How to Calculate Antacid Neutralising Capacity from Excess Acid

Two antacid tablets are dropped into equal volumes of hydrochloric acid. One fizzes vigorously. The other barely bubbles. Which tablet neutralises more acid?

You can’t tell from the bubbles. To find an antacid’s neutralising capacity, add a known excess of hydrochloric acid (HCl), measure how much acid remains using sodium hydroxide (NaOH), and subtract:

\[
\boxed{\text{Acid consumed}=\text{Acid added}-\text{Acid left}}
\]

The result tells you how many moles of HCl the antacid neutralised. You can report it as millimoles of HCl per tablet or per gram of antacid.

Here’s a useful prediction. If two tablets receive exactly the same amount of HCl, which one should need more NaOH in the final titration?

The tablet that neutralised less acid. It leaves more HCl behind, so more NaOH is needed to neutralise that leftover acid.

That reversal is the key to interpreting excess-acid data. The NaOH titre measures acid remaining, not acid consumed by the tablet.

01Why measure leftover acid instead of titrating the antacid directly?

Suppose you want to find the neutralising capacity of a crushed antacid tablet.

One approach would be to add HCl gradually to the solid while monitoring pH. You might expect a clear endpoint when the tablet has neutralised all the acid it can.

The problem is that antacid solids don’t necessarily react instantly.

Calcium carbonate, a common antacid ingredient, reacts with hydrochloric acid:

\[
\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}
\]

The solid gradually reacts, and carbon dioxide bubbles escape. Some particles may remain while the solution’s pH is changing.

If you stop adding acid at a particular indicator colour, you may not know whether the active solid has completely reacted. The apparent endpoint can depend on mixing, reaction time, and dissolved carbon dioxide.

A more useful experimental design is to add more HCl than the tablet is expected to consume, allow the reaction to finish, and measure the excess HCl.

This is called a back titration.

Follow the acid through the apparatus

The procedure follows a clear sequence.

1. Weigh the antacid sample and place it in a conical flask.
2. Add an accurately measured volume of HCl of known concentration, using a suitable volumetric pipette or other calibrated apparatus.
3. Swirl and allow the antacid to react completely. HCl must still be present in excess.
4. Deal appropriately with carbon dioxide produced by carbonate ingredients, so dissolved carbon dioxide does not distort the remaining-acid measurement.
5. Titrate the remaining HCl using NaOH of known concentration and a suitable endpoint method.
6. Calculate the HCl remaining from the NaOH titre, then subtract it from the HCl originally added.

The titration reaction is:

\[
\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}
\]

Its mole ratio is \(1:1\). One mole of NaOH neutralises one mole of HCl.

That is why the amount of NaOH delivered from the burette can tell us the amount of HCl remaining.

The method depends on the endpoint measuring the residual strong acid accurately. Carbonate-containing samples require particular care because dissolved carbon dioxide can also react with NaOH. We’ll examine that limitation later.

02Separate the three amounts before calculating

The most reliable way to approach excess-acid data is to treat the acid as three different quantities:

• Added acid: the total HCl originally delivered to the sample.
• Leftover acid: the HCl still present after the sample has reacted.
• Consumed acid: the HCl used up by the sample.

These amounts are connected, but they aren’t interchangeable.

Consider this illustrative acid balance:

The capacity is 9.30 mmol HCl for that sample, not 3.20 mmol.

The 3.20 mmol is the acid the tablet didn’t neutralise.

The calculation method

Use:

\[
n=cV
\]

where:

• \(n\) is the amount of substance in moles (mol).
• \(c\) is the concentration in moles per litre (\(\mathrm{mol\,L^{-1}}\)).
• \(V\) is the solution volume in litres (L).

Let \(n_a\) be the initial amount of HCl and \(n_l\) be the amount left after the antacid reacts.

Let \(c_a\) and \(V_a\) represent the concentration and volume of HCl added. Let \(c_b\) and \(V_b\) represent the concentration and titre volume of NaOH.

Assuming the entire remaining acid solution is titrated and the NaOH measures only the leftover HCl:

\[
\begin{aligned}
n_a &= c_aV_a\\
n_l &= c_bV_b\\
q &= n_a-n_l
\end{aligned}
\]

Here, \(q\) is the amount of HCl consumed by the antacid, in mol.

