How to Calculate Chloride Concentration from AgCl Mass

A water sample is treated with silver nitrate, and a white solid forms. After filtration, the solid and filter assembly have gained 0.45 g. Can you use that mass to calculate the chloride concentration?

Not yet. The solid is wet, so its measured mass includes water and potentially dissolved impurities. You need the mass of pure, dry silver chloride before the calculation means anything.

The method is called precipitation gravimetry. You convert dissolved chloride ions into solid silver chloride, collect and dry the solid, then use its mass to calculate how much chloride was originally present.

The central calculation is:

\[ \boxed{ c(\mathrm{Cl^-})= \frac{m(\mathrm{AgCl})} {143.32\times V} } \]

Here, \(m(\mathrm{AgCl})\) is the net mass of pure, dry silver chloride in grams, and \(V\) is the original sample volume represented by that precipitate, in litres. The resulting concentration is in \(\mathrm{mol\,L^{-1}}\).

The important part is obtaining the correct silver chloride mass and using the correct sample volume. Let’s work through both.

01Why does silver chloride tell us the amount of chloride?

Silver nitrate provides silver ions, which react with dissolved chloride ions to form solid silver chloride.

The net ionic equation is:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

Silver chloride, \(\ce{AgCl}\), is a white precipitate. The dissolved silver and chloride ions combine to form an ionic solid that can be separated from the solution.

Notice the mole ratio:

\[ \mathrm{Ag^+ : Cl^- : AgCl}=1:1:1 \]

This is the key to the calculation.

Every mole of silver chloride contains one mole of chloride ions. Therefore, if you know the number of moles of silver chloride collected, you also know the number of moles of chloride originally present in the sample.

You are effectively using silver chloride as a measurable form of chloride.

Calculate the molar mass of silver chloride

Before converting mass into moles, you need the molar mass of \(\ce{AgCl}\).

Using the periodic table:

Therefore:

\[ \begin{aligned} M(\ce{AgCl})&=107.87+35.45\\ &=143.32\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The relationship between mass and amount of substance is:

\[ n=\frac{m}{M} \]

Since the reaction has a 1:1 mole ratio:

\[ \boxed{ n(\mathrm{Cl^-})= \frac{m(\mathrm{AgCl})}{143.32} } \]

This equation assumes that the collected solid is pure silver chloride and that precipitation is sufficiently complete.

02Follow the chloride through the experiment

The calculation is relatively short. Getting a reliable mass requires several stages.

Step 1: Collect a representative sample

Start with a representative sample of the solution being analysed.

If the solution is not uniform, taking a small portion may give an inaccurate result. Mix the sample thoroughly before measuring an aliquot using a volumetric pipette.

An aliquot is a precisely measured portion of a larger sample.

There is also an important distinction when analysing environmental water:

• If you are measuring dissolved chloride, any required filtration of suspended particles must happen before taking the aliquot.
• If you are measuring total chloride, removing suspended material without an appropriate extraction procedure could remove some of the chloride you intend to measure.

The sample preparation must match what you are trying to measure.

Step 2: Precipitate the chloride ions

Transfer the measured aliquot into a beaker or conical flask.

Add a controlled excess of silver nitrate solution, \(\ce{AgNO3}\), while mixing.

The silver ions react with the chloride ions to form white silver chloride.

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

The excess silver nitrate helps drive the precipitation towards completion.

The other dissolved ions that do not form insoluble products remain in the solution, which is called the supernatant or mother liquor.

Step 3: Filter and wash the precipitate

The silver chloride must now be separated from the remaining solution.

Use filtration to collect the solid in a suitable pre-weighed filter crucible.

The solid is retained, while the liquid passes through into the receiving flask.

However, the collected precipitate can still contain unwanted material.

For example, the liquid trapped between the particles may contain:

• Excess silver nitrate.
• Dissolved nitrate ions.
• Other soluble salts from the original sample.

Wash the precipitate using small portions of an appropriate chloride-free wash liquid.

This removes dissolved impurities without introducing additional chloride. The wash procedure must also minimise the loss of silver chloride.

Step 4: Dry the precipitate to constant mass

This is where many students make their biggest mistake.

A freshly filtered precipitate contains water. If you weigh it immediately, the balance measures:

\[ \text{Mass measured}= \text{Mass of AgCl}+\text{Mass of retained liquid} \]

That is not the mass you need.

Dry the precipitate, allow it to cool in a desiccator, and weigh it.

