How to Use Two Stoichiometric Ratios in Back Titration

You add 5.00 mmol of sodium hydroxide to a fertiliser sample. After the sample reacts, the sodium hydroxide left over requires 1.00 mmol of sulfuric acid to neutralise it.

Would you subtract 1.00 mmol from 5.00 mmol to find how much sodium hydroxide the sample consumed?

That seems reasonable, but it gives the wrong answer. One mole of sulfuric acid neutralises two moles of sodium hydroxide. The 1.00 mmol of sulfuric acid therefore represents 2.00 mmol of leftover sodium hydroxide.

The sample consumed \(5.00-2.00=3.00\) mmol of sodium hydroxide.

Now suppose the substance being measured is ammonium chloride, \(\ce{NH4Cl}\). Each mole of ammonium chloride consumes one mole of sodium hydroxide, so the sample contained 3.00 mmol of ammonium chloride.

That’s how you track two different stoichiometric ratios in a back titration:

1. Use the titration reaction ratio to calculate the amount of reagent left over.
2. Subtract that amount from the reagent originally added, keeping both amounts in moles of the same reagent.
3. Use the sample reaction ratio to convert the reagent consumed into the amount of substance being measured.

The two mole ratios belong to different chemical reactions. You must apply each one at the correct stage.

01Why this back titration needs two reactions

Consider a fertiliser containing ammonium chloride mixed with an inert material such as sodium chloride.

You want to measure the ammonium chloride content.

Adding sulfuric acid directly to the dissolved sample won’t give a useful neutralisation titre for ammonium chloride. The ammonium ion, \(\ce{NH4+}\), is already an acid. There isn’t a straightforward reaction in which each ammonium ion consumes a fixed amount of added strong acid.

Instead, use a known excess of sodium hydroxide.

Ammonium ions react with hydroxide ions to form ammonia and water:

\[ \ce{NH4+ + OH- -> NH3 + H2O} \]

For ammonium chloride, the complete equation is:

\[ \ce{NH4Cl + NaOH -> NH3 + NaCl + H2O} \]

Each mole of ammonium chloride consumes one mole of sodium hydroxide.

This is the sample reaction, and its mole ratio is:

\[ \ce{NH4Cl}:\ce{NaOH}=1:1 \]

But we don’t directly measure how much sodium hydroxide disappears. We measure how much remains.

How the apparatus makes that possible

The experimental sequence is important.

1. Weigh the fertiliser sample into a suitable reaction flask.
2. Add an accurately measured volume of standard sodium hydroxide solution. The amount must exceed what is needed to react with all the ammonium chloride.
3. Warm the mixture under appropriate supervised laboratory conditions. Ammonia gas is released and removed from the reaction solution, using suitable ventilation and gas-trapping equipment.
4. Allow the remaining solution to cool. It contains sodium chloride, any other non-interfering dissolved salts, and leftover sodium hydroxide.
5. Titrate the leftover sodium hydroxide with standard sulfuric acid from a burette.

Why remove the ammonia before titrating?

Predict what happens if ammonia is still dissolved when sulfuric acid is added.

Sulfuric acid can react with the ammonia as well as the leftover hydroxide. Some of the acid titre would then be measuring ammonia rather than just sodium hydroxide.

That would make the calculated amount of leftover sodium hydroxide too large and the calculated ammonium chloride content too small.

Removing the ammonia gives the second reaction a clearer purpose: measuring the excess sodium hydroxide.

In the idealised calculations that follow, assume all ammonium chloride reacts, the ammonia is removed completely, no alkaline solution is lost, and no other substance consumes sodium hydroxide unless specifically stated.

The second reaction: measuring leftover sodium hydroxide

Sulfuric acid reacts with sodium hydroxide:

\[ \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} \]

The mole ratio is:

\[ \ce{H2SO4}:\ce{NaOH}=1:2 \]

This is different from the ammonium chloride reaction.

In a typical titration, the cooled solution initially contains excess sodium hydroxide, making phenolphthalein pink. Sulfuric acid is added from the burette until the endpoint is reached.

The volume and concentration of sulfuric acid reveal the amount of sodium hydroxide remaining.