To express this as millimoles per gram:

\[
C_m=\frac{1000q}{m}
\]

where \(C_m\) is neutralising capacity in \(\mathrm{mmol\,g^{-1}}\), and \(m\) is the sample mass in grams.

Multiplying by 1000 converts mol into mmol.

One essential condition: both acid amounts in the subtraction must refer to the entire sample reaction. If only an aliquot is titrated, the remaining acid must first be scaled to the whole reaction mixture.

04Why a larger NaOH titre doesn’t always mean a weaker antacid

There’s a useful shortcut when every sample receives the same amount of acid: the smaller the NaOH titre, the greater the acid consumption.

But suppose the samples receive different amounts of HCl.

Now the titre by itself isn’t enough. You must calculate each acid balance separately.

05Neutralising capacity is not the same as ingredient mass

An antacid may contain calcium carbonate, magnesium hydroxide, or more than one acid-neutralising ingredient.

Both calcium carbonate and magnesium hydroxide react with two moles of HCl per mole of active compound:

\[
\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}
\]

\[
\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}
\]

But their molar masses are different:

One mole of either substance neutralises two moles of HCl, but one mole of magnesium hydroxide has a considerably smaller mass.

This means antacids can have different capacities per gram even if they have similar capacities per mole of active ingredient.

The gas is another clue, but it must be interpreted carefully.

Calcium carbonate produces carbon dioxide. Magnesium hydroxide doesn’t. So bubbling can help identify carbonate chemistry, but the intensity of bubbling does not reliably establish neutralising capacity.

A back titration measures total acid consumption. If multiple ingredients react with the HCl, it measures their combined effect under the experimental conditions.

This is why a result such as \(10.0\ \mathrm{mmol\ HCl}\) per tablet cannot automatically be converted into a mass of calcium carbonate.

You can express the result as a calcium carbonate equivalent, but that is not proof of the tablet’s actual calcium carbonate content.

For the separate task of determining calcium carbonate mass from a back titration, see Back Titration of an Antacid: Calculating Calcium Carbonate.

06What if the blank titration doesn’t agree with the acid you added?

A good experiment may include a blank: the same procedure carried out without an antacid sample.

The blank helps reveal whether handling, heating, transfer, or other parts of the method have altered the amount of acid available for measurement.

Suppose 16.00 mmol of HCl is originally dispensed into each flask.

After identical handling:

If the difference between the dispensed amount and the blank measurement is caused by the same reproducible acid loss in both flasks, the corrected neutralising capacity is:

\[
\begin{aligned}
q&=15.60-7.20\\
&=8.40\ \mathrm{mmol}
\end{aligned}
\]

Using the nominal 16.00 mmol instead would give 8.80 mmol, an overestimate of 0.40 mmol.

A blank is useful only if it controls the same relevant effects in the sample experiment. It cannot automatically correct for acid loss that occurs differently when the antacid is present.

There is also an important distinction between an acid-loss correction and a chemical interference. For example, dissolved carbon dioxide can consume some NaOH during a titration. A blank containing no carbonate may not reveal that problem.

07How to decide whether a capacity result is valid

Before accepting a calculated neutralising capacity, consider what the experiment actually measured.

For the idealised model, with HCl genuinely in excess:

\[
0<n_l\leq n_a
\]

The leftover acid must be greater than zero and cannot exceed the original acid amount.

If your calculation gives a negative acid consumption, don’t report a negative neutralising capacity. Recheck units, sample fractions, concentrations, and possible interfering reactions.

Also remember that neutralising capacity is measured under particular conditions. A result obtained after two minutes may differ from one obtained after thirty minutes if the solid reacts slowly.

A laboratory capacity measurement doesn’t, by itself, establish how quickly a product works in the stomach or its clinical effectiveness.

08Questions and solutions

The questions below use original, illustrative data rather than measurements from commercial products. They progress from the basic acid balance to interpreting competing measurements and experimental limitations.

Question 1

A 1.10 g antacid tablet is reacted completely with 20.00 mL of \(0.600\ \mathrm{mol\,L^{-1}}\) HCl.

The remaining HCl requires 21.00 mL of \(0.200\ \mathrm{mol\,L^{-1}}\) NaOH for neutralisation.

Calculate the tablet’s neutralising capacity in mmol HCl per tablet and mmol HCl per gram.

