Concentration of Solutions

Lessons

Senior 3 · Chemistry

Concentration of Solutions

5 lessons

Lesson 3 of 5

Making Standard Solutions and Dilutions

25 min

Exactly the right strength

A farmer mixing a concentrated crop spray must follow the label to the millilitre: too strong and it scorches the leaves and endangers the person spraying; too weak and the pests survive. A chemistry laboratory has the same problem, only stricter. Every titration compares an unknown solution with one whose concentration is known exactly, a standard solution, so the answer can be no more accurate than the way that solution was made.

Definition

Standard solution: a solution whose concentration is known accurately. It is made by dissolving an accurately weighed mass of a pure solid and making the solution up to an exact volume in a volumetric flask, a flask with a single etched mark on its neck that shows one precise volume (for example $250.0\,\mathrm{cm^3}$ ). Example: 2.65 g of anhydrous sodium carbonate made up to $250.0\,\mathrm{cm^3}$ gives a standard $0.100\,\mathrm{mol\,dm^{-3}}$ solution.

Making a standard solution from a solid

Calculate the mass needed: $n = c \times V$ , then $m = n \times M$ .
Weigh the solid accurately, for example in a weighing bottle on a balance reading to 0.01 g.
Dissolve it in a small volume of distilled water in a beaker, stirring with a glass rod. Never dissolve it in the volumetric flask itself.
Transfer the solution through a funnel into the volumetric flask. Rinse the beaker, rod and funnel with distilled water and add every rinsing to the flask, so that no solute is left behind.
Make up to the mark with distilled water, adding the last few drops with a dropping pipette, until the bottom of the meniscus sits on the mark (Figure 3.2).
Stopper and invert the flask several times so the solution has the same concentration throughout.

Worked example

Problem. What mass of anhydrous sodium carbonate, $\mathrm{Na_2CO_3}$ , is needed to make $250.0\,\mathrm{cm^3}$ of a $0.100\,\mathrm{mol\,dm^{-3}}$ standard solution? (Na = 23, C = 12, O = 16)

Solution. $n = c \times V = 0.100 \times 0.2500 = 0.0250\,\mathrm{mol}$

$M(\mathrm{Na_2CO_3}) = 46 + 12 + 48 = 106\,\mathrm{g\,mol^{-1}}$

$m = 0.0250 \times 106 = 2.65\,\mathrm{g}$

Anhydrous sodium carbonate is used rather than sodium hydroxide because it is pure and stable. Solid sodium hydroxide absorbs water and carbon dioxide from the air, so a weighed sample is never quite pure NaOH.

Worked example

Problem. A technician needs $500\,\mathrm{cm^3}$ of $0.050\,\mathrm{mol\,dm^{-3}}$ copper(II) sulfate solution, made from blue crystals of copper(II) sulfate-5-water, $\mathrm{CuSO_4{\cdot}5H_2O}$ . What mass should be weighed? (Cu = 63.5, S = 32, O = 16, H = 1)

Solution. $n = 0.050 \times 0.500 = 0.025\,\mathrm{mol}$ . The water of crystallisation counts in the molar mass:

$M = 63.5 + 32 + 64 + 5 \times 18 = 249.5\,\mathrm{g\,mol^{-1}}$

$m = 0.025 \times 249.5 = 6.24\,\mathrm{g}$ (using $M = 159.5$ for anhydrous $\mathrm{CuSO_4}$ would give a solution that is too weak).

Making a solution by dilution

A weaker solution is made from a stronger stock solution by adding water. Measure a volume of the stock with a pipette or burette, run it into a volumetric flask, and make it up to the mark. Adding water does not change the amount of solute; it only spreads it through a larger volume (Figure 3.3). So the moles before and after are equal:

$c_{1} V_{1} = c_{2} V_{2}$

where $c_{1}$ and $V_{1}$ are the concentration and volume of stock used, and $c_{2}$ and $V_{2}$ those of the diluted solution. Both volumes may be in $\mathrm{cm^3}$ , as long as they are in the same unit.

Definition

Dilution: lowering the concentration of a solution by adding more solvent, without changing the amount of solute. The dilution factor is $\frac{V_2}{V_1}$ . Example: 25.0 cm³ made up to 250 cm³ is a tenfold dilution, so $2.0\,\mathrm{mol\,dm^{-3}}$ becomes $0.20\,\mathrm{mol\,dm^{-3}}$ .

Worked example

Problem. (a) What volume of $2.0\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid must be diluted to make $250\,\mathrm{cm^3}$ of $0.10\,\mathrm{mol\,dm^{-3}}$ acid? (b) $25.0\,\mathrm{cm^3}$ of $0.80\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide is made up to $100\,\mathrm{cm^3}$ . What is the new concentration?

