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Water quality indicators (SACE Stage 2 Chemistry)

Syllabus dot point

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

8 min read

Quick answer

Water quality is judged from several measurable indicators: dissolved oxygen (DO), biochemical oxygen demand (BOD), pH, turbidity, hardness, and the concentrations of specific ions and pollutants. Each is reported with appropriate units (often ppm\text{ppm} or mol L−1\text{mol L}^{-1}), and together they reveal whether water is healthy for aquatic life and safe for use. You must define each indicator, explain the chemistry behind it, and convert between concentration units.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. The key indicators
  4. Units of concentration
  5. Linking indicators to chemistry
  6. Choosing the right technique per indicator
  7. Why it matters for monitoring
  8. Exam-style questions

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 ppm\text{ppm}, mg L−1\text{mg L}^{-1} and mol L−1\text{mol L}^{-1}.

Lead worked calculation

Converting a nitrate reading between units

A river sample contains NO3−\text{NO}_3^- at 8.0 ppm8.0\ \text{ppm}. Convert this to mol L−1\text{mol L}^{-1}. (M(NO3−)=62.0 g mol−1M(\text{NO}_3^-) = 62.0\ \text{g mol}^{-1}; density =1.00 g mL−1= 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 g mL−11.00\ \text{g mL}^{-1}, one litre of solution has a mass of exactly 1 kg1\ \text{kg}, so the mass ratio simplifies: 1 ppm1\ \text{ppm} is exactly 1 mg L−11\ \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 ppm=8.0 mg L−1=8.0×10−3 g L−18.0\ \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=mM=8.0×10−3 g62.0 g mol−1=1.29×10−4 moln = \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 1.29×10−4 mol1.29 \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(NO3−)=1.3×10−4 mol L−1c(\text{NO}_3^-) = 1.3 \times 10^{-4}\ \text{mol L}^{-1}

Final answer: c(NO3−)=1.3×10−4 mol L−1c(\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

Dissolved oxygen and BOD

Dissolved oxygen (DO) is the concentration of O2\text{O}_2 molecules dissolved in water, vital for fish and aerobic organisms; cold, fast-flowing water holds more. Biochemical oxygen demand (BOD) is the mass of oxygen consumed by micro-organisms breaking down organic matter in a sample over a set time. High BOD means heavy organic pollution, which depletes DO and can suffocate aquatic life.

  • pH measures acidity; most aquatic life needs roughly 6.56.5 to 8.58.5. 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 Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+}, which form scale and precipitate soap.
  • Specific ions such as NO3−\text{NO}_3^-, PO43−\text{PO}_4^{3-} (nutrients causing eutrophication) and heavy-metal ions such as Pb2+\text{Pb}^{2+} are measured by titration, AAS or chromatography depending on concentration.

Units of concentration

Parts per million

ppm=mass of solute (mg)mass of solution (kg)=mass of solutemass of solution×106\text{ppm} = \frac{\text{mass of solute (mg)}}{\text{mass of solution (kg)}} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6}

For dilute aqueous solutions where the density is close to 1.00 g mL−11.00\ \text{g mL}^{-1}, 1 ppm≈1 mg L−11\ \text{ppm} \approx 1\ \text{mg L}^{-1}. To reach mol L−1\text{mol L}^{-1}, convert the mass to moles with n=m/Mn = m/M and divide by the volume in litres.

Linking indicators to chemistry

The indicators are not independent. Excess nutrients (NO3−\text{NO}_3^-, PO43−\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.

Confusing DO and BOD, or mishandling ppm density

DO is the oxygen present now; BOD is the oxygen that will be consumed. They move in opposite directions when organic pollution rises. With ppm, remember it is a mass-per-mass ratio: the 1 ppm=1 mg L−11\ \text{ppm} = 1\ \text{mg L}^{-1} shortcut only holds when the solution density is 1.00 g mL−11.00\ \text{g mL}^{-1}. Always convert milligrams to grams before using n=m/Mn = m/M.

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 ppm\text{ppm} and mol L−1\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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2021-style4 marks
A 250 mL250\ \text{mL} water sample was found to contain 3.5 mg3.5\ \text{mg} of dissolved calcium ions. Express this concentration in ppm\text{ppm} and in mol L−1\text{mol L}^{-1}. (M(Ca)=40.08 g mol−1M(\text{Ca}) = 40.08\ \text{g mol}^{-1}; assume the density of the sample is 1.00 g mL−11.00\ \text{g mL}^{-1}.)
Show worked answer →

Step 1 (ppm): ppm=mg solutekg solution\text{ppm} = \dfrac{\text{mg solute}}{\text{kg solution}}. The sample mass is 250 g=0.250 kg250\ \text{g} = 0.250\ \text{kg}, so ppm=3.50.250=14 ppm\text{ppm} = \dfrac{3.5}{0.250} = 14\ \text{ppm}. (2 marks)

Step 2 (mol L−1^{-1}): n(Ca2+)=3.5×10−340.08=8.73×10−5 moln(\text{Ca}^{2+}) = \dfrac{3.5 \times 10^{-3}}{40.08} = 8.73 \times 10^{-5}\ \text{mol}. (1 mark)

Step 3: c=nV=8.73×10−50.250=3.5×10−4 mol L−1c = \dfrac{n}{V} = \dfrac{8.73 \times 10^{-5}}{0.250} = 3.5 \times 10^{-4}\ \text{mol L}^{-1}. (1 mark)

SACE 2019-style3 marks
Two 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 O2\text{O}_2: organic matter+O2→CO2+H2O\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 55 days at 20 ∘C20\ ^\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)

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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