Define concentration in mol per dm cubed and g per dm cubed, interconvert the two, and carry out titration calculations to find an unknown concentration

What this dot point is asking

SEAB wants you to define the concentration of a solution in two ways (moles per cubic decimetre and grams per cubic decimetre), convert between them, and carry out a titration calculation to find an unknown concentration. The titration calculation is a guaranteed feature of the written and practical papers, so a clean, repeatable method is essential.

The answer

Defining concentration

Concentration measures how much solute is dissolved in a given volume of solution. There are two units:

• Moles per cubic decimetre ($\text{mol/dm}^3$): the amount of solute in moles per $1\ \text{dm}^3$ of solution. This is the more useful unit for calculations.
• Grams per cubic decimetre ($\text{g/dm}^3$): the mass of solute in grams per $1\ \text{dm}^3$.

The key equation, with volume in cubic decimetres, is:

where $n$ is the amount in moles, $c$ the concentration in $\text{mol/dm}^3$ and $V$ the volume in $\text{dm}^3$. Remember $1\ \text{dm}^3 = 1000\ \text{cm}^3$, so divide a volume in $\text{cm}^3$ by $1000$ before using it.

Converting between the two units

To convert from moles per cubic decimetre to grams per cubic decimetre, multiply by the relative molecular mass:

To go the other way, divide by $M_r$. So a $0.10\ \text{mol/dm}^3$ solution of sodium hydroxide ($M_r = 40$) has a concentration of $0.10 \times 40 = 4.0\ \text{g/dm}^3$.

What a titration does

A titration measures the volume of one solution that exactly reacts with a known volume of another, so an unknown concentration can be found. A pipette delivers a fixed volume of one solution into a conical flask, an indicator is added, and the other solution is run in from a burette until the indicator just changes colour, the end point. The burette reading gives the volume that reacted. Concordant titres (within $0.10\ \text{cm}^3$) are averaged before calculating.

The three-step titration calculation

Every titration calculation follows the same route:

1. Moles of the known solution: use $n = c \times V$ for the solution whose concentration and volume you know.
2. Mole ratio: use the balanced equation to find the moles of the other reactant.
3. Unknown concentration: divide those moles by the volume (in $\text{dm}^3$) of the unknown solution, $c = n/V$.

Examples in context

Example 1. Standardising an acid. To find the exact concentration of a hydrochloric acid, it is titrated against a sodium carbonate solution of known concentration. The titre and the balanced equation give the acid concentration, after which the acid can itself be used as a known standard, showing how titration underpins quantitative analysis.

Example 2. Checking the strength of vinegar. The ethanoic acid content of vinegar is measured by titrating a measured sample against standard sodium hydroxide. Converting the result from mol per dm cubed to g per dm cubed gives the mass of acid per litre, a practical food-analysis use of the same calculation.

Try this

Q1. Calculate the amount, in moles, of solute in $50.0\ \text{cm}^3$ of a $0.200\ \text{mol/dm}^3$ solution. [2 marks]

• Cue. $V = 0.0500\ \text{dm}^3$; $n = cV = 0.200 \times 0.0500 = 0.0100\ \text{mol}$.

Q2. A solution of sodium chloride has a concentration of $0.50\ \text{mol/dm}^3$. Calculate its concentration in g per dm cubed. (Mr of NaCl = 58.5.) [2 marks]

• Cue. $0.50 \times 58.5 = 29.3\ \text{g/dm}^3$.

Q3. State the order of the three steps in a titration calculation. [1 mark]

• Cue. Moles of the known solution ($n = cV$), then the mole ratio from the balanced equation, then the unknown concentration ($c = n/V$).