Then repeat the drying, cooling, and weighing process until successive measurements agree within the balance’s measurement resolution.

This is called drying to constant mass.

It gives you confidence that further drying is not causing a measurable loss of mass.

Silver chloride is also photosensitive. It can darken and decompose when exposed to light, so protect the precipitate from strong light during handling.

Step 5: Calculate the net mass of silver chloride

You cannot simply use the final mass of the crucible and precipitate together.

You must subtract the mass of the empty crucible.

\[ \boxed{ m(\mathrm{AgCl})= m(\text{crucible + dry AgCl}) - m(\text{empty crucible}) } \]

For example:

Therefore:

\[ \begin{aligned} m(\mathrm{AgCl}) &=24.8254-24.6821\\ &=0.1433\ \mathrm{g} \end{aligned} \]

This net mass is what you use in the chemical calculation.

03Convert silver chloride mass into chloride concentration

Once you have the correct mass, the calculation follows three steps.

Step 1: Calculate the moles of silver chloride.

\[ n(\mathrm{AgCl})= \frac{m(\mathrm{AgCl})}{143.32} \]

Step 2: Use the mole ratio to find chloride moles.

\[ n(\mathrm{Cl^-})=n(\mathrm{AgCl}) \]

Step 3: Divide by the volume of the original sample represented by the precipitate.

\[ \boxed{ c(\mathrm{Cl^-})= \frac{n(\mathrm{Cl^-})}{V} } \]

Remember to convert the sample volume into litres.

For example:

\[ 25.00\ \mathrm{mL}=0.02500\ \mathrm{L} \]

What if the question asks for chloride concentration in mg/L?

Sometimes the question wants a mass concentration rather than a molar concentration.

You can convert the silver chloride mass into chloride mass using the molar mass ratio:

\[ m(\mathrm{Cl^-})= m(\mathrm{AgCl}) \times \frac{35.45}{143.32} \]

The fraction represents the proportion of silver chloride’s mass contributed by chlorine.

Therefore:

\[ \boxed{ [\mathrm{Cl^-}]\ (\mathrm{mg/L})= \frac{m(\mathrm{AgCl})\times (35.45/143.32)\times1000} {V} } \]

Here, the silver chloride mass is in grams and the sample volume is in litres.

The factor of 1000 converts grams into milligrams.

Watch out for dilution

The volume in the concentration calculation is not always the volume in the final flask.

There are two different situations.

Adding water after taking an aliquot does not change the number of chloride ions in that aliquot.

However, diluting the original sample before taking an aliquot reduces the chloride concentration in the solution being analysed.

For a dilution:

\[ c_1V_1=c_2V_2 \]

where:

• \(c_1\) is the original concentration.
• \(V_1\) is the original sample volume taken.
• \(c_2\) is the diluted concentration.
• \(V_2\) is the final diluted volume.

This distinction becomes important in the second worked example.

04Worked examples

05Experimental errors and limitations

Even if your arithmetic is perfect, the calculated concentration can still be wrong.

The result depends on whether the precipitate represents the chloride in the original sample accurately.

Consider the following common problems.

Constant mass does not automatically mean pure AgCl

This is a particularly important distinction.

Drying to constant mass tells you that the mass is no longer changing significantly during further drying.

It does not prove that everything in the crucible is silver chloride.

For example, imagine that some dissolved sodium nitrate remains trapped in the precipitate after filtration.

During drying, the water evaporates, but the sodium nitrate remains as a solid.

The crucible could reach a perfectly stable mass while still containing two different solids.

The measured mass would then be:

\[ m_{\mathrm{measured}}= m(\mathrm{AgCl})+ m(\mathrm{NaNO_3}) \]

If you treat the entire mass as AgCl, you overestimate the amount of chloride.

This is why washing is a chemical separation step, not just a way to make the precipitate look cleaner.

What if another ion also forms a precipitate?

Silver ions do not react exclusively with chloride.

For example, bromide ions also form an insoluble silver salt:

\[ \ce{Ag+(aq) + Br-(aq) -> AgBr(s)} \]

Silver bromide is a different solid with a different molar mass.

If both chloride and bromide are present, the collected solid contains both \(\ce{AgCl}\) and \(\ce{AgBr}\).

You cannot automatically assume that the entire precipitate mass represents silver chloride.

This is a problem of selectivity: the reaction must distinguish the substance being measured from other substances in the sample.

