Back Titration of an Antacid: Calculating Calcium Carbonate

Imagine crushing an antacid tablet, adding a measured excess of hydrochloric acid, and then using sodium hydroxide to measure the acid left over. If one sample requires more sodium hydroxide than another, which sample contained more calcium carbonate? Predict before reading on: if both samples started with the same amount of hydrochloric acid, the sample requiring less sodium hydroxide contained more calcium carbonate, provided calcium carbonate was the only ingredient consuming the acid. It used more of the original acid, leaving less behind.

01Why measure the acid left over?

Calcium carbonate, \(\ce{CaCO3}\), is a solid. Rather than trying to measure its amount directly, you can react it with a known excess of hydrochloric acid, \(\ce{HCl}\), and then measure how much acid remains.

This is called a back titration. You measure what is left after a reaction, then use the difference to calculate what reacted.

The logic is simple:

• Start with a known amount of hydrochloric acid.
• Allow the antacid to react with some of that acid.
• Use sodium hydroxide, \(\ce{NaOH}\), to measure the acid remaining.
• Subtract the leftover acid from the starting amount.
• Convert the acid consumed into the mass of calcium carbonate.

The important step is the subtraction. The sodium hydroxide titre does not directly tell you how much calcium carbonate was present. It tells you how much acid was left.

02Two reactions determine the calculation

First, calcium carbonate reacts with hydrochloric acid:

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

The reaction produces calcium chloride, carbon dioxide, and water.

Notice the mole ratio:

\[ 1\ \text{mol }\ce{CaCO3} : 2\ \text{mol }\ce{HCl} \]

Every mole of calcium carbonate consumes two moles of hydrochloric acid.

After the reaction, the hydrochloric acid remaining in solution is titrated with sodium hydroxide:

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

This reaction has a \(1:1\) mole ratio. One mole of sodium hydroxide neutralises one mole of hydrochloric acid.

That gives us two ratios to use:

• The titration tells us the moles of hydrochloric acid remaining.
• The carbonate reaction tells us that two moles of hydrochloric acid were consumed for every mole of calcium carbonate.

03The calculation in five steps

Let \(n\) represent amount in moles, \(c\) concentration in \(\mathrm{mol\,L^{-1}}\), \(V\) volume in litres, and \(M\) molar mass in \(\mathrm{g\,mol^{-1}}\).

Step 1: Calculate the initial moles of hydrochloric acid

Use the concentration and volume of the acid added to the antacid.

\[ n_{\text{initial}}(\ce{HCl}) =c_{\ce{HCl}}V_{\ce{HCl}} \]

Convert the volume from millilitres to litres before substituting.

Step 2: Calculate the moles of hydrochloric acid remaining

For a titration of the entire remaining acid solution, the \(1:1\) reaction ratio gives:

\[ n_{\text{remaining}}(\ce{HCl}) =c_{\ce{NaOH}}V_{\ce{NaOH}} \]

The sodium hydroxide concentration and titre give the moles of sodium hydroxide used, which equal the moles of hydrochloric acid remaining.

If you titrate only a portion of the solution, called an aliquot, you must scale this amount up to represent the entire solution. We will work through that shortly.

Step 3: Subtract to find the acid consumed

\[ \begin{aligned} n_{\text{reacted}}(\ce{HCl}) ={}&n_{\text{initial}}(\ce{HCl})\\ &-n_{\text{remaining}}(\ce{HCl}) \end{aligned} \]

Both quantities must refer to the same whole solution.

Step 4: Convert the acid consumed into moles of calcium carbonate

The reaction requires two moles of hydrochloric acid per mole of calcium carbonate.

\[ n(\ce{CaCO3}) =\frac{n_{\text{reacted}}(\ce{HCl})}{2} \]

Step 5: Convert moles into mass

The molar mass of calcium carbonate is approximately \(100.09\ \mathrm{g\,mol^{-1}}\).

\[ m(\ce{CaCO3})=n(\ce{CaCO3})M(\ce{CaCO3}) \]

The units check out:

\[ \mathrm{mol}\times \mathrm{g\,mol^{-1}}=\mathrm{g} \]

If the original antacid sample’s mass is known, you can also calculate the mass percentage of calcium carbonate.

06What can go wrong?

The subtraction method works only when the measured quantities represent the chemistry you think they do.

For the simplified calculations above, we assumed that:

• Hydrochloric acid was present in excess, so all the calcium carbonate reacted.
• Calcium carbonate was the only ingredient consuming hydrochloric acid.
• No hydrochloric acid was lost before the titration.
• Carbon dioxide did not contribute to the sodium hydroxide titre.
• Any aliquot was scaled correctly to represent the full solution.

Each assumption has a reason.