Assume that HCl remains in excess, the whole reaction solution is titrated, and no other substances consume NaOH.

Solution 1

The tablet neutralises \(7.80\ \mathrm{mmol}\) of HCl, equivalent to \(7.09\ \mathrm{mmol\,g^{-1}}\).

First calculate the HCl added:

\[
\begin{aligned}
n_a&=cV\\
&=0.600(0.02000)\\
&=0.01200\ \mathrm{mol}\\
&=12.00\ \mathrm{mmol}
\end{aligned}
\]

Now calculate NaOH used:

\[
\begin{aligned}
n(\ce{NaOH})&=0.200(0.02100)\\
&=0.004200\ \mathrm{mol}\\
&=4.200\ \mathrm{mmol}
\end{aligned}
\]

Because the NaOH:HCl mole ratio is \(1:1\), the remaining HCl is 4.200 mmol.

Subtract:

\[
\begin{aligned}
q&=12.00-4.200\\
&=7.800\ \mathrm{mmol}
\end{aligned}
\]

Divide by the tablet mass:

\[
C_m=\frac{7.800}{1.10}
=7.09\ \mathrm{mmol\,g^{-1}}
\]

The important distinction is that the 4.200 mmol measured by NaOH is the acid remaining. The 7.800 mmol difference is the acid consumed by the tablet.

Question 2

Two antacid tablets, P and Q, are tested using separate excess-HCl back titrations.

A matched blank is prepared for each tablet under identical conditions, except that the blank contains no antacid.

All leftover acid is titrated with \(0.200\ \mathrm{mol\,L^{-1}}\) NaOH.

Assume each blank accurately represents the acid remaining after the same reproducible handling losses experienced by its tablet sample. The antacid reactions are complete, and no titration interferences occur.

Determine:

a) The corrected capacity of each tablet in mmol HCl.

b) Which tablet neutralises more acid per tablet and which neutralises more per gram.

c) How using the original dispensed HCl amounts instead of the blanks would affect the results.

d) Why the larger tablet-sample titre alone does not establish which tablet has the greater neutralising capacity.

Solution 2

Tablet Q has the greater corrected capacity per tablet, at \(10.40\ \mathrm{mmol}\), but tablet P has the greater capacity per gram, at \(10.5\ \mathrm{mmol\,g^{-1}}\). Using the nominal acid additions would overestimate both capacities by 0.50 mmol.

Part a: Corrected capacities

The blank measures the acid available after handling.

For P:

\[
\begin{aligned}
n_{\rm blank,P}
&=0.200(0.06000)\\
&=12.00\ \mathrm{mmol}\\[4pt]
n_{\rm sample,P}
&=0.200(0.01800)\\
&=3.600\ \mathrm{mmol}
\end{aligned}
\]

Therefore:

\[
q_P=12.00-3.600
=8.400\ \mathrm{mmol}
\]

For Q:

\[
\begin{aligned}
n_{\rm blank,Q}
&=0.200(0.07500)\\
&=15.00\ \mathrm{mmol}\\[4pt]
n_{\rm sample,Q}
&=0.200(0.02300)\\
&=4.600\ \mathrm{mmol}
\end{aligned}
\]

Therefore:

\[
q_Q=15.00-4.600
=10.400\ \mathrm{mmol}
\]

Part b: Compare per tablet and per gram

Per tablet, Q has the larger capacity because \(10.40>8.400\).

Per gram:

\[
\begin{aligned}
C_{m,P}&=\frac{8.400}{0.800}\\
&=10.5\ \mathrm{mmol\,g^{-1}}\\[4pt]
C_{m,Q}&=\frac{10.400}{1.300}\\
&=8.00\ \mathrm{mmol\,g^{-1}}
\end{aligned}
\]

P is more effective per gram under these experimental conditions, while Q provides greater total capacity per tablet.

Part c: Effect of ignoring the blanks

Using the dispensed acid amounts instead gives:

\[
\begin{aligned}
q_{P,\rm nominal}&=12.50-3.600\\
&=8.900\ \mathrm{mmol}\\[4pt]
q_{Q,\rm nominal}&=15.50-4.600\\
&=10.900\ \mathrm{mmol}
\end{aligned}
\]

Both exceed the corrected results by 0.500 mmol.