It becomes especially important when interpreting environmental analysis results.

06How is this different from a precipitation titration?

Both methods use the same silver chloride precipitation reaction, but they measure different things.

In gravimetry, your key measurement is mass.

In titration, your key measurement is volume.

Both rely on the balanced equation and its 1:1 mole ratio.

If you want to practise the alternative method, see Precipitation Titrations: Calculate an Unknown Concentration.

07Questions and solutions

The following questions move from a straightforward gravimetric calculation towards problems involving dilution, experimental reliability, and competing precipitation reactions.

Assume that precipitation is quantitative and the collected solids are stable unless a question states otherwise.

Use:

\[ M(\ce{AgCl})=143.32\ \mathrm{g\,mol^{-1}} \]

\[ M(\ce{Cl})=35.45\ \mathrm{g\,mol^{-1}} \]

Question 1

A \(25.00\ \mathrm{mL}\) water sample is analysed using precipitation gravimetry.

The empty filter crucible has a mass of \(31.2264\ \mathrm{g}\). After precipitation, filtration, washing, and drying to constant mass, the crucible and precipitate have a combined mass of \(31.3697\ \mathrm{g}\).

Calculate:

a) The mass of silver chloride collected.

b) The number of moles of chloride ions in the original sample.

c) The chloride concentration in \(\mathrm{mol\,L^{-1}}\).

d) The chloride concentration in \(\mathrm{mg\,L^{-1}}\).

Solution 1

The chloride concentration is \(0.04000\ \mathrm{mol\,L^{-1}}\), equivalent to approximately \(1418\ \mathrm{mg\,L^{-1}}\).

a) Calculate the silver chloride mass.

Subtract the empty crucible mass:

\[ \begin{aligned} m(\mathrm{AgCl}) &=31.3697-31.2264\\ &=0.1433\ \mathrm{g} \end{aligned} \]

b) Calculate the moles of chloride ions.

First calculate the moles of silver chloride:

\[ \begin{aligned} n(\mathrm{AgCl}) &=\frac{0.1433}{143.32}\\ &=1.000\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The reaction has a 1:1 mole ratio.

Therefore:

\[ n(\mathrm{Cl^-})=1.000\times10^{-3}\ \mathrm{mol} \]

c) Calculate the molar concentration.

Convert the sample volume:

\[ 25.00\ \mathrm{mL}=0.02500\ \mathrm{L} \]

Then:

\[ \begin{aligned} c(\mathrm{Cl^-}) &=\frac{1.000\times10^{-3}}{0.02500}\\ &=0.04000\ \mathrm{mol\,L^{-1}} \end{aligned} \]

d) Calculate the mass concentration.

The mass of chloride in the sample is:

\[ \begin{aligned} m(\mathrm{Cl^-}) &=0.1433\times\frac{35.45}{143.32}\\ &=0.03545\ \mathrm{g}\\ &=35.45\ \mathrm{mg} \end{aligned} \]

Therefore:

\[ \begin{aligned} [\mathrm{Cl^-}] &=\frac{35.45}{0.02500}\\ &=1418\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The important point is that the mass of silver chloride must first be converted into chloride moles or chloride mass. Using the silver chloride mass directly as the chloride mass would give an incorrect result.

Question 2

An environmental water sample is diluted by transferring \(12.50\ \mathrm{mL}\) of the original sample into a \(250.0\ \mathrm{mL}\) volumetric flask and making it up to the calibration mark.

A \(50.00\ \mathrm{mL}\) aliquot of the diluted solution is transferred into a beaker. The student adds a further \(75.0\ \mathrm{mL}\) of distilled water before adding excess silver nitrate.

The following measurements are recorded:

Calculate the original chloride concentration in:

a) \(\mathrm{mol\,L^{-1}}\).

b) \(\mathrm{mg\,L^{-1}}\).

Explain why neither the wet precipitate mass nor the total volume in the beaker should be used directly.

Solution 2

The original chloride concentration is \(0.4000\ \mathrm{mol\,L^{-1}}\), equivalent to \(1.418\times10^4\ \mathrm{mg\,L^{-1}}\).

Step 1: Determine the correct dry precipitate mass.

The final two measurements agree, so the constant mass is \(27.5249\ \mathrm{g}\).

\[ \begin{aligned} m(\mathrm{AgCl}) &=27.5249-27.3816\\ &=0.1433\ \mathrm{g} \end{aligned} \]

The initial wet mass is not suitable because it includes retained liquid.