More than one ingredient can consume acid

Actual commercial antacid formulations may contain calcium carbonate, magnesium hydroxide, aluminium hydroxide, sodium hydrogen carbonate, or other ingredients that react with acid.

If more than one ingredient consumes hydrochloric acid, the back titration measures their combined acid-neutralising effect. It does not automatically reveal the mass of calcium carbonate alone.

For example, magnesium hydroxide also consumes two moles of hydrochloric acid per mole of magnesium hydroxide:

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

If both ingredients react completely, the total acid consumption depends on both amounts. Without additional composition data or an independent measurement, you cannot separate their individual contributions.

Carbon dioxide can interfere with the titration

The reaction between calcium carbonate and hydrochloric acid produces carbon dioxide. Some of this gas may remain dissolved in the solution.

Dissolved carbon dioxide can react with sodium hydroxide during the titration, consuming additional base. If you incorrectly treat every mole of sodium hydroxide as evidence of one mole of leftover hydrochloric acid, you overestimate the amount of acid remaining.

That means you underestimate the amount of hydrochloric acid consumed by the antacid, and consequently underestimate the calcium carbonate mass.

A suitable experimental procedure must account for this interference, for example by using a validated step to remove dissolved carbon dioxide without losing hydrochloric acid.

The general rule is to check what the titrant actually reacts with before assigning all of its measured moles to one substance.

07Questions and solutions

Question 1

A sample of antacid is treated with \(40.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid. The remaining acid requires \(18.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide for neutralisation.

Assuming calcium carbonate is the only ingredient that consumes acid, calculate the mass of calcium carbonate in the sample.

Use \(M(\ce{CaCO3})=100.09\ \mathrm{g\,mol^{-1}}\).

Solution 1

The sample contains \(0.1101\ \mathrm{g}\) of calcium carbonate. The calculation starts with the acid added and subtracts the acid remaining.

First, calculate the initial hydrochloric acid:

\[ \begin{aligned} n_{\text{initial}}(\ce{HCl}) &=0.1000\times0.04000\\ &=0.004000\ \mathrm{mol} \end{aligned} \]

Next, calculate the remaining hydrochloric acid from the sodium hydroxide titre:

\[ \begin{aligned} n_{\text{remaining}}(\ce{HCl}) &=0.1000\times0.01800\\ &=0.001800\ \mathrm{mol} \end{aligned} \]

The acid consumed is:

\[ \begin{aligned} n_{\text{reacted}}(\ce{HCl}) &=0.004000-0.001800\\ &=0.002200\ \mathrm{mol} \end{aligned} \]

Two moles of hydrochloric acid react with one mole of calcium carbonate:

\[ n(\ce{CaCO3})=\frac{0.002200}{2} =0.001100\ \mathrm{mol} \]

Finally:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.001100\times100.09\\ &=0.1101\ \mathrm{g} \end{aligned} \]

The answer is \(0.1101\ \mathrm{g}\). The sodium hydroxide titre measures the leftover acid, so it must be subtracted before applying the \(2:1\) acid-to-carbonate ratio.

Question 2

A \(0.6000\ \mathrm{g}\) antacid sample is treated with \(50.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid. After the reaction, the solution is diluted to \(250.0\ \mathrm{mL}\).

A \(25.00\ \mathrm{mL}\) aliquot requires \(7.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

Assuming calcium carbonate is the only acid-consuming ingredient, calculate:

(a) the mass of calcium carbonate in the sample;

(b) the mass percentage of calcium carbonate.

Solution 2

The sample contains \(0.150\ \mathrm{g}\) of calcium carbonate, corresponding to \(25.0\%\) by mass. The essential step is scaling the aliquot result to the full solution.

(a) Mass of calcium carbonate

Initial moles of hydrochloric acid:

\[ \begin{aligned} n_{\text{initial}}(\ce{HCl}) &=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]

Moles of sodium hydroxide used for the aliquot:

\[ \begin{aligned} n(\ce{NaOH}) &=0.1000\times0.00700\\ &=0.000700\ \mathrm{mol} \end{aligned} \]

This is the amount of hydrochloric acid in the \(25.00\ \mathrm{mL}\) aliquot.

The total solution is ten times the aliquot volume:

\[ \frac{250.0}{25.00}=10.00 \]

Therefore, the total remaining acid is:

\[ n_{\text{remaining}}(\ce{HCl}) =0.000700\times10.00 =0.00700\ \mathrm{mol} \]

The acid consumed is:

\[ \begin{aligned} n_{\text{reacted}}(\ce{HCl}) &=0.01000-0.00700\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

Convert this to moles of calcium carbonate:

\[ n(\ce{CaCO3})=\frac{0.00300}{2} =0.00150\ \mathrm{mol} \]

Calculate the mass:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.00150\times100.09\\ &=0.150\ \mathrm{g} \end{aligned} \]

(b) Mass percentage

\[ \begin{aligned} \%\ce{CaCO3} &=\frac{0.150}{0.6000}\times100\\ &=25.0\% \end{aligned} \]

The aliquot contains only one-tenth of the solution’s total remaining acid. Forgetting the factor of ten would make the acid balance invalid.