The blank data show that the available acid is 0.500 mmol less than the nominal addition in each test. Under the stated matched-loss assumption, attributing that missing acid to the antacid would overstate its capacity.

Part d: Why raw titres mislead

Q requires more NaOH because more acid remains in its sample flask. But Q also started with a larger acid dose.

Comparing 18.00 mL with 23.00 mL ignores the different starting amounts. The relevant calculation is the acid lost between each matching blank and tablet sample.

Question 3

An antacid tablet has a total mass of 0.500 g and contains only calcium carbonate, magnesium hydroxide, and inert binder.

A back titration with excess HCl gives a neutralising capacity of 12.0 mmol HCl per tablet.

In a separate experiment on an identical tablet, all the carbon dioxide produced by the reaction is collected in a gas syringe. The measured gas volume is 48.0 mL.

At the experimental conditions, assume the molar gas volume is \(24.0\ \mathrm{L\,mol^{-1}}\).

Use:

• \(M(\ce{CaCO3})=100.09\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Mg(OH)2})=58.32\ \mathrm{g\,mol^{-1}}\)

Assume the reactions go to completion and that calcium carbonate is the only gas-producing ingredient.

a) Determine the mass of each active ingredient and the mass of inert binder.

b) Suppose the gas-collection system actually loses 25% of the carbon dioxide before measurement, but the acid-capacity experiment remains accurate. Explain how this affects the inferred calcium carbonate, magnesium hydroxide, and binder masses.

c) Does the gas loss change the neutralising capacity measured by the separate back titration?

Solution 3

The accurate data imply approximately 0.200 g of calcium carbonate, 0.233 g of magnesium hydroxide, and 0.0665 g of inert binder. Losing 25% of the carbon dioxide would underestimate calcium carbonate, overestimate magnesium hydroxide and binder, but leave the independently measured acid capacity unchanged.

Part a: Determine the ingredients

First convert the gas volume:

\[
48.0\ \mathrm{mL}=0.0480\ \mathrm{L}
\]

Then:

\[
\begin{aligned}
n(\ce{CO2})&=\frac{V}{V_m}\\
&=\frac{0.0480}{24.0}\\
&=0.00200\ \mathrm{mol}
\end{aligned}
\]

Here, \(V_m\) is the given molar gas volume.

Calcium carbonate produces carbon dioxide in a \(1:1\) mole ratio:

\[
n(\ce{CaCO3})=0.00200\ \mathrm{mol}
\]

Its mass is:

\[
\begin{aligned}
m(\ce{CaCO3})&=nM\\
&=0.00200(100.09)\\
&=0.20018\ \mathrm{g}
\end{aligned}
\]

This carbonate consumes twice its amount in HCl:

\[
q_{\ce{CaCO3}}=4.00\ \mathrm{mmol}
\]

The remaining acid consumption must come from magnesium hydroxide:

\[
q_{\ce{Mg(OH)2}}=12.0-4.00
=8.00\ \mathrm{mmol}
\]

Since each mole of magnesium hydroxide consumes two moles of HCl:

\[
n(\ce{Mg(OH)2})=\frac{8.00}{2}
=4.00\ \mathrm{mmol}
\]

Its mass is:

\[
\begin{aligned}
m(\ce{Mg(OH)2})
&=0.00400(58.32)\\
&=0.23328\ \mathrm{g}
\end{aligned}
\]

The remaining tablet mass is inert binder:

\[
\begin{aligned}
m_{\rm binder}
&=0.500-0.20018-0.23328\\
&=0.06654\ \mathrm{g}
\end{aligned}
\]

So the binder mass is approximately 0.0665 g.

Part b: Effect of missing carbon dioxide

If only 75% of the gas is collected, the observed amount is:

\[
n_{\rm measured}(\ce{CO2})
=0.75(2.00)
=1.50\ \mathrm{mmol}
\]

The analyst would incorrectly infer only 1.50 mmol of calcium carbonate.

That carbonate would account for 3.00 mmol HCl, leaving:

\[
12.0-3.00=9.00\ \mathrm{mmol}
\]

attributed to magnesium hydroxide.