Step 2: Calculate the moles of chloride.

\[ \begin{aligned} n(\mathrm{AgCl}) &=\frac{0.1433}{143.32}\\ &=1.000\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Using the 1:1 mole ratio:

\[ n(\mathrm{Cl^-})=1.000\times10^{-3}\ \mathrm{mol} \]

Step 3: Calculate the concentration in the diluted sample.

The aliquot was \(50.00\ \mathrm{mL}\), or \(0.05000\ \mathrm{L}\).

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{1.000\times10^{-3}}{0.05000}\\ &=0.02000\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The additional \(75.0\ \mathrm{mL}\) of water does not change the number of chloride moles in the aliquot.

Step 4: Correct for the original dilution.

The dilution factor is:

\[ \frac{250.0}{12.50}=20.00 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=0.02000\times20.00\\ &=0.4000\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 5: Convert to mg/L.

\[ \begin{aligned} [\mathrm{Cl^-}] &=0.4000\times35.45\times1000\\ &=14180\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The two volume errors to avoid are:

• Using the wet precipitate mass, which includes water and overestimates the amount of AgCl.
• Using the total beaker volume of \(125.0\ \mathrm{mL}\) instead of the original \(50.00\ \mathrm{mL}\) aliquot volume when calculating the diluted concentration.

The original dilution must then be corrected separately.

Question 3

A student analyses a \(50.00\ \mathrm{mL}\) chloride solution using a pre-weighed crucible.

The following results are obtained:

The student argues that the first mass recorded after drying should be used because the precipitate has already been heated once.

a) Determine the correct mass of silver chloride.

b) Calculate the chloride concentration in \(\mathrm{mol\,L^{-1}}\).

c) Calculate the chloride concentration in \(\mathrm{mg\,L^{-1}}\).

d) Calculate the concentration that would have been reported if the student had incorrectly used the wet precipitate mass. Explain why this result is unreliable.

Solution 3

The correct chloride concentration is approximately \(0.0251\ \mathrm{mol\,L^{-1}}\), or \(889\ \mathrm{mg\,L^{-1}}\).

a) Determine the correct silver chloride mass.

The final two measurements are effectively constant.

Therefore, use \(22.6606\ \mathrm{g}\).

\[ \begin{aligned} m(\mathrm{AgCl}) &=22.6606-22.4810\\ &=0.1796\ \mathrm{g} \end{aligned} \]

The first drying measurement is not suitable because the mass continues to decrease during subsequent drying.

b) Calculate the chloride concentration.

First determine the amount of silver chloride:

\[ \begin{aligned} n(\mathrm{AgCl}) &=\frac{0.1796}{143.32}\\ &=1.253\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The chloride amount is identical:

\[ n(\mathrm{Cl^-})=1.253\times10^{-3}\ \mathrm{mol} \]

Convert the sample volume:

\[ 50.00\ \mathrm{mL}=0.05000\ \mathrm{L} \]

Then:

\[ \begin{aligned} c(\mathrm{Cl^-}) &=\frac{1.253\times10^{-3}}{0.05000}\\ &=0.02506\ \mathrm{mol\,L^{-1}} \end{aligned} \]

c) Calculate the mass concentration.

\[ \begin{aligned} [\mathrm{Cl^-}] &=0.02506\times35.45\times1000\\ &=888.5\ \mathrm{mg\,L^{-1}} \end{aligned} \]

d) Investigate the incorrect wet-mass calculation.

The student would calculate the apparent silver chloride mass as:

\[ \begin{aligned} m_{\mathrm{apparent}} &=22.9320-22.4810\\ &=0.4510\ \mathrm{g} \end{aligned} \]

Treating this as pure AgCl gives:

\[ \begin{aligned} n_{\mathrm{apparent}} &=\frac{0.4510}{143.32}\\ &=3.147\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The apparent concentration would be:

\[ \begin{aligned} c_{\mathrm{apparent}} &=\frac{3.147\times10^{-3}}{0.05000}\\ &=0.06294\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This is approximately 2.51 times the correct concentration.

The mistake is assuming that everything contributing to the wet mass is silver chloride.

It is not. Retained water and dissolved material contribute to the balance reading but do not represent additional chloride ions.

The correct endpoint is constant mass, not simply the first time the sample has been heated.

Question 4

A student precipitates chloride from a \(100.0\ \mathrm{mL}\) sample using excess silver nitrate.

