NSWChemistrySyllabus dot point

Inquiry Question 1: How are the ions present in the environment identified and measured?

Conduct investigations to measure the concentration of cations and anions in solution using gravimetric analysis and precipitation titrations

A focused answer to the HSC Chemistry Module 8 dot point on quantitative wet-chemistry analysis. The full gravimetric workflow (precipitate, filter, dry, weigh), worked sulfate-as-barium-sulfate calculation, the Mohr precipitation titration of chloride with silver nitrate, sources of error, and worked HSC past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to use a precipitation reaction to measure concentration two ways: by weighing the dried precipitate (gravimetric analysis) or by titrating a known precipitant to a colour-change endpoint (precipitation titration, typically the Mohr method for chloride). You should know the workflow, the calculations, and the most common sources of error.

The answer

Gravimetric analysis: the workflow

Weigh the sample accurately on an analytical balance.
Dissolve in a measured volume of water, and acidify with the appropriate dilute acid to remove carbonate and other interferents.
Add excess precipitating reagent to drive the precipitation to completion.
Digest the precipitate by warming, which grows crystals and reduces co-precipitation.
Filter through pre-weighed filter paper (ashless) or a sintered glass crucible.
Wash the precipitate with a small volume of distilled water (and a dilute solution of a common ion to suppress dissolution).
Dry to constant mass in an oven (or ignite to a known oxide).
Weigh the dried precipitate.

The mass of the precipitate gives moles, which converts back to moles (and mass) of the original ion via the stoichiometry of the precipitation equation.

The canonical example: sulfate as barium sulfate

For a sample of mass $m$ that gives a precipitate of mass $m_{BaSO_4}$ :

n(BaSO_4) = \frac{m_{BaSO_4}}{233.4}

n(SO_4^{2-}) = n(BaSO_4) \quad \text{(1:1)}

\%SO_4^{2-} = \frac{n(SO_4^{2-}) \times 96.1}{m} \times 100\%

Acidify with dilute $HCl$ first; this removes carbonate (which would otherwise precipitate as $BaCO_3$ ) but does not dissolve $BaSO_4$ .

Other common gravimetric pairs

Target ion	Precipitate weighed	Acid used
IMATH_13	IMATH_14	dilute IMATH_15
IMATH_16	IMATH_17	dilute IMATH_18
IMATH_19	IMATH_20 (then ignite to $Mg_2P_2O_7$ )	IMATH_22 / $NH_4Cl$ buffer
IMATH_24	IMATH_25 or IMATH_26	dilute IMATH_27
IMATH_28	IMATH_29 (or ignite to $CaO$ )	acetate buffer

Precipitation titration: the Mohr method

Use when you want speed and do not need part-per-billion precision. The classic Mohr titration measures $Cl^-$ :

Titrant: standardised $AgNO_3$ (commonly 0.1 mol/L).
Indicator: a few drops of $K_2CrO_4$ .
End-point: the first persistent red-brown colour of $Ag_2CrO_4$ .

The chemistry has two stages. While free chloride remains:

Ag^+_{(aq)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} \quad \text{(white)}

When chloride is exhausted, the next drop of $Ag^+$ reacts with chromate to give a red-brown precipitate, signalling the end-point:

2Ag^+_{(aq)} + CrO_4^{2-}_{(aq)} \rightarrow Ag_2CrO_{4(s)} \quad \text{(red-brown)}

$Ag_2CrO_4$ has a higher $K_{sp}$ than $AgCl$ , so $AgCl$ precipitates first. The chromate stays in solution until all chloride is consumed.

Calculation pattern for a precipitation titration

For volume of titrant $V_t$ and concentration $c_t$ :

n(Ag^+) = c_t \times V_t

n(Cl^-) = n(Ag^+) \quad \text{(1:1)}

c(Cl^-) = \frac{n(Cl^-)}{V_{sample}}

Convert to g/L by multiplying by 35.5 g/mol.

pH window for the Mohr method

The titration must be run at pH 7 to 9.5.

Below pH 6.5: chromate protonates to dichromate, which is soluble with silver. No coloured end-point.
Above pH 10: $AgOH$ and brown $Ag_2O$ precipitate, consuming titrant and falsifying the result.

Adjust with $NaHCO_3$ or a phosphate buffer if needed.

Sources of error

Gravimetric:

Incomplete precipitation if insufficient precipitant is added (use a clear excess).
Co-precipitation of impurities (acidify to remove carbonate; dilute the sample to reduce inclusion).
Particle loss through the filter (digest the precipitate first to grow larger crystals).
Incomplete drying (dry to constant mass; reweigh after a second drying cycle).
Hygroscopic precipitates absorb moisture during weighing (cool in a desiccator).

Precipitation titration:

pH out of range distorts the end-point.
Slow precipitate formation makes the end-point hard to spot; swirl thoroughly.
Indicator concentration: too much chromate masks the white-to-red change.
Coloured samples (sea water with biological matter, for example) hide the end-point.

