How to Check a Back-Titration Answer Against Sample Mass
Learn how to check whether a calculated analyte mass is physically possible by comparing it with the original sample mass. Work through acid balance calculations, aliquot corrections, common errors, and examples involving mixtures.
A 0.3000 g powder sample undergoes a back titration. After working through the mole calculations, you report that it contains 0.451 g of sodium carbonate.
Something is wrong. Your answer claims that sodium carbonate makes up 150% of the original sample. A component cannot have more mass than the entire mixture it came from.
This is exactly why you should check a back-titration answer against the original sample mass before reporting it.
The basic check is:
\[ \%\text{ analyte}= \frac{m_{\text{analyte}}}{m_{\text{original sample}}} \times100 \]
The calculated analyte mass must not exceed the original sample mass. If the calculated percentage is greater than 100%, your result is physically impossible under the stated assumptions.
But there is an important catch: a percentage below 100% does not automatically mean your answer is correct. You also need to check that you’ve separated the acid added, acid left over, and acid consumed by the sample correctly