The crucible has a mass of \(20.0000\ \mathrm{g}\). After filtration and drying to constant mass, the crucible and solid have a combined mass of \(20.3000\ \mathrm{g}\).

However, a matched control experiment establishes that \(0.0100\ \mathrm{g}\) of soluble reagent residue remains in the unwashed solid.

The student argues that drying to constant mass proves that the precipitate is pure silver chloride.

a) Explain why this argument is incorrect.

b) Calculate the corrected mass of silver chloride.

c) Calculate the corrected chloride concentration in \(\mathrm{mol\,L^{-1}}\).

d) Calculate the percentage by which the uncorrected concentration overestimates the true concentration.

Assume the \(0.0100\ \mathrm{g}\) residue is entirely non-volatile material other than silver chloride, and that the control provides a valid correction.

Solution 4

The corrected chloride concentration is approximately \(0.02023\ \mathrm{mol\,L^{-1}}\). The uncorrected result overestimates the concentration by approximately \(3.45\%\).

a) Explain why constant mass does not prove purity.

Drying to constant mass establishes that further drying produces no significant change in the measured mass.

It does not establish the chemical identity of every solid in the crucible.

For example, soluble salts trapped in the precipitate may remain after the water evaporates. These salts contribute to the final mass even though they are not silver chloride.

The sample must therefore be washed adequately before drying.

b) Calculate the corrected silver chloride mass.

First subtract the empty crucible mass:

\[ m_{\mathrm{apparent}} =20.3000-20.0000 =0.3000\ \mathrm{g} \]

Now subtract the known impurity mass:

\[ \begin{aligned} m(\mathrm{AgCl}) &=0.3000-0.0100\\ &=0.2900\ \mathrm{g} \end{aligned} \]

c) Calculate the corrected chloride concentration.

\[ \begin{aligned} n(\mathrm{AgCl}) &=\frac{0.2900}{143.32}\\ &=2.023\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ n(\mathrm{Cl^-})=2.023\times10^{-3}\ \mathrm{mol} \]

The sample volume is \(0.1000\ \mathrm{L}\).

\[ \begin{aligned} c(\mathrm{Cl^-}) &=\frac{2.023\times10^{-3}}{0.1000}\\ &=0.02023\ \mathrm{mol\,L^{-1}} \end{aligned} \]

d) Calculate the percentage overestimate.

First calculate the uncorrected concentration:

\[ \begin{aligned} c_{\mathrm{uncorrected}} &=\frac{0.3000/143.32}{0.1000}\\ &=0.02093\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The percentage overestimate is calculated relative to the corrected value:

\[ \begin{aligned} \%\text{ overestimate} &=\frac{0.02093-0.02023}{0.02023} \times100\\ &=3.45\% \end{aligned} \]

The important distinction is that a stable mass is not necessarily a chemically pure mass.

A reliable gravimetric result requires both adequate drying and adequate separation of impurities.

Question 5

An environmental laboratory receives a sample containing chloride and bromide ions, but no other ions that form precipitates with silver ions under the experimental conditions.

A \(20.00\ \mathrm{mL}\) portion of the original sample is diluted to \(200.0\ \mathrm{mL}\).

A \(25.00\ \mathrm{mL}\) aliquot of the diluted solution is treated with excess silver nitrate. The resulting precipitate is filtered, washed, and dried to constant mass.

The total mass of the pure mixed precipitate is:

\[ m_{\mathrm{total}}=0.3210\ \mathrm{g} \]

An independent, selective measurement establishes that the bromide concentration in the diluted solution is:

\[ [\mathrm{Br^-}]=0.0100\ \mathrm{mol\,L^{-1}} \]

Use:

\[ M(\ce{AgCl})=143.32\ \mathrm{g\,mol^{-1}} \]

\[ M(\ce{AgBr})=187.77\ \mathrm{g\,mol^{-1}} \]

a) Calculate the mass of silver bromide in the mixed precipitate.

b) Calculate the mass of silver chloride in the mixed precipitate.

c) Determine the original chloride concentration in \(\mathrm{mol\,L^{-1}}\).

d) A student argues that repeating the experiment with a \(50.00\ \mathrm{mL}\) aliquot instead of a \(25.00\ \mathrm{mL}\) aliquot would independently reveal the chloride concentration, even without measuring bromide.

Evaluate this claim. Explain why more precise mass measurements or a larger aliquot cannot necessarily solve the problem.