02Keep added, leftover, and consumed reagent separate

Think of sodium hydroxide as a measured supply. The sample uses some, and the titration accounts for what is still available.

The first calculation doesn’t tell you how much ammonium chloride was present. It tells you how much sodium hydroxide survived the first reaction.

Only after subtracting that leftover amount can you use the ammonium chloride mole ratio.

The two ratios have different jobs

The central relationship is:

\[ n_C=n_A-n_L \]

Here:

• \(n_A\) is the amount of sodium hydroxide originally added, in mol.
• \(n_L\) is the amount of sodium hydroxide left over, in mol.
• \(n_C\) is the amount of sodium hydroxide consumed by the sample, in mol.

All three quantities refer to sodium hydroxide.

The consumed amount is positive. The subtraction assumes the sample is the only substance responsible for the decrease in sodium hydroxide.

If sulfuric acid is the titrant, the leftover amount is:

\[ n_L=2n_T \]

where \(n_T\) is the amount of sulfuric acid delivered by the burette, in mol.

For an ammonium chloride sample:

\[ n(\ce{NH4Cl})=n_C \]

This last equality comes from the separate \(1:1\) sample reaction.

A practical calculation order

Use this sequence whenever an excess reagent is measured by a different titrant:

For solutions, calculate the amount of substance using:

\[ n=cV \]

where \(n\) is the amount in mol, \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is volume in L.

Convert millilitres into litres before substituting.

The most dangerous mistake is skipping Stage 3 and subtracting moles of titrant directly from moles of the original reagent.

03Worked calculations

Look at the balance before moving on

The original 0.01000 mol of sodium hydroxide has been accounted for:

This check is useful because the subtraction must conserve the amount of sodium hydroxide.

A result where leftover sodium hydroxide exceeds the amount originally added would mean the assumed reaction model, measurements, or calculations cannot all be correct.

04When the numbers look right but the chemistry is wrong

A back titration is only as reliable as the assumptions connecting the two reactions.

Mistake 1: Using the titrant’s moles as the leftover reagent’s moles

If sulfuric acid neutralises sodium hydroxide, 0.00200 mol of sulfuric acid represents 0.00400 mol of sodium hydroxide.

If hydrochloric acid neutralises sodium hydroxide, 0.00200 mol of hydrochloric acid represents 0.00200 mol of sodium hydroxide.

The same numerical titre in moles doesn’t necessarily mean the same amount of leftover reagent.

Mistake 2: Forgetting that ammonia can react with the titrant

Suppose ammonium chloride has reacted with excess sodium hydroxide, but some ammonia remains dissolved.

When sulfuric acid is added, it can protonate the ammonia:

\[ \ce{NH3 + H+ -> NH4+} \]

Now the acid titre includes acid consumed by the ammonia as well as the leftover sodium hydroxide.

The apparent amount of leftover sodium hydroxide is too high.

Subtracting it gives an ammonium chloride result that is too low.

This explains why the experimental order matters.

Heating, cooling, and titrating aren’t interchangeable operations. The unwanted ammonia needs to be removed before the remaining sodium hydroxide is measured.

Mistake 3: Assuming a zero titre means zero sample

Imagine adding too little sodium hydroxide.

Every hydroxide ion might react with the ammonium chloride, leaving no sodium hydroxide to titrate.

That doesn’t mean the sample contains no ammonium chloride. It means the excess-reagent condition was not achieved.

A back titration requires some of the deliberately added reagent to remain. Otherwise, you cannot tell how much more of that reagent the sample might have consumed.

Mistake 4: Assuming any substance that consumes sodium hydroxide is the target

If another component of the sample also consumes sodium hydroxide, the subtraction counts both reactions.

The calculation measures the total sodium hydroxide consumed, not automatically the amount of ammonium chloride.

To identify a particular component, the experimental chemistry must make its contribution distinguishable.

That’s an important distinction between a mathematically correct answer and a chemically justified answer.

05Questions and solutions

The first question checks the calculation sequence. The later questions test whether the assumptions behind that sequence are actually satisfied.

Unless stated otherwise, assume reactions go to completion, the solutions are well mixed, and reported titres represent ideal stoichiometric endpoints. Use the molar masses supplied in the questions.

