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SAChemistrySyllabus dot point

How do chemists measure and judge the quality of water?

Describe the key indicators of water quality, including dissolved oxygen, BOD, pH, turbidity and ion concentrations.

Dissolved oxygen, BOD, pH, turbidity, hardness and ion concentrations as water-quality indicators, with the chemistry behind each measurement and worked SACE-style ppm and dissolved-oxygen calculations.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Lead worked calculation
The key indicators
Units of concentration
Linking indicators to chemistry
Choosing the right technique per indicator
Why it matters for monitoring

What this dot point is asking

SACE expects you to interpret these indicators in context, link them to chemical processes, and perform unit conversions, especially between $\text{ppm}$ , $\text{mg L}^{-1}$ and $\text{mol L}^{-1}$ .

Lead worked calculation

Converting a nitrate reading between units

A river sample contains $\text{NO}_3^-$ at $8.0\ \text{ppm}$ . Convert this to $\text{mol L}^{-1}$ . ( $M(\text{NO}_3^-) = 62.0\ \text{g mol}^{-1}$ ; density $= 1.00\ \text{g mL}^{-1}$ .)

Step 1: Interpret what ppm means for a dilute aqueous sample

The abbreviation ppm stands for "parts per million" by mass, which is milligrams of solute per kilogram of solution. However, when the solution density is $1.00\ \text{g mL}^{-1}$ , one litre of solution has a mass of exactly $1\ \text{kg}$ , so the mass ratio simplifies: $1\ \text{ppm}$ is exactly $1\ \text{mg L}^{-1}$ . This shortcut only works at this particular density, which is why the question states it. You can now write the nitrate concentration as a mass concentration:

8.0\ \text{ppm} = 8.0\ \text{mg L}^{-1} = 8.0 \times 10^{-3}\ \text{g L}^{-1}

Step 2: Convert the mass concentration to moles per litre

To get from grams of nitrate to moles of nitrate, divide by the molar mass. This is the definition of molar mass: it converts between the macroscopic mass you can measure and the chemist's counting unit, the mole. Because the mass is already expressed per litre, the result is directly a molar concentration.

n = \frac{m}{M} = \frac{8.0 \times 10^{-3}\ \text{g}}{62.0\ \text{g mol}^{-1}} = 1.29 \times 10^{-4}\ \text{mol}

Step 3: State the final concentration and interpret it

The $1.29 \times 10^{-4}\ \text{mol}$ was calculated from the mass present in one litre, so the molar concentration is exactly the number from Step 2, rounded appropriately to two significant figures to match the given data.

c(\text{NO}_3^-) = 1.3 \times 10^{-4}\ \text{mol L}^{-1}

Final answer: $c(\text{NO}_3^-) = 1.3 \times 10^{-4}\ \text{mol L}^{-1}$ . This level exceeds the typical drinking-water guideline, signalling possible fertiliser or sewage run-off affecting the catchment.

The key indicators

pH measures acidity; most aquatic life needs roughly $6.5$ to $8.5$ . Acid mine drainage or acid rain lowers pH; algal blooms can raise it.
Turbidity is the cloudiness caused by suspended particles. High turbidity blocks light for aquatic plants and can carry adsorbed pollutants. It is measured by light scattering (nephelometry).
Hardness is the total concentration of $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ , which form scale and precipitate soap.
Specific ions such as $\text{NO}_3^-$ , $\text{PO}_4^{3-}$ (nutrients causing eutrophication) and heavy-metal ions such as $\text{Pb}^{2+}$ are measured by titration, AAS or chromatography depending on concentration.

Units of concentration

Linking indicators to chemistry

The indicators are not independent. Excess nutrients ( $\text{NO}_3^-$ , $\text{PO}_4^{3-}$ ) feed algal blooms; when the algae die, bacteria decomposing them raise BOD and crash the DO, killing fish. This chain, eutrophication, shows why a single number cannot describe water quality and why monitoring tracks several indicators together. Temperature also matters: warmer water dissolves less oxygen and accelerates microbial activity, so the same organic load is more damaging in summer.

Choosing the right technique per indicator

Each indicator has a matched measurement method. Dissolved oxygen is measured by the Winkler redox titration or an oxygen probe; pH by a calibrated pH meter or indicator; turbidity by light scattering; hardness and major ions by complexometric or acid-base titration; and trace metals or nutrients by AAS or chromatography. Matching the method to the expected concentration matters: titration suits indicators present at high concentration, while AAS and chromatography reach the trace levels (parts per billion) needed for toxic metals and pesticides.

Why it matters for monitoring

Water-quality indicators turn a vague idea of "clean water" into measurable, comparable data. Regulatory limits are set in these units, so analysts must both measure the indicators accurately and convert between $\text{ppm}$ and $\text{mol L}^{-1}$ to compare results against guidelines and across sites. Tracking several indicators together, rather than a single number, is what reveals the underlying chemistry, such as the eutrophication chain, and lets authorities act before aquatic ecosystems are damaged.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20214 marksA

250\ \text{mL}

water sample was found to contain

3.5\ \text{mg}

of dissolved calcium ions. Express this concentration in

\text{ppm}

and in

\text{mol L}^{-1}

. (

M(\text{Ca}) = 40.08\ \text{g mol}^{-1}

; assume the density of the sample is

1.00\ \text{g mL}^{-1}

Show worked answer →

Step 1 (ppm): $\text{ppm} = \dfrac{\text{mg solute}}{\text{kg solution}}$ . The sample mass is $250\ \text{g} = 0.250\ \text{kg}$ , so $\text{ppm} = \dfrac{3.5}{0.250} = 14\ \text{ppm}$ . (2 marks)

Step 2 (mol L $^{-1}$ ): $n(\text{Ca}^{2+}) = \dfrac{3.5 \times 10^{-3}}{40.08} = 8.73 \times 10^{-5}\ \text{mol}$ . (1 mark)

Step 3: $c = \dfrac{n}{V} = \dfrac{8.73 \times 10^{-5}}{0.250} = 3.5 \times 10^{-4}\ \text{mol L}^{-1}$ . (1 mark)

SACE 20193 marksTwo water samples are taken from the same river, one upstream and one downstream of a sewage outfall. The downstream sample has a much higher biochemical oxygen demand (BOD) and a much lower dissolved oxygen (DO) reading. Explain, in terms of the chemistry occurring, why these two measurements move in opposite directions.

Show worked answer →

The sewage adds biodegradable organic matter to the water. (1 mark)

Aerobic micro-organisms oxidise this organic matter for energy, consuming dissolved $\text{O}_2$ : $\text{organic matter} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ . BOD measures the amount of oxygen consumed over a fixed period (often $5$ days at $20\ ^\circ\text{C}$ ), so more organic load gives a higher BOD. (1 mark)

Because the bacteria draw oxygen out of the water faster than it can dissolve back in from the air, the dissolved oxygen falls. High BOD and low DO therefore go together and signal organic pollution. (1 mark)