Solution 5

The original chloride concentration is approximately \(0.765\ \mathrm{mol\,L^{-1}}\). The student’s claim is incorrect because increasing the aliquot size does not provide an independent measurement of the two precipitates.

This question is more difficult because the collected solid is not entirely silver chloride.

a) Calculate the mass of silver bromide.

Start with the bromide concentration in the diluted solution:

\[ [\mathrm{Br^-}]=0.0100\ \mathrm{mol\,L^{-1}} \]

The aliquot volume is:

\[ 25.00\ \mathrm{mL}=0.02500\ \mathrm{L} \]

Calculate the bromide amount:

\[ \begin{aligned} n(\mathrm{Br^-}) &=cV\\ &=0.0100\times0.02500\\ &=2.50\times10^{-4}\ \mathrm{mol} \end{aligned} \]

The reaction is:

\[ \ce{Ag+(aq) + Br-(aq) -> AgBr(s)} \]

The mole ratio is 1:1.

Therefore:

\[ n(\mathrm{AgBr})=2.50\times10^{-4}\ \mathrm{mol} \]

Calculate its mass:

\[ \begin{aligned} m(\mathrm{AgBr}) &=nM\\ &=(2.50\times10^{-4})(187.77)\\ &=0.04694\ \mathrm{g} \end{aligned} \]

b) Calculate the mass of silver chloride.

The total precipitate contains both silver chloride and silver bromide.

Therefore:

\[ \begin{aligned} m(\mathrm{AgCl}) &=m_{\mathrm{total}}-m(\mathrm{AgBr})\\ &=0.3210-0.04694\\ &=0.2741\ \mathrm{g} \end{aligned} \]

Only this remaining mass represents silver chloride.

c) Calculate the original chloride concentration.

First calculate the moles of silver chloride:

\[ \begin{aligned} n(\mathrm{AgCl}) &=\frac{0.2741}{143.32}\\ &=1.912\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The 1:1 reaction ratio gives:

\[ n(\mathrm{Cl^-})=1.912\times10^{-3}\ \mathrm{mol} \]

Calculate the chloride concentration in the diluted solution:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{1.912\times10^{-3}}{0.02500}\\ &=0.07649\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Now correct for the original dilution.

The original \(20.00\ \mathrm{mL}\) sample was diluted to \(200.0\ \mathrm{mL}\).

\[ \text{Dilution factor}=\frac{200.0}{20.00}=10.00 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=0.07649\times10.00\\ &=0.7649\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Hence:

\[ \boxed{ [\mathrm{Cl^-}]_{\mathrm{original}} =0.765\ \mathrm{mol\,L^{-1}} } \]

d) Evaluate the student’s claim.

The student’s claim is incorrect.

The central problem is that the measured mass represents two different compounds.

The total precipitate mass is:

\[ m_{\mathrm{total}}= n(\mathrm{Cl^-})M(\ce{AgCl}) + n(\mathrm{Br^-})M(\ce{AgBr}) \]

There are two unknown amounts:

• The amount of chloride.
• The amount of bromide.

One total mass measurement cannot independently determine both quantities.

The independent bromide measurement provides the additional information needed to calculate the bromide contribution and subtract it from the total mass.

Now consider the proposed \(50.00\ \mathrm{mL}\) aliquot.

Both chloride and bromide amounts would double because the aliquot contains twice as much of the same diluted solution.

Consequently, the total precipitate mass would also double, but the relative contributions of AgCl and AgBr would remain unchanged.

The second measurement would simply reproduce the same relationship on a larger scale. It would not provide an independent equation that separates the two ions.

This is the important distinction between precision and selectivity.

A more precise balance could measure the combined mass more accurately. It still could not identify which proportion of that mass came from silver chloride and which came from silver bromide.

The experiment needs an additional selective measurement, not merely a larger sample or a more precise balance.

08Where to go next

Once you have calculated the chloride concentration, the next step is deciding how to report and evaluate the analytical result.

For environmental analysis, this often means converting between \(\mathrm{mol\,L^{-1}}\) and \(\mathrm{mg\,L^{-1}}\), accounting for dilution, and checking whether the method has introduced systematic errors.

You can build on this calculation using How to Convert Gravimetric Results into mg/L.

The key principle to carry forward is simple: the mass of a precipitate only tells you the amount of the analyte when you know what the precipitate contains, how much was recovered, and which original sample volume it represents.