Question 1

A fertiliser sample containing ammonium chloride and inert filler is treated with 20.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) sodium hydroxide.

After the ammonia is completely removed, the leftover sodium hydroxide requires 8.00 mL of 0.0500 \(\mathrm{mol\,L^{-1}}\) sulfuric acid.

Calculate the amount, in mol, and mass, in g, of ammonium chloride in the sample.

Use \(M(\ce{NH4Cl})=53.49\ \mathrm{g\,mol^{-1}}\).

Solution 1

The sample contains 0.00120 mol, or 0.0642 g, of ammonium chloride.

First calculate the sodium hydroxide added:

\[ \begin{aligned} n_A&=0.1000\times0.02000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

Calculate the sulfuric acid titre:

\[ \begin{aligned} n_T&=0.0500\times0.00800\\ &=0.000400\ \mathrm{mol} \end{aligned} \]

The \(\ce{H2SO4}:\ce{NaOH}\) ratio is \(1:2\), so the leftover sodium hydroxide is:

\[ \begin{aligned} n_L&=2(0.000400)\\ &=0.000800\ \mathrm{mol} \end{aligned} \]

Subtract:

\[ \begin{aligned} n_C&=0.002000-0.000800\\ &=0.001200\ \mathrm{mol} \end{aligned} \]

Ammonium chloride reacts \(1:1\) with sodium hydroxide, so its amount is 0.001200 mol.

Its mass is:

\[ \begin{aligned} m&=0.001200\times53.49\\ &=0.0642\ \mathrm{g} \end{aligned} \]

The tempting mistake is to subtract 0.000400 mol directly from 0.002000 mol. That would mix moles of sulfuric acid with moles of sodium hydroxide, ignoring the \(1:2\) titration ratio.

Question 2

Two identical fertiliser samples contain the same amount of ammonium chloride. Each receives 0.01000 mol of excess sodium hydroxide.

In Trial A, all the ammonia produced is removed before titration.

In Trial B, the reaction is complete, but some of the ammonia remains dissolved in the solution.

Both solutions are titrated with 0.1000 \(\mathrm{mol\,L^{-1}}\) sulfuric acid to an acidic endpoint at which all retained ammonia is protonated.

The results are:

Assume retained ammonia is the only difference between the trials.

Explain which trial gives the valid back-titration result for ammonium chloride. Calculate the amount of ammonia retained in Trial B and the percentage of the originally produced ammonia that this represents.

Explain the effect of incorrectly treating Trial B’s titre as a measurement of leftover sodium hydroxide alone.

Solution 2

Trial A gives the valid ammonium chloride result. Trial B retains 0.000500 mol of ammonia, equivalent to approximately 7.96% of the ammonia originally produced. Incorrectly interpreting Trial B’s entire titre as leftover sodium hydroxide would underestimate the ammonium chloride content.

Trial A measures only the leftover sodium hydroxide.

The amount of sulfuric acid used is:

\[ \begin{aligned} n_T&=0.1000\times0.01860\\ &=0.001860\ \mathrm{mol} \end{aligned} \]

Thus:

\[ \begin{aligned} n_L&=2(0.001860)\\ &=0.003720\ \mathrm{mol} \end{aligned} \]

The amount of sodium hydroxide consumed by the ammonium chloride is:

\[ \begin{aligned} n_C&=0.01000-0.003720\\ &=0.006280\ \mathrm{mol} \end{aligned} \]

Because ammonium chloride and sodium hydroxide react \(1:1\), 0.006280 mol of ammonium chloride reacted and produced 0.006280 mol of ammonia.

Now consider Trial B.

The extra sulfuric acid used is:

\[ \begin{aligned} \Delta n_T &=0.1000(0.02110-0.01860)\\ &=0.000250\ \mathrm{mol} \end{aligned} \]

One mole of sulfuric acid supplies two moles of hydrogen ions for neutralisation.