The inferred magnesium hydroxide amount would be:

\[
n(\ce{Mg(OH)2})=\frac{9.00}{2}
=4.50\ \mathrm{mmol}
\]

The incorrect ingredient masses would therefore be:

\[
\begin{aligned}
m(\ce{CaCO3})&=0.1501\ \mathrm{g}\\
m(\ce{Mg(OH)2})&=0.2624\ \mathrm{g}
\end{aligned}
\]

The inferred binder mass would be:

\[
0.500-0.1501-0.2624
=0.0875\ \mathrm{g}
\]

approximately, depending on rounding.

The inferred binder is too large because some of the heavier calcium carbonate has been incorrectly replaced in the calculation by an equal number of moles of lighter magnesium hydroxide.

Part c: Effect on capacity

The neutralising capacity remains \(12.0\ \mathrm{mmol}\) HCl per tablet.

The gas-collection error changes the inferred composition, not the result of the separate, accurate back titration.

Both active compounds consume two moles of HCl per mole, so the total acid consumption can be measured correctly even while the division between ingredients is wrong.

This is why neutralising capacity and composition must not be treated as the same measurement.

Question 4

A researcher is trying to determine the neutralising capacity of a slowly reacting antacid formulation.

Identical 0.950 g tablets are tested separately, using different initial amounts of HCl and different reaction times.

The measured residual HCl values are:

Assume the measurements are accurate, there are no acid losses or titration interferences, and only the antacid consumes HCl.

a) Calculate the acid consumed in each trial.

b) A student argues that trials B and C prove the antacid’s maximum capacity is 15.0 mmol because both give the same result. Explain why this conclusion fails.

c) State the best-supported capacity after 90 minutes and explain what the measurements do and do not establish about the tablet’s ultimate capacity.

d) Predict the HCl remaining if an identical tablet is treated with 30.0 mmol HCl for 90 minutes, assuming the reaction is complete by that time and acid consumption is independent of the initial excess.

Solution 4

Trials D and E support a measured capacity of \(18.0\ \mathrm{mmol}\) HCl after 90 minutes, but the data do not prove that the tablet could never neutralise more acid under longer reaction conditions. If complete reaction occurs by 90 minutes, the predicted HCl remaining from a 30.0 mmol dose is 12.0 mmol.

Part a: Acid consumed

Use \(q=n_a-n_l\) for each trial.

Trial A has no measured HCl remaining, so it does not satisfy the requirement that acid remain in excess. Its 12.0 mmol result measures all the acid supplied, not necessarily the tablet’s full capacity.

Part b: Why the apparent 15.0 mmol plateau is misleading

Trials B and C both have excess acid after 10 minutes, and both show that 15.0 mmol was consumed.

It is tempting to conclude that adding more acid cannot increase the consumption, so the reaction must be complete.

But trial D disproves that conclusion.

The same tablet formulation consumes 18.0 mmol after 45 minutes using 24.0 mmol of initial HCl.

The missing condition in B and C was sufficient reaction time.

Adding more acid does not guarantee that a slow solid has finished reacting. A consistent result at one fixed reaction time can be a kinetic limitation rather than the maximum chemical capacity.

Part c: Best-supported capacity

The best-supported result is:

\[
q=24.0-6.00
=18.0\ \mathrm{mmol}
\]

The 45-minute and 90-minute results agree, supporting the conclusion that the reaction may have reached completion by 45 minutes.

However, matching results at two times do not logically exclude an even slower later reaction.

To call 18.0 mmol the ultimate capacity, we need the additional assumption or independent evidence that all acid-consuming ingredients have reacted completely.

Part d: Prediction with 30.0 mmol HCl

Under the stated complete-reaction assumption, the tablet consumes 18.0 mmol regardless of the larger initial excess.

\[
\begin{aligned}
n_l&=n_a-q\\
&=30.0-18.0\\
&=12.0\ \mathrm{mmol}
\end{aligned}
\]

The larger acid dose would increase the acid remaining, not the tablet’s capacity.

This distinction matters because adding acid until some remains is necessary for back titration, but it is not sufficient to prove that the antacid reaction is complete.

Question 5

An antacid tablet of mass 0.800 g contains only calcium carbonate, magnesium hydroxide, and inert binder.

Two identical tablets are tested separately. Each receives 25.00 mL of \(0.800\ \mathrm{mol\,L^{-1}}\) HCl.

Both react completely, HCl remains in excess, and no HCl is lost from either experiment.