Solution. (a) $V_1 = \frac{c_2 V_2}{c_1} = \frac{0.10 \times 250}{2.0} = 12.5\,\mathrm{cm^3}$ . Run 12.5 cm³ from a burette into a 250 cm³ volumetric flask containing some water, and make up to the mark.

(b) $c_2 = \frac{c_1 V_1}{V_2} = \frac{0.80 \times 25.0}{100} = 0.20\,\mathrm{mol\,dm^{-3}}$ .

Warning

Common mistakes and safety. (1) If you go past the mark, you cannot pour some out: the solution is already diluted, so start again. (2) Forgetting to invert the flask leaves a strong layer at the bottom and a weak one at the top. (3) $c_{1} V_{1} = c_{2} V_{2}$ is for dilution only; for a reaction such as a titration you must use the mole ratio from the equation. (4) When the stock is a concentrated acid, put some water in the flask first and add the acid to it, never water to acid. If any acid touches your skin, rinse with plenty of running water for at least 15 minutes.

Tip

Water adds volume, never moles. In a dilution the number of moles of solute stays the same, which is all $c_{1} V_{1} = c_{2} V_{2}$ says.

Royal Society of Chemistry: Practical skills, making a standard solution. Watch for the rinsing of the beaker and funnel, the last drops added to the mark, and the inverting of the flask.

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Practice

What is a standard solution? Why is a volumetric flask, not a beaker, used to make one?
What mass of sodium hydroxide is needed to make $100\,\mathrm{cm^3}$ of a $0.200\,\mathrm{mol\,dm^{-3}}$ solution? (NaOH = 40)
What mass of potassium manganate(VII), $\mathrm{KMnO_4}$ , is needed for $250\,\mathrm{cm^3}$ of a $0.020\,\mathrm{mol\,dm^{-3}}$ solution? ( $\mathrm{KMnO_4} = 158$ )
Why are the beaker, rod and funnel rinsed into the flask?
$10.0\,\mathrm{cm^3}$ of $1.5\,\mathrm{mol\,dm^{-3}}$ sulfuric acid is made up to $250\,\mathrm{cm^3}$ . Find the new concentration.
What volume of $18.4\,\mathrm{mol\,dm^{-3}}$ sulfuric acid is needed to make $500\,\mathrm{cm^3}$ of $1.0\,\mathrm{mol\,dm^{-3}}$ acid?
To what final volume must $50\,\mathrm{cm^3}$ of $0.40\,\mathrm{mol\,dm^{-3}}$ salt solution be diluted to give $0.10\,\mathrm{mol\,dm^{-3}}$ ? How much water is added?
A student makes up a standard solution but adds water past the mark, then pours a little out to bring it back. Evaluate what she did.

Show answersHide answers

A solution whose concentration is known accurately. A volumetric flask holds one exact volume at its mark; a beaker's markings are only approximate.
$n = 0.200 \times 0.100 = 0.0200\,\mathrm{mol}$ ; $m = 0.0200 \times 40 = 0.80\,\mathrm{g}$ .
$n = 0.020 \times 0.250 = 0.0050\,\mathrm{mol}$ ; $m = 0.0050 \times 158 = 0.79\,\mathrm{g}$ .
So that all the weighed solute ends up in the flask; any left on the glass would make the solution weaker than calculated.
$c_2 = \frac{1.5 \times 10.0}{250} = 0.060\,\mathrm{mol\,dm^{-3}}$ .
$V_1 = \frac{1.0 \times 500}{18.4} = 27.2\,\mathrm{cm^3}$ (added slowly to water already in the flask).
$V_2 = \frac{0.40 \times 50}{0.10} = 200\,\mathrm{cm^3}$ ; water added $= 200 - 50 = 150\,\mathrm{cm^3}$ .
Wrong method. Once she passed the mark the solution was already too dilute; pouring some out removes solute as well as water, so the concentration stays too low. She must start again.

Sources and credits

Photo: Brand volumetric flask 100ml.jpg, Lucasbosch, licensed under CC BY-SA 3.0, via Wikimedia Commons.
Video: Practical skills assessment video, titration, standard solution, Royal Society of Chemistry, embedded from YouTube.
Written by Ganzaa. The diagrams, worked examples and Practice items were drawn and written by Ganzaa.

1Exactly the right strength

2Making a standard solution from a solid

3Making a solution by dilution

Exactly the right strength

Making a standard solution from a solid

Making a solution by dilution