Each mole of ammonia accepts one hydrogen ion:

\[ \ce{NH3 + H+ -> NH4+} \]

Therefore:

\[ \begin{aligned} n(\ce{NH3})_{\mathrm{retained}} &=2(0.000250)\\ &=0.000500\ \mathrm{mol} \end{aligned} \]

The percentage retained is:

\[ \begin{aligned} \%\text{ retained} &=\frac{0.000500}{0.006280}\times100\\ &=7.96\% \end{aligned} \]

If Trial B’s titre were incorrectly treated as measuring only sodium hydroxide:

\[ \begin{aligned} n_{L,\mathrm{apparent}} &=2(0.1000)(0.02110)\\ &=0.004220\ \mathrm{mol} \end{aligned} \]

The apparent sodium hydroxide consumption would be:

\[ \begin{aligned} n_{C,\mathrm{apparent}} &=0.01000-0.004220\\ &=0.005780\ \mathrm{mol} \end{aligned} \]

That’s lower than the correct 0.006280 mol.

The important point is that more acid doesn’t necessarily mean more sodium hydroxide remained. The additional acid might have reacted with another substance.

Here, retained ammonia makes the acid titre larger while the actual ammonium chloride content stays unchanged.

Question 3

A fertiliser mixture contains only ammonium chloride and inert sodium chloride.

Two separate 0.5000 g portions of the mixture are analysed.

In the first trial, 20.00 mL of 0.2000 \(\mathrm{mol\,L^{-1}}\) sodium hydroxide is added. After heating and removal of ammonia, no leftover sodium hydroxide is detected.

A student concludes that the sample originally contained exactly 0.004000 mol of ammonium chloride.

In the second trial, another 0.5000 g portion receives 50.00 mL of 0.2000 \(\mathrm{mol\,L^{-1}}\) sodium hydroxide. After reaction and removal of ammonia, the solution requires 20.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) sulfuric acid.

Determine whether the student’s first conclusion was justified, calculate the actual ammonium chloride percentage by mass, and explain why the two trials are consistent.

Use \(M(\ce{NH4Cl})=53.49\ \mathrm{g\,mol^{-1}}\).

Solution 3

The first conclusion was not justified. The fertiliser actually contains 0.006000 mol of ammonium chloride per 0.5000 g portion, giving 64.2% ammonium chloride by mass.

The first trial did not establish that sodium hydroxide was in excess.

The amount initially added was:

\[ \begin{aligned} n_A&=0.2000\times0.02000\\ &=0.004000\ \mathrm{mol} \end{aligned} \]

With no leftover sodium hydroxide detected, it is possible that all 0.004000 mol reacted while some ammonium chloride remained.

Under the stated complete-reaction assumptions, the observation establishes a lower bound of 0.004000 mol of ammonium chloride, not its exact amount.

Now calculate the result from Trial 2.

Sodium hydroxide added:

\[ \begin{aligned} n_A&=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]

Sulfuric acid used:

\[ \begin{aligned} n_T&=0.1000\times0.02000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

Leftover sodium hydroxide:

\[ \begin{aligned} n_L&=2(0.002000)\\ &=0.004000\ \mathrm{mol} \end{aligned} \]

Sodium hydroxide consumed:

\[ \begin{aligned} n_C&=0.01000-0.004000\\ &=0.006000\ \mathrm{mol} \end{aligned} \]

The ammonium chloride reaction is \(1:1\), so:

\[ n(\ce{NH4Cl})=0.006000\ \mathrm{mol} \]

Calculate the mass:

\[ \begin{aligned} m&=0.006000\times53.49\\ &=0.32094\ \mathrm{g} \end{aligned} \]

Percentage by mass:

\[ \begin{aligned} \%\ce{NH4Cl} &=\frac{0.32094}{0.5000}\times100\\ &=64.2\% \end{aligned} \]

The trials are consistent because the first trial supplied only 0.004000 mol of sodium hydroxide, while the sample contained 0.006000 mol of ammonium chloride.

There wasn’t enough hydroxide to react with all the ammonium chloride.

Approximately 0.002000 mol of ammonium chloride remained unreacted after the first trial, assuming complete consumption of the available hydroxide.

The second trial supplied sufficient sodium hydroxide for all the ammonium chloride to react, with 0.004000 mol left over.