When to choose which

Need	Gravimetric	Precipitation titration
Highest precision	Yes (0.1% or better)	Moderate (1%)
Fast turnaround	No (hours to days)	Yes (minutes per sample)
Many samples	No	Yes
Trace analysis (ppb)	No (both unsuitable, use AAS or UV-vis)	No

Common traps

Forgetting to acidify. The most common mark loss is leaving carbonate in solution, which precipitates with $Ba^{2+}$ or $Ag^+$ and inflates the result.

Confusing moles of precipitate with moles of target ion. Always go through the formula. For example, in the magnesium pyrophosphate ignition method, one mole of $Mg_2P_2O_7$ contains two moles of $PO_4^{3-}$ -derived phosphorus.

Running the Mohr titration outside pH 7 to 9.5. Memorise the window and the chemical reason for both ends.

Reading the Mohr end-point too soon. Faint pink that disappears on swirling is not the end-point. Wait for a colour that persists for 30 seconds.

Forgetting to dry to constant mass. The first dry weighing is rarely the true mass; reweigh after a second drying cycle to confirm.

In one sentence

Gravimetric analysis weighs a dried precipitate to back-calculate the moles of the target ion (sulfate as $BaSO_4$ , chloride as $AgCl$ ), while precipitation titration delivers a standardised titrant to a colour-change end-point ( $Ag^+$ titrating $Cl^-$ with chromate indicator at pH 7 to 9.5), trading some precision for much faster throughput.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC5 marksA 1.20 g sample of fertiliser was dissolved in water and acidified with dilute HCl. Excess barium chloride solution was added. The precipitate was filtered, washed, dried and weighed. Its mass was 0.583 g. Calculate the percentage by mass of sulfate in the fertiliser. Identify two sources of error in this gravimetric analysis.

Show worked answer →

A 5 mark answer needs the stoichiometry, the mass calculation, the percentage and at least two specific error sources.

Step 1: Identify the precipitate. Sulfate plus barium gives $BaSO_4$ :

Ba^{2+}_{(aq)} + SO_4^{2-}_{(aq)} \rightarrow BaSO_{4(s)}

Molar mass of $BaSO_4$ = 137.3 + 32.1 + 4(16.0) = 233.4 g/mol.

Step 2: Moles of $BaSO_4$ . $n = 0.583 / 233.4 = 2.498 \times 10^{-3}$ mol.

Step 3: Moles of sulfate. 1:1 ratio, so $n(SO_4^{2-}) = 2.498 \times 10^{-3}$ mol.

Step 4: Mass of sulfate. Molar mass of $SO_4^{2-}$ = 32.1 + 4(16.0) = 96.1 g/mol. Mass = $2.498 \times 10^{-3} \times 96.1 = 0.240$ g.

Step 5: Percentage. $(0.240 / 1.20) \times 100\% = 20.0\%$ sulfate by mass.

Error sources (any two of):

Incomplete precipitation if insufficient $BaCl_2$ is added; mass of $BaSO_4$ is too low.
Co-precipitation of impurities (carbonate, phosphate) gives high mass unless the solution is acidified first.
Loss of fine particles through the filter; mass is too low. Use ashless filter paper and double-filter if needed.
Incomplete drying; residual water mass is too high.
Loss during transfer between beakers and filter funnel.

Markers reward (1) the equation, (2) correct moles, (3) the percentage, (4) two valid error sources each with the direction of the error.

2020 HSC4 marksA 25.00 mL sample of seawater was titrated with 0.100 mol/L $AgNO_3$ using potassium chromate as the indicator (Mohr method). The endpoint was reached when 22.40 mL of titrant had been delivered. Calculate the chloride concentration in g/L and explain why this method requires a neutral to slightly basic solution.

Show worked answer →

A 4 mark answer needs the titration calculation in mol/L, conversion to g/L, and the chemical reason for the pH constraint.

Step 1: Moles of $AgNO_3$ . $n = 0.100 \times 0.02240 = 2.240 \times 10^{-3}$ mol.

Step 2: Moles of $Cl^-$ . 1:1 reaction $Ag^+ + Cl^- \rightarrow AgCl$ , so $n(Cl^-) = 2.240 \times 10^{-3}$ mol.

Step 3: Concentration in mol/L. $c = 2.240 \times 10^{-3} / 0.02500 = 0.0896$ mol/L.

Step 4: Concentration in g/L. Molar mass of $Cl^-$ = 35.5 g/mol. Concentration = $0.0896 \times 35.5 = 3.18$ g/L.

Why neutral to slightly basic. The Mohr method relies on a sharp end-point when the second precipitate, red $Ag_2CrO_4$ , forms after all $Cl^-$ is consumed. Two pH constraints apply:

Too acidic ( $pH < 6.5$ ): chromate protonates to dichromate, $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$ , which does not form an insoluble silver salt. The end-point is lost.
Too basic ( $pH > 10$ ): silver hydroxide $AgOH$ (and then dark $Ag_2O$ ) precipitates before the chromate end-point, consuming titrant and giving a high result.

Typical buffered range is pH 7 to 9.5.

Markers reward (1) correct moles of titrant, (2) 1:1 stoichiometry, (3) g/L conversion, (4) the chemical reason for both pH limits.