Question 3

Two \(0.8000\ \mathrm{g}\) antacid samples, A and B, contain calcium carbonate as their only acid-consuming ingredient.

Sample A receives \(40.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid and requires \(16.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide to titrate the remaining acid.

Sample B receives \(30.00\ \mathrm{mL}\) of the same hydrochloric acid and requires \(8.00\ \mathrm{mL}\) of the same sodium hydroxide solution.

A student claims that B contains more calcium carbonate because it requires less sodium hydroxide.

Calculate the mass of calcium carbonate in each sample and evaluate the claim.

Solution 3

Sample A contains \(0.3203\ \mathrm{g}\) of calcium carbonate, while sample B contains \(0.2602\ \mathrm{g}\). Therefore, A contains \(0.0601\ \mathrm{g}\) more calcium carbonate, and the student’s claim is incorrect.

The trap is comparing the sodium hydroxide titres without accounting for the different initial amounts of hydrochloric acid.

Sample A

Initial acid:

\[ n_{\text{initial}}(\ce{HCl}) =0.2000\times0.04000 =0.008000\ \mathrm{mol} \]

Remaining acid:

\[ n_{\text{remaining}}(\ce{HCl}) =0.1000\times0.01600 =0.001600\ \mathrm{mol} \]

Acid consumed:

\[ 0.008000-0.001600 =0.006400\ \mathrm{mol} \]

Calcium carbonate mass:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.006400}{2}\\ &=0.003200\ \mathrm{mol} \end{aligned} \]

\[ m(\ce{CaCO3}) =0.003200\times100.09 =0.3203\ \mathrm{g} \]

Sample B

Initial acid:

\[ n_{\text{initial}}(\ce{HCl}) =0.2000\times0.03000 =0.006000\ \mathrm{mol} \]

Remaining acid:

\[ n_{\text{remaining}}(\ce{HCl}) =0.1000\times0.00800 =0.0008000\ \mathrm{mol} \]

Acid consumed:

\[ 0.006000-0.0008000 =0.005200\ \mathrm{mol} \]

Calcium carbonate mass:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.005200}{2}\\ &=0.002600\ \mathrm{mol} \end{aligned} \]

\[ m(\ce{CaCO3}) =0.002600\times100.09 =0.2602\ \mathrm{g} \]

The difference is:

\[ 0.3203-0.2602=0.0601\ \mathrm{g} \]

The smaller titre for B does not mean that B contains more calcium carbonate. B started with less hydrochloric acid, so the two titres cannot be compared directly.

Question 4

An antacid contains calcium carbonate, magnesium hydroxide, and inert filler. A \(0.5000\ \mathrm{g}\) sample is treated with \(40.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

The remaining acid requires \(18.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide for neutralisation.

Assume both active ingredients react completely, and each mole of calcium carbonate or magnesium hydroxide consumes two moles of hydrochloric acid.

(a) Calculate the total amount of hydrochloric acid consumed.

(b) Can the mass of calcium carbonate be determined uniquely from these data? Explain your reasoning and identify what additional information would be needed.

Solution 4

The sample consumes \(0.006200\ \mathrm{mol}\) of hydrochloric acid, but its calcium carbonate mass cannot be determined uniquely from the titration alone. Both active ingredients consume acid, so the result represents their combined contribution.

(a) Total acid consumed

Initial hydrochloric acid:

\[ \begin{aligned} n_{\text{initial}}(\ce{HCl}) &=0.2000\times0.04000\\ &=0.008000\ \mathrm{mol} \end{aligned} \]

Remaining hydrochloric acid:

\[ \begin{aligned} n_{\text{remaining}}(\ce{HCl}) &=0.1000\times0.01800\\ &=0.001800\ \mathrm{mol} \end{aligned} \]

Subtract:

\[ \begin{aligned} n_{\text{reacted}}(\ce{HCl}) &=0.008000-0.001800\\ &=0.006200\ \mathrm{mol} \end{aligned} \]

(b) Can we identify the calcium carbonate mass?

No. Since both ingredients consume two moles of hydrochloric acid per mole of active ingredient:

\[ \begin{aligned} n(\ce{CaCO3})+n(\ce{Mg(OH)2}) &=\frac{0.006200}{2}\\ &=0.003100\ \mathrm{mol} \end{aligned} \]

This equation gives the sum of their amounts, not the amount of either ingredient individually.