The trap is assuming that a zero acid titre is simply an ordinary back-titration result. In reality, the absence of leftover reagent means the experiment has lost the information needed to determine the full amount of sample.

Question 4

A 0.23912 g fertiliser sample consists entirely of ammonium chloride, \(\ce{NH4Cl}\), and ammonium sulfate, \(\ce{(NH4)2SO4}\).

The sample is treated with 35.00 mL of 0.2000 \(\mathrm{mol\,L^{-1}}\) sodium hydroxide.

After complete reaction and removal of ammonia, the remaining sodium hydroxide requires 15.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) sulfuric acid.

Calculate the mass of each ammonium salt in the original sample.

Use:

• \(M(\ce{NH4Cl})=53.49\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{(NH4)2SO4})=132.14\ \mathrm{g\,mol^{-1}}\)

Explain why the measured sample mass is essential for distinguishing the two salts.

Solution 4

The sample contains 0.10698 g of ammonium chloride and 0.13214 g of ammonium sulfate. The back titration measures the total amount of ammonium ions, and the sample mass supplies the second independent relationship needed to distinguish the salts.

First calculate the sodium hydroxide added:

\[ \begin{aligned} n_A&=0.2000\times0.03500\\ &=0.007000\ \mathrm{mol} \end{aligned} \]

The sulfuric acid titre is:

\[ \begin{aligned} n_T&=0.1000\times0.01500\\ &=0.001500\ \mathrm{mol} \end{aligned} \]

Using the \(1:2\) sulfuric acid-to-sodium hydroxide ratio:

\[ \begin{aligned} n_L&=2(0.001500)\\ &=0.003000\ \mathrm{mol} \end{aligned} \]

Therefore, the sodium hydroxide consumed by the sample is:

\[ \begin{aligned} n_C&=0.007000-0.003000\\ &=0.004000\ \mathrm{mol} \end{aligned} \]

Now account for both salts.

Let:

• \(x\) be the amount of ammonium chloride, in mol.
• \(y\) be the amount of ammonium sulfate, in mol.

Ammonium chloride consumes one mole of sodium hydroxide per mole of salt.

Ammonium sulfate contains two ammonium ions and consumes two moles of sodium hydroxide per mole of salt.

Therefore:

\[ x+2y=0.004000 \]

This is our first equation.

The second equation comes from the sample’s total mass.

Using the supplied molar masses:

\[ 53.49x+132.14y=0.23912 \]

The molar-mass coefficients are in \(\mathrm{g\,mol^{-1}}\), so each term is a mass in g.

Rearrange the first equation:

\[ x=0.004000-2y \]

Substitute into the mass equation:

\[ \begin{aligned} 53.49(0.004000-2y)&\\ +132.14y&=0.23912\\ 0.21396+25.16y&=0.23912\\ 25.16y&=0.02516\\ y&=0.001000\ \mathrm{mol} \end{aligned} \]

Then:

\[ \begin{aligned} x&=0.004000-2(0.001000)\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

Calculate the masses:

\[ \begin{aligned} m(\ce{NH4Cl}) &=0.002000\times53.49\\ &=0.10698\ \mathrm{g} \end{aligned} \]

\[ \begin{aligned} m(\ce{(NH4)2SO4}) &=0.001000\times132.14\\ &=0.13214\ \mathrm{g} \end{aligned} \]

The masses add to 0.23912 g, matching the original sample.

A tempting approach would be to divide the total sodium hydroxide consumption by a single coefficient and treat the result as the total moles of fertiliser salts.

That fails because the two salts have different amounts of ammonium per formula unit.

The titration measures 0.004000 mol of ammonium ions in total. By itself, that number cannot distinguish one salt from the other.

The known sample mass provides an additional constraint, allowing the two amounts to be determined.

If an unknown mass of inert filler were also present, there would be three unknown amounts but only two independent equations. The individual salt masses would no longer be uniquely determined from these data.

Question 5

A fertiliser contains ammonium chloride and an inert filler.

A sample is treated with 0.01000 mol of excess sodium hydroxide. All the ammonium chloride reacts, and all ammonia is removed.