The entire remaining solution in each experiment is diluted to 250 mL for titration, without changing the amount of HCl. No aliquots are taken.

The two experiments differ in how they handle carbon dioxide.

Trial D – carbon dioxide removed

All carbon dioxide is removed using a validated procedure that does not change the HCl amount.

The remaining acid requires 30.00 mL of \(0.200\ \mathrm{mol\,L^{-1}}\) NaOH.

Trial R – carbon dioxide retained

Some carbon dioxide remains dissolved in the liquid. Gas that leaves the liquid is collected in a syringe.

The collected gas contains 1.20 mmol of carbon dioxide. All other carbon dioxide produced remains dissolved in the solution.

The titration requires 39.00 mL of \(0.200\ \mathrm{mol\,L^{-1}}\) NaOH.

For this idealised question, assume the chosen titration endpoint causes each mole of dissolved carbon dioxide to consume exactly one additional mole of NaOH:

\[
\ce{CO2(aq) + OH^-(aq) -> HCO3^-(aq)}
\]

All other titration interferences are negligible.

Use:

• \(M(\ce{CaCO3})=100.09\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Mg(OH)2})=58.32\ \mathrm{g\,mol^{-1}}\)

Determine:

a) The true neutralising capacity per tablet.

b) The amount of dissolved carbon dioxide in Trial R and the total amount of calcium carbonate per tablet.

c) The amount of magnesium hydroxide and mass of inert binder.

d) Why calculating capacity directly from the Trial R titre gives a different result.

e) If Trial D had never been performed, could the Trial R titre and collected gas measurement alone uniquely determine the true neutralising capacity? Support your conclusion with an alternative possible tablet composition.

Solution 5

The true capacity is \(14.0\ \mathrm{mmol}\) HCl per tablet. The composition is 3.00 mmol calcium carbonate, 4.00 mmol magnesium hydroxide, and approximately 0.266 g inert binder. Trial R gives an incorrect capacity if all its NaOH consumption is attributed to leftover HCl, because dissolved carbon dioxide also consumes NaOH.

More importantly, without Trial D, the remaining measurements do not uniquely establish the capacity.

Part a: True neutralising capacity

Calculate the initial HCl:

\[
\begin{aligned}
n_a&=0.800(0.02500)\\
&=0.02000\ \mathrm{mol}\\
&=20.00\ \mathrm{mmol}
\end{aligned}
\]

In Trial D, carbon dioxide has been removed without changing the HCl amount.

The 30.00 mL NaOH titre therefore measures only remaining HCl:

\[
\begin{aligned}
n_l&=0.200(0.03000)\\
&=0.006000\ \mathrm{mol}\\
&=6.000\ \mathrm{mmol}
\end{aligned}
\]

So the true capacity is:

\[
\begin{aligned}
q&=20.00-6.000\\
&=14.00\ \mathrm{mmol}
\end{aligned}
\]

Part b: Determine the carbon dioxide and calcium carbonate

Trial R uses:

\[
\begin{aligned}
n(\ce{NaOH})&=0.200(0.03900)\\
&=0.007800\ \mathrm{mol}\\
&=7.800\ \mathrm{mmol}
\end{aligned}
\]

Of that, 6.000 mmol is needed to neutralise the leftover HCl.

The additional NaOH must have reacted with dissolved carbon dioxide under the supplied endpoint model:

\[
\begin{aligned}
n(\ce{CO2})_{\rm dissolved}
&=7.800-6.000\\
&=1.800\ \mathrm{mmol}
\end{aligned}
\]

But the syringe also collected 1.20 mmol carbon dioxide.

So total carbon dioxide produced was:

\[
\begin{aligned}
n(\ce{CO2})_{\rm total}
&=1.800+1.20\\
&=3.00\ \mathrm{mmol}
\end{aligned}
\]

Each mole of calcium carbonate produces one mole of carbon dioxide:

\[
n(\ce{CaCO3})=3.00\ \mathrm{mmol}
\]

Part c: Magnesium hydroxide and binder

The calcium carbonate consumes two moles of HCl per mole:

\[
q_{\ce{CaCO3}}=2(3.00)
=6.00\ \mathrm{mmol}
\]

The tablet neutralised 14.00 mmol HCl in total.