If you incorrectly assume that all the acid was consumed by calcium carbonate, you would calculate:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.003100\times100.09\\ &=0.3103\ \mathrm{g} \end{aligned} \]

However, \(0.3103\ \mathrm{g}\) is only the calcium carbonate equivalent of the measured acid consumption. It is not necessarily the actual calcium carbonate mass.

To determine the calcium carbonate mass uniquely, you would need independent information, such as a known amount of magnesium hydroxide or a validated method that distinguishes the two ingredients. Knowing the total sample mass is not enough when the proportions of active ingredients and inert filler are unknown.

Question 5

An antacid sample contains only calcium carbonate. A measured \(50.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid is added, and all the calcium carbonate reacts.

Immediately after the reaction, \(0.00300\ \mathrm{mol}\) of hydrochloric acid remains.

Before the remaining acid is titrated, two things happen:

• \(0.00020\ \mathrm{mol}\) of hydrochloric acid is lost from the warm solution.
• Dissolved carbon dioxide remains in the solution and consumes an additional \(0.00040\ \mathrm{mol}\) of hydroxide ions by the chosen titration endpoint.

A student ignores both effects and assumes that every mole of sodium hydroxide used corresponds to one mole of hydrochloric acid remaining.

Calculate:

(a) the mass of calcium carbonate the student reports;

(b) the true mass of calcium carbonate;

(c) the magnitude and direction of the error.

Assume no other effects occur.

Solution 5

The student reports \(0.3403\ \mathrm{g}\) of calcium carbonate, while the true mass is \(0.3503\ \mathrm{g}\). The result is therefore underestimated by \(0.0100\ \mathrm{g}\).

The subtle point is that the two experimental errors push the result in opposite directions. You must combine them before deciding the net effect.

Step 1: Calculate the initial amount of hydrochloric acid.

\[ \begin{aligned} n_{\text{initial}}(\ce{HCl}) &=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]

Step 2: Determine the amount of sodium hydroxide measured in the titration.

After the hydrochloric acid loss, the remaining amount is:

\[ 0.00300-0.00020 =0.00280\ \mathrm{mol} \]

The dissolved carbon dioxide consumes an additional \(0.00040\ \mathrm{mol}\) of hydroxide ions.

Therefore, the total amount of sodium hydroxide used is equivalent to:

\[ \begin{aligned} n(\ce{NaOH}) &=0.00280+0.00040\\ &=0.00320\ \mathrm{mol} \end{aligned} \]

The student incorrectly interprets this entire amount as the hydrochloric acid remaining.

Step 3: Calculate the student’s reported mass.

The student calculates the apparent amount of hydrochloric acid consumed:

\[ \begin{aligned} n_{\text{reacted, apparent}}(\ce{HCl}) &=0.01000-0.00320\\ &=0.00680\ \mathrm{mol} \end{aligned} \]

The apparent amount of calcium carbonate is:

\[ n_{\text{apparent}}(\ce{CaCO3}) =\frac{0.00680}{2} =0.00340\ \mathrm{mol} \]

Therefore:

\[ \begin{aligned} m_{\text{apparent}}(\ce{CaCO3}) &=0.00340\times100.09\\ &=0.3403\ \mathrm{g} \end{aligned} \]

Step 4: Calculate the true mass.

Before either experimental error occurs, the hydrochloric acid consumed by calcium carbonate is:

\[ \begin{aligned} n_{\text{reacted, true}}(\ce{HCl}) &=0.01000-0.00300\\ &=0.00700\ \mathrm{mol} \end{aligned} \]

Thus:

\[ n_{\text{true}}(\ce{CaCO3}) =\frac{0.00700}{2} =0.00350\ \mathrm{mol} \]

The true mass is:

\[ \begin{aligned} m_{\text{true}}(\ce{CaCO3}) &=0.00350\times100.09\\ &=0.3503\ \mathrm{g} \end{aligned} \]

Step 5: Calculate the error and explain its direction.

The difference is:

\[ \begin{aligned} \text{Error} &=0.3503-0.3403\\ &=0.0100\ \mathrm{g} \end{aligned} \]

The student underestimates the calcium carbonate mass by \(0.0100\ \mathrm{g}\).

Why?

• Losing hydrochloric acid makes the measured titre smaller. If this happened alone, the student would think less acid remained and overestimate the calcium carbonate mass.
• Dissolved carbon dioxide consumes extra sodium hydroxide. This makes the measured titre larger, so the student thinks more hydrochloric acid remained and underestimates the calcium carbonate mass.

The carbon dioxide effect is larger than the hydrochloric acid loss. The net apparent leftover acid is \(0.00020\ \mathrm{mol}\) greater than the true leftover amount, causing the final underestimate.

This is the key experimental lesson: a back titration is only as reliable as the chemical species being measured. Before trusting the final mass, establish whether the titre represents hydrochloric acid alone and whether the acid balance identifies calcium carbonate specifically or only the total acid-neutralising capacity of the sample.