However, before titration, some of the leftover sodium hydroxide absorbs carbon dioxide from the air:

\[ \ce{CO2 + 2NaOH -> Na2CO3 + H2O} \]

The final solution contains free sodium hydroxide and sodium carbonate, \(\ce{Na2CO3}\). Assume no alkaline liquid is lost and there are no other acid-reacting substances.

The solution is titrated with 0.1000 \(\mathrm{mol\,L^{-1}}\) sulfuric acid while its pH is monitored.

Two cumulative titre readings are recorded:

At Endpoint A, all free hydroxide has been neutralised, and carbonate ions have been converted to hydrogen carbonate ions:

\[ \ce{CO3^2- + H+ -> HCO3-} \]

At Endpoint B, the hydrogen carbonate has also been completely acidified:

\[ \ce{HCO3- + H+ -> CO2 + H2O} \]

Treat these endpoint conversions as complete for the calculation.

A student uses Endpoint A, doubles the moles of sulfuric acid to find the leftover sodium hydroxide, and subtracts from the original amount.

Another student argues that Endpoint B must give the wrong result because carbon dioxide has already consumed some of the original sodium hydroxide.

Determine:

• The amounts of free sodium hydroxide and sodium carbonate present before titration.
• The correct amount and mass of ammonium chloride in the original sample.
• Which student’s reasoning is incorrect, and precisely why.
• Whether the correct result would still be guaranteed if some alkaline solution had been lost before titration.

Use \(M(\ce{NH4Cl})=53.49\ \mathrm{g\,mol^{-1}}\).

Solution 5

The original sample contained 0.006000 mol, or 0.32094 g, of ammonium chloride. Before titration, 0.002000 mol of free sodium hydroxide and 0.001000 mol of sodium carbonate were present.

Endpoint B provides the correct acid-equivalent balance under the stated assumptions. Endpoint A undercounts the basic capacity still present in the carbonate-containing solution.

The surprising point is that sodium hydroxide can disappear as a chemical species without losing all its ability to neutralise acid.

Step 1: Calculate the acid equivalents at both endpoints.

At Endpoint A:

\[ \begin{aligned} n_T&=0.1000\times0.01500\\ &=0.001500\ \mathrm{mol} \end{aligned} \]

Each mole of sulfuric acid can supply two moles of hydrogen ions for neutralisation.

Therefore, Endpoint A represents:

\[ 2(0.001500)=0.003000\ \mathrm{mol} \]

of hydrogen-ion equivalents.

At Endpoint B:

\[ \begin{aligned} n_T&=0.1000\times0.02000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

The total hydrogen-ion equivalents are:

\[ 2(0.002000)=0.004000\ \mathrm{mol} \]

These are cumulative quantities. The extra acid between the two endpoints is not a separate titre to add to the second reading.

Step 2: Separate the two substances that consume acid.

Let:

• \(b\) be the amount of free sodium hydroxide, in mol.
• \(c\) be the amount of sodium carbonate, in mol.

At Endpoint A, the acid neutralises all free hydroxide and supplies one hydrogen ion per carbonate ion to form hydrogen carbonate.

Therefore:

\[ b+c=0.003000 \]

At Endpoint B, each carbonate ion has consumed two hydrogen ions in total.

Therefore:

\[ b+2c=0.004000 \]

Subtract the first equation from the second:

\[ \begin{aligned} c&=0.004000-0.003000\\ &=0.001000\ \mathrm{mol} \end{aligned} \]

Substitute back:

\[ \begin{aligned} b&=0.003000-0.001000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

The final reaction solution therefore contains 0.002000 mol of free sodium hydroxide and 0.001000 mol of sodium carbonate.

Step 3: Account for where the original sodium hydroxide went.

This is where the ordinary back-titration subtraction needs more careful interpretation.

The original sodium hydroxide had three possible destinations:

1. It reacted with the ammonium chloride.
2. It remained as free sodium hydroxide.
3. It reacted with carbon dioxide to form sodium carbonate.

The carbon dioxide reaction is:

\[ \ce{CO2 + 2NaOH -> Na2CO3 + H2O} \]

Each mole of sodium carbonate represents two moles of the original sodium hydroxide.