Therefore, magnesium hydroxide accounts for:

\[
\begin{aligned}
q_{\ce{Mg(OH)2}}
&=14.00-6.00\\
&=8.00\ \mathrm{mmol}
\end{aligned}
\]

It also consumes two moles of HCl per mole:

\[
n(\ce{Mg(OH)2})=\frac{8.00}{2}
=4.00\ \mathrm{mmol}
\]

Convert both ingredient amounts to mass:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.00300(100.09)\\
&=0.30027\ \mathrm{g}\\[4pt]
m(\ce{Mg(OH)2})
&=0.00400(58.32)\\
&=0.23328\ \mathrm{g}
\end{aligned}
\]

Then:

\[
\begin{aligned}
m_{\rm binder}
&=0.800-0.30027-0.23328\\
&=0.26645\ \mathrm{g}
\end{aligned}
\]

The binder mass is approximately 0.266 g.

Part d: Why Trial R gives an incorrect capacity

If we incorrectly assume all 7.800 mmol NaOH neutralised HCl, we get:

\[
\begin{aligned}
q_{\rm apparent}
&=20.00-7.800\\
&=12.200\ \mathrm{mmol}
\end{aligned}
\]

That result is 1.800 mmol below the true capacity.

The usual \(1:1\) HCl:NaOH calculation assumes that the NaOH titre measures leftover HCl only.

Trial R breaks this assumption.

The NaOH neutralises leftover HCl and reacts with dissolved carbon dioxide. Consequently, the larger titre does not mean that the tablet consumed less HCl.

This is a chemical interference, not a change in the tablet’s true neutralising capacity.

Part e: Could Trial R alone determine capacity?

No. The amount of dissolved carbon dioxide would be unknown, so the Trial R titre could not be divided uniquely between leftover HCl and carbon dioxide.

To see why, consider a different hypothetical 0.800 g tablet containing:

• 5.00 mmol calcium carbonate.
• 3.00 mmol magnesium hydroxide.
• The remaining mass as inert binder.

Its total acid consumption would be:

\[
q=2(5.00)+2(3.00)
=16.00\ \mathrm{mmol}
\]

It would leave:

\[
20.00-16.00=4.00\ \mathrm{mmol}
\]

of HCl.

Its calcium carbonate would produce 5.00 mmol carbon dioxide. If the syringe collected the same 1.20 mmol, the amount remaining dissolved would be:

\[
5.00-1.20=3.80\ \mathrm{mmol}
\]

At the stated endpoint, the NaOH requirement would be:

\[
4.00+3.80=7.80\ \mathrm{mmol}
\]

This is exactly the same Trial R titre as the original tablet, even though the true capacities are different.

The alternative composition also fits the 0.800 g tablet mass:

\[
\begin{aligned}
m_{\rm active}
&=0.00500(100.09)\\
&\quad+0.00300(58.32)\\
&=0.67541\ \mathrm{g}
\end{aligned}
\]

The remaining 0.125 g is inert binder, so there is no mass contradiction.

Thus two chemically different tablets could produce the same uncorrected NaOH titre and collected gas amount.

Trial D supplies the missing independent measurement: the actual leftover HCl.

That breaks the ambiguity and makes the capacity and composition calculations possible.

In a real titration, the exact NaOH consumption caused by dissolved carbon dioxide depends on the endpoint and acid-base equilibria. The specified \(1:1\) conversion in this question is an idealisation, not a universal correction to apply to every carbonate titration.

09What to do with an excess-acid result

When reporting an antacid’s neutralising capacity, make clear what is being measured.

A useful result gives the acid consumed in mmol HCl, the amount or mass of sample tested, and enough experimental conditions to make the comparison meaningful. For example, reaction time and endpoint method can affect what was actually measured.

The most important decision is always which amount belongs in the subtraction. The acid added belongs to the whole reaction, and the leftover acid must also represent the whole reaction. NaOH used in the final titration is simply the measurement that lets you calculate the leftover amount.

The next complication is often that only part of the reacted solution is titrated. In that situation, the acid balance stays the same, but you must reconstruct the total leftover HCl before subtracting.

For that specific calculation, see How to Correct for Diluted Aliquots in Back Titration.

Once the acid balance is correct, interpreting neutralising capacity becomes a question of chemistry rather than bookkeeping: which substances consumed the acid, how completely did they react, and what does the measured result actually establish?