrm{mol}/\mathrm{L}$ : there is three times more of the dilute solution, so the mixture is closer to $0.10\,\mathrm{mol}/\mathrm{L}$ . It is the average weighted by the volumes.

26.7 Problem: The Drip Bag

Problem 26.1

Weekend problem — how much of a concentrated salt solution goes into one bag of “0.9 %” saline?

A hospital pharmacy laboratory keeps a concentrated sterile solution of sodium chloride containing $200\,\mathrm{g}/\mathrm{L}$ (on its label: “20 %”). It must prepare bags of $500\,\mathrm{mL}$ of the “0.9 %” solution used in drips, by diluting the concentrated solution with sterile water.

Part I — What 0.9 % means.

The label “0.9 %” means $0.9\,\mathrm{g}$ of salt per $100\,\mathrm{mL}$ of solution. Give the mass concentration in $\mathrm{g}/\mathrm{L}$ .
Express it also in milligrams per millilitre.
What mass of sodium chloride does a $500\,\mathrm{mL}$ bag contain?
Sodium chloride dissolves up to about $360\,\mathrm{g}$ per kilogram of water. Is the solution of the bag far from saturation?
A patient receives $1.0\,\mathrm{L}$ of this solution in a day. What mass of salt is that?

Part II — In moles.

Compute the molar mass of sodium chloride.
Compute the molar concentration of the solution of the bag.
What amount of sodium chloride does one bag contain?
What are the molar concentrations of sodium ions and of chloride ions in the bag?
How many sodium ions does one bag contain?

Part III — From the concentrated solution.

Compute the dilution factor between the concentrated solution and the solution of the bag.
Which mass of sodium chloride must come from the concentrated solution for one bag? Deduce the volume of concentrated solution to take.
Check this volume with the dilution factor.
Compute the molar concentration of the concentrated solution, and check that $c_0 V_0 = c_1 V_1$ .
About what volume of sterile water is then added (assume the volumes add)?

Part IV — Glassware and accuracy.

No volumetric pipette of this volume exists. Which piece of glassware, graduated in tenths of a millilitre, could deliver it?
In which vessel should the final volume be made up, to be accurate?
If $23.0\,\mathrm{mL}$ were taken instead of the right volume, what mass concentration would the bag have? By what percentage would it be wrong?
State the final answer: what volume of the concentrated solution goes into one $500\,\mathrm{mL}$ bag?

Solution

Solution of Problem 26.1.

1. $0.9\,\mathrm{g}$ per $0.100\,\mathrm{L}$ : $C_m = 9.0\,\mathrm{g}/\mathrm{L}$ .

2. $9.0\,\mathrm{g}$ per $1000\,\mathrm{mL}$ , that is $9.0\,\mathrm{mg}$ per millilitre.

3. $m = 9.0 \times 0.500 = 4.5\,\mathrm{g}$ .

4. Yes: about $9\,\mathrm{g}$ per litre against some $360\,\mathrm{g}$ per kilogram of water at saturation, about forty times less.

5. $9.0 \times 1.0 = 9.0\,\mathrm{g}$ of salt.

6. $M(\ce{NaCl}) = 23.0 + 35.5 = 58.5\,\mathrm{g}/\mathrm{mol}$ .

7. $c = 9.0 / 58.5 = 0.154\,\mathrm{mol}/\mathrm{L}$ .

8. $n = 0.154 \times 0.500 = 0.0769\,\mathrm{mol}$ (or $4.5 / 58.5$ ).

9. Each $\ce{NaCl}$ gives one $\ce{Na+}$ and one $\ce{Cl-}$ : both at $0.154\,\mathrm{mol}/\mathrm{L}$ .

10. $0.0769 \times 6.02 \times 10^{23} = 4.63 \times 10^{22}$ sodium ions.

11. $F = 200 / 9.0 = 22.2$ .

12. The $4.5\,\mathrm{g}$ of the bag all come from the concentrated solution: $V_0 = 4.5\,\mathrm{g} / 200\,\mathrm{g}/\mathrm{L} = 0.0225\,\mathrm{L} = 22.5\,\mathrm{mL}$ .

13. $V_0 = V_1 / F = 500 / 22.2 = 22.5\,\mathrm{mL}$ .

14. $c_0 = 200 / 58.5 = 3.42\,\mathrm{mol}/\mathrm{L}$ ; $c_0 V_0 = 3.42 \times 0.0225 = 0.0769\,\mathrm{mol}$ and $c_1 V_1 = 0.154 \times 0.500 = 0.0769\,\mathrm{mol}$ : equal.

15. About $500 - 22.5 = 478\,\mathrm{mL}$ of sterile water.

16. A graduated pipette of $25\,\mathrm{mL}$ (or a burette), read to $0.1\,\mathrm{mL}$ .

17. In a $500.0\,\mathrm{mL}$ volumetric flask, made up to the mark.

18. $C_m = 200 \times 23.0 / 500 = 9.20\,\mathrm{g}/\mathrm{L}$ , too concentrated by $0.20 / 9.0 \approx 2.2\,\%$ .

19. $22.5\,\mathrm{mL}$ of the concentrated (“20 %”) solution for one $500\,\mathrm{mL}$ bag.

Terms defined in this chapter

See all 852 terms in the glossary