Therefore, the amount of original sodium hydroxide converted into carbonate is:

\[ \begin{aligned} n_{\mathrm{carbonate\ use}} &=2(0.001000)\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

The ammonium chloride must have consumed:

\[ \begin{aligned} n_C&=0.01000-0.002000\\ &\quad-0.002000\\ &=0.006000\ \mathrm{mol} \end{aligned} \]

Because ammonium chloride consumes sodium hydroxide in a \(1:1\) ratio:

\[ n(\ce{NH4Cl})=0.006000\ \mathrm{mol} \]

Its mass is:

\[ \begin{aligned} m&=0.006000\times53.49\\ &=0.32094\ \mathrm{g} \end{aligned} \]

Step 4: Explain why Endpoint B works.

At first, carbon dioxide appears to have caused a problem. It has consumed 0.002000 mol of sodium hydroxide that the fertiliser didn’t consume.

You might expect this to make the ammonium chloride content appear too high.

But consider what happens to the carbonate during the full acid titration.

The carbonate formed from two moles of sodium hydroxide can consume two moles of hydrogen ions when it is completely acidified.

The sodium hydroxide molecules have disappeared, but their two acid-neutralising equivalents are now represented by the carbonate ion.

At Endpoint B, the acid neutralises:

• 0.002000 mol of free hydroxide.
• Two hydrogen-ion equivalents for each of the 0.001000 mol of carbonate.

The total is:

\[ \begin{aligned} n_{\mathrm{equiv}} &=0.002000+2(0.001000)\\ &=0.004000\ \mathrm{mol} \end{aligned} \]

So the final endpoint measures the same number of base equivalents as the original sodium hydroxide that was not consumed by ammonium chloride.

Subtracting that from the initial sodium hydroxide gives:

\[ 0.01000-0.004000 =0.006000\ \mathrm{mol} \]

This works because the carbonate remains in the solution and is fully accounted for by the acid titration.

The second student’s claim is therefore incorrect under these assumptions. Carbon dioxide consumption does not automatically cause an error in the final acid-equivalent balance.

Step 5: Explain why Endpoint A gives the wrong result.

The first student treats Endpoint A as though it accounts for all the leftover basic material.

The apparent consumption would be:

\[ \begin{aligned} n_{C,\mathrm{apparent}} &=0.01000-0.003000\\ &=0.007000\ \mathrm{mol} \end{aligned} \]

That overestimates the ammonium chloride by 0.001000 mol.

At Endpoint A, each carbonate ion has accepted only one hydrogen ion. Its remaining capacity to accept another hydrogen ion has not yet been counted.

This is why the usual rule, ‘convert titre to leftover reagent and subtract’, cannot be applied blindly when the leftover reagent has changed chemical form.

The titration endpoint determines what is actually being measured.

Step 6: Consider what happens if alkaline solution is lost.

The correct result is no longer guaranteed.

If some alkaline liquid is lost before titration, some of the original sodium hydroxide or carbonate is no longer present to react with the sulfuric acid.

The final titre would then underrepresent the original base equivalents not consumed by the fertiliser.

Subtracting it from the initial amount could falsely attribute the lost base to ammonium chloride.

The central principle is conservation of amount, not merely the appearance of a balanced equation.

The experiment can only account for substances that remain in the measured system, or whose losses are independently known.

06The next decision: Which chemical quantity does the titre measure?

When solving a back titration, don’t start by deciding whether to multiply or divide by two.

Start by identifying what the burette actually measures.

For the ammonium chloride experiment:

• The initial reagent is sodium hydroxide.
• Sulfuric acid measures the sodium hydroxide left over.
• Subtraction gives the sodium hydroxide consumed by the sample.
• The sample reaction converts that consumption into moles of ammonium chloride.

Those conversions are connected, but each has its own chemical justification.

For an acid-first version of the same approach, see Back Titration of an Antacid: Calculating Calcium Carbonate. The direction of the excess reagent changes, but the separation between added, leftover, and consumed amounts remains the same.

Once you’ve identified the correct two reaction ratios, the next useful check is whether the calculated sample composition is physically possible. The method in How to Check a Back-Titration Answer Against Sample Mass helps catch results that satisfy the arithmetic but contradict the amount of material originally weighed.