Gravimetric analysis and precipitation titrations explained: HSC Chemistry Module 8

Syllabus dot point

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

11 min readUpdated 22 Jun 2026

Quick answer

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.

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In plain English

Both techniques answer the same question, "how much of this ion is in solution?", by turning an invisible dissolved ion into something you can physically measure. Gravimetric analysis is the slow, careful method: you tip in enough of a partner ion to lock every target ion into a solid, then you filter it out, dry it bone-dry, and weigh it on a balance, like counting a jar of coins by weighing them instead of counting one by one. Precipitation titration is the fast method: you drip in a measured "partner" solution from a burette until a colour-change dye tells you the exact moment every target ion has just been used up, like adding drops of dye to a swimming pool until the whole pool visibly changes colour, then reading off how many drops it took. Both methods rely on the same core idea, a 1:1 (or known-ratio) reaction between the target ion and the reagent you add, so the amount of reagent used tells you the amount of target ion that was there.

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.

Core relationships and checkpoints

The bridge equation: $n = m/M$ applied to the WEIGHED PRECIPITATE, then carried across via the balanced equation's mole ratio to the target ion.
Excess reagent, always. Add a clear excess of the precipitating reagent so precipitation of the target ion is complete; a deficiency of reagent under-reports the target ion.
Acidify before precipitating to remove carbonate/other interferents that would otherwise co-precipitate and inflate the mass, without dissolving the target precipitate itself.
Dry to CONSTANT mass, meaning two consecutive weighings agree, not just "dry once and stop".
Mohr endpoint: first PERSISTENT red-brown $Ag_2CrO_4$ colour, valid only over pH 7 to 9.5.

The apparatus and workflow for the gravimetric method are shown below:

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 $BaSO_4$ .

Worked gravimetric calculation: sulfate as barium sulfate

A $1.000$ g sample of a soluble fertiliser is dissolved in water, acidified with dilute $HCl$ , and treated with excess barium chloride solution. The precipitate is filtered, washed, dried and weighed as $0.640$ g of $BaSO_4$ . Calculate the percentage by mass of sulfate in the fertiliser.

Step 1 - moles of precipitate: $n(BaSO_4) = \dfrac{0.640}{233.4} = 2.74 \times 10^{-3}\ \text{mol}$
Step 2 - moles of sulfate (1:1 ratio): $n(SO_4^{2-}) = n(BaSO_4) = 2.74 \times 10^{-3}\ \text{mol}$
Step 3 - mass of sulfate: $m(SO_4^{2-}) = 2.74 \times 10^{-3} \times 96.1 = 0.264\ \text{g}$
Step 4 - percentage by mass: $\%(SO_4^{2-}) = \dfrac{0.264}{1.000} \times 100 = 26.4\%$ (3 s.f.)

Marker's note: full marks require the 1:1 mole ratio stated explicitly, correct molar masses ( $BaSO_4 = 233.4$ , $SO_4^{2-} = 96.1$ g/mol), units carried through every line, and a final answer to an appropriate number of significant figures. The classic error is dividing the precipitate mass by the sulfate molar mass directly - always go via moles of the weighed precipitate.

Other common gravimetric pairs

Target ion	Precipitate weighed	Acid used
$SO_4^{2-}$	$BaSO_4$	dilute $HCl$
$Cl^-$	$AgCl$	dilute $HNO_3$
$PO_4^{3-}$	$MgNH_4PO_4 \cdot 6H_2O$ (then ignite to $Mg_2P_2O_7$ )	$NH_3$ / $NH_4Cl$ buffer
$Pb^{2+}$	$PbSO_4$ or $PbCrO_4$	dilute $H_2SO_4$
$Ca^{2+}$	$CaC_2O_4 \cdot H_2O$ (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.

An illustrative titration curve shows why the endpoint is sharp:

Exam technique for gravimetric and titration calculations

Always show the SAME four steps for any gravimetric or precipitation-titration calculation: (1) the balanced precipitation equation with the mole ratio, (2) moles of what you actually weighed or titrated, (3) moles of the target ion using that ratio, (4) the final answer with correct units and significant figures matching the given data. For a titration, always convert volumes to litres (mL divided by 1000) before using $n = cV$ . When a question gives a graph, read the equivalence point volume directly off the steepest part of the curve, do not estimate from the axis scale alone.

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.

Examples in context

Example 1. Sulfate in mine drainage at Cobar. EPA contractors monitoring acid mine drainage from the Cobar copper mine in central NSW use gravimetric analysis to measure sulfate. A 50.0 mL aliquot is acidified with HCl and treated with excess $BaCl_2$ to precipitate $BaSO_4$ . The precipitate is filtered through pre-weighed filter paper, washed, dried at 110 degrees C overnight and weighed. A typical mass of 0.282 g corresponds to $n(SO_4^{2-}) = 0.282 / 233.39 = 1.21 \times 10^{-3}$ mol, giving $[SO_4^{2-}] = 24$ mmol L $^{-1}$ or 2300 mg L $^{-1}$ , far above the 250 mg L $^{-1}$ aesthetic guideline. The result feeds discharge licence compliance.

Example 2. Mohr titration of chloride in NSW pool water. Aquatic centre managers at the Sydney Olympic Park Aquatic Centre run weekly Mohr titrations to confirm chloride concentration in saltwater pools. A 25.0 mL pool sample is titrated against standardised 0.100 mol L $^{-1}$ silver nitrate using potassium chromate as indicator. The first persistent red tint of $Ag_2CrO_4$ marks the end point. A titre of 12.50 mL gives $n(Cl^-) = 0.100 \times 0.01250 = 1.25 \times 10^{-3}$ mol, hence $[Cl^-] = 0.0500$ mol L $^{-1}$ or 1775 mg L $^{-1}$ , sitting in the operating range for salt chlorination. The HSC Mohr titration framework is the same chemistry the pool operator runs in the plant room.

Try this

Q1. Outline the steps of a gravimetric analysis for sulfate ions in water. [3 marks]

Cue. Acidify, add excess $BaCl_2$ , filter the $BaSO_4$ precipitate, wash, dry to constant mass, weigh, calculate moles and concentration.

Q2. A 100 mL water sample yields 0.247 g of $AgCl$ after addition of excess $AgNO_3$ . Calculate $[Cl^-]$ in mol L $^{-1}$ . [3 marks]

Cue. $n(AgCl) = 0.247 / 143.32 = 1.72 \times 10^{-3}$ mol; $n(Cl^-) = 1.72 \times 10^{-3}$ mol; $[Cl^-] = 0.0172$ mol L $^{-1}$ .

Q3. Mohr's method titrates $Cl^-$ with $AgNO_3$ . (a) Write the equation for the indicator end point. (b) Explain why the pH must be between 7 and 9.5. (c) State one source of error in the Mohr titration. [2+2+1 marks]

Cue. (a) $2Ag^+ + CrO_4^{2-} \rightarrow Ag_2CrO_{4(s, red)}$ . (b) Acidic conditions protonate chromate to dichromate; basic conditions precipitate AgOH. (c) Co-precipitation of bromide or iodide; impure water blank.

Exam-style questions

Questions in the style of NESA 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.

2022 HSC-style5 marks

A 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 HSC-style4 marks

A 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.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

State the chemical formula of the precipitate formed, and the acid used to pre-treat the sample, for a gravimetric determination of (a) sulfate ions and (b) chloride ions.

Show worked solution →

A 3-mark identify needs the correct precipitate formula AND the correct pre-treatment acid for each ion.

(a) Sulfate. Precipitate: $BaSO_4$ . Pre-treatment acid: dilute $HCl$ (removes carbonate without dissolving $BaSO_4$ ).

(b) Chloride. Precipitate: $AgCl$ . Pre-treatment acid: dilute $HNO_3$ (prevents precipitation of other silver salts, such as silver carbonate or silver phosphate, that would co-precipitate and inflate the mass).

Marking criteria: 1 mark for the correct sulfate precipitate and acid, 1 mark for the correct chloride precipitate and acid, 1 mark for correctly linking each acid to the impurity it suppresses.

foundation4 marks

Outline, in order, the eight steps of a gravimetric analysis procedure from receiving a solid sample to obtaining a final mass, naming the purpose of the digestion and washing steps.

Show worked solution →

The eight-step workflow.

Weigh the sample accurately.
Dissolve in a measured volume of water and acidify to remove interferents.
Add excess precipitating reagent to drive precipitation to completion.
Digest (warm) the precipitate.
Filter through pre-weighed ashless paper or a sintered crucible.
Wash the precipitate with a small volume of distilled water or a common-ion wash solution.
Dry to constant mass.
Weigh the dried precipitate.

Purpose of digestion. Warming grows larger, purer crystals, which reduces co-precipitation of impurities and reduces loss of fine particles through the filter.

Purpose of washing. Removes soluble impurities (mother liquor) adhering to the precipitate surface without dissolving an appreciable amount of the precipitate itself.

Marking criteria: 1 mark for the steps in the correct order (accept minor step-combination), 1 mark for identifying excess reagent as necessary for complete precipitation, 1 mark for the purpose of digestion, 1 mark for the purpose of washing.

core5 marks

A 0.850 g sample of an ionic chloride salt is dissolved in water and treated with excess silver nitrate solution. The precipitate of silver chloride is filtered, washed, dried and weighed, giving a mass of 1.146 g. Calculate the percentage by mass of chloride in the original sample, to 3 significant figures. (

M(AgCl) = 143.3\ \text{g mol}^{-1}

M(Cl) = 35.5\ \text{g mol}^{-1}

Show worked solution →

Step 1: write the precipitation equation (1:1 stoichiometry).

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

Step 2: moles of $AgCl$ precipitate.

n(AgCl) = \frac{m}{M} = \frac{1.146\ \text{g}}{143.3\ \text{g mol}^{-1}} = 7.998 \times 10^{-3}\ \text{mol}

Step 3: moles of chloride (1:1 ratio).

n(Cl^-) = 7.998 \times 10^{-3}\ \text{mol}

Step 4: mass of chloride.

m(Cl^-) = n \times M = 7.998 \times 10^{-3}\ \text{mol} \times 35.5\ \text{g mol}^{-1} = 0.2839\ \text{g}

Step 5: percentage by mass in the original sample.

\%Cl^- = \frac{0.2839\ \text{g}}{0.850\ \text{g}} \times 100\% = 33.4\%

Rounded to 3 significant figures (matching the 3 s.f. in the 0.850 g and 1.146 g data): $\%Cl^- = 33.4\%$ .

Marking criteria: 1 mark for the correct 1:1 stoichiometric equation, 1 mark for correct moles of $AgCl$ , 1 mark for correctly carrying the 1:1 ratio to chloride, 1 mark for the mass of chloride, 1 mark for the correct percentage to 3 significant figures with a percent sign. Note the true experimental percentage would likely differ slightly from any theoretical value if the sample were of known formula, due to the error sources discussed in the text (co-precipitation, incomplete drying, particle loss).

core5 marks

The titration curve below is an illustrative plot of pAg (that is,

-\log_{10}[Ag^+]

) versus volume of

0.100\ \text{mol L}^{-1}

AgNO_3

added to a 25.00 mL chloride sample, showing a sharp inflection (equivalence point) at 18.00 mL. (a) Identify the equivalence point volume from the curve and explain why the curve shows a sharp jump there rather than a gradual change. (b) Calculate the original chloride concentration in mol/L.

Show worked solution →

(a) Equivalence point and shape. The equivalence point is read from the steepest part of the curve, where the graph shows the inflection at 18.00 mL. Before the equivalence point, added $Ag^+$ is almost completely consumed by the still-plentiful $Cl^-$ , so free $[Ag^+]$ (and hence pAg) barely changes. Right at the equivalence point, one more drop of titrant causes a large relative change in free $[Ag^+]$ because there is essentially no $Cl^-$ left to consume it, producing the sharp vertical jump in pAg. Past the equivalence point, pAg falls more gradually again as excess $Ag^+$ simply dilutes into the growing volume.

(b) Concentration calculation.

n(Ag^+) = c \times V = 0.100\ \text{mol L}^{-1} \times 0.01800\ \text{L} = 1.80 \times 10^{-3}\ \text{mol}

n(Cl^-) = n(Ag^+) = 1.80 \times 10^{-3}\ \text{mol} \quad \text{(1:1 stoichiometry)}

c(Cl^-) = \frac{n}{V_{sample}} = \frac{1.80 \times 10^{-3}\ \text{mol}}{0.02500\ \text{L}} = 0.0720\ \text{mol L}^{-1}

Marking criteria: (a) 1 mark for correctly reading 18.00 mL from the graph, 1 mark for explaining the sharp jump in terms of $Cl^-$ being exhausted so free $Ag^+$ rises sharply. (b) 1 mark for correct moles of $Ag^+$ , 1 mark for the 1:1 stoichiometric ratio, 1 mark for the correct final concentration with units.

core6 marks

A student determines chloride in a bore-water sample two ways: gravimetrically (precipitating and weighing

AgCl

) and by Mohr titration. Compare the two methods in terms of precision, speed and suitability for testing 40 bore-water samples in a single day, and justify which method the student should choose.

Show worked solution →

Precision: Gravimetric analysis is inherently more precise (typically 0.1% or better) because it relies on a direct mass measurement on an analytical balance, which is a highly precise instrument, and avoids the subjectivity of judging a colour-change endpoint by eye. The Mohr titration is somewhat less precise (around 1%) because the endpoint judgement (first persistent red-brown tinge) introduces small operator-dependent variation.
Speed: Gravimetric analysis is slow: it requires precipitation, digestion, filtration and drying to constant mass, which typically takes many hours to more than a day per sample (including cooling in a desiccator and possibly a second drying cycle to confirm constant mass). The Mohr titration is fast, taking only minutes per sample once the titrant is standardised.
Suitability for 40 samples in one day: Gravimetric analysis is impractical for 40 samples in a single day because of the drying time alone; even with several ovens running in parallel, achieving constant mass for that many crucibles in one working day is unrealistic. The Mohr titration can comfortably process 40 samples in a single day, since each titration takes only a few minutes plus rinsing and refilling the burette.
Justification: The student should choose the Mohr precipitation titration for this task. While it sacrifices some precision compared with the gravimetric method, the difference (around 1% versus 0.1%) is small relative to typical variation in bore-water chloride levels, and the enormous time saving makes it the only realistic option for processing 40 samples in one day. Gravimetric analysis would instead be justified if only one or two samples needed the highest achievable accuracy, for example for a regulatory reference measurement.

Marking criteria: 1 mark for correctly comparing precision, 1 mark for correctly comparing speed, 1 mark for correctly reasoning that gravimetric analysis cannot realistically process 40 samples in a day, 1 mark for correctly reasoning that Mohr titration can, 1 mark for a clear final recommendation, 1 mark for the recommendation being properly justified by weighing precision against practicality rather than just restating the comparison.

exam8 marks

A council water lab reports a chloride result from a Mohr titration that is noticeably higher than the result later obtained by gravimetric analysis of

AgCl

on the same sample. Evaluate THREE plausible sources of this discrepancy, explain the direction each would push the Mohr result, and recommend the single most likely cause given that the sample is river water with a naturally cloudy brown tint and a measured pH of 6.2.

Show worked solution →

This is an 8-mark EVALUATE: markers reward a judgement that uses the specific stimulus (turbid, tinted, pH 6.2), not a generic list of errors.

Band 6 PLAN.

State three candidate error sources for the Mohr result specifically, each with the direction of the effect (too high or too low).
Use the given evidence (pH 6.2, brown tint/cloudiness) to argue which is most likely.
End with an explicit recommendation, not a neutral list.

Candidate 1: pH out of range (too acidic): The Mohr method requires pH 7 to 9.5. At pH 6.2, some chromate indicator is protonated to dichromate ( $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$ ), which does not form a red silver precipitate. The titrator must add MORE titrant than the true equivalence point before enough unprotonated chromate remains to give a visible colour change, so this error pushes the Mohr result HIGHER than the true value, consistent with the observation.
Candidate 2: obscured endpoint from colour/turbidity: The natural brown tint and cloudiness of river water make the pale pink-to-red colour change harder to see against the background colour. An operator waiting for a colour that is more visible against the brown background is likely to over-titrate, again giving a titre (and hence a chloride result) that is HIGHER than the true value.
Candidate 3: co-precipitation with other halides or ions: If bromide or iodide is present, they also precipitate with $Ag^+$ before chloride is exhausted, at a similar or even lower silver concentration, but $Ag^+$ reacts with them as if they were chloride, mole for mole. This makes the apparent chloride result too HIGH, since some of the titrant reacted with other anions, not just $Cl^-$ . (Gravimetric analysis by mass would report a similarly inflated combined-halide mass, so this candidate is less able to explain a MOHR-SPECIFIC discrepancy relative to gravimetric.)
Recommendation: Given the stated evidence, the pH of 6.2 is the most likely dominant cause: it falls clearly below the required 7 to 9.5 window and directly interferes with the chromate indicator chemistry itself, whereas the brown tint mainly makes an already-correct endpoint harder to judge (a smaller effect) and co-precipitation of other halides would likely also elevate the gravimetric result, which was NOT reported as elevated. The lab should buffer the sample to pH 7 to 9.5 (for example with dilute $NaHCO_3$ ) before repeating the Mohr titration to test this explanation.

Marker's note: top-band answers (1) give the correct direction (too high, not too low) for at least two candidates with a chemical mechanism, not just an assertion, (2) explicitly connect the pH 6.2 evidence to the chromate protonation equilibrium, (3) use the cloudy/tinted evidence to justify a SEPARATE, smaller contributing error rather than conflating it with the pH cause, and (4) end with a specific, testable recommendation rather than a vague "check the method" statement.

Practise this

Sources & how we know this

Chemistry Stage 6 Syllabus (2017) — NESA
Chemistry HSC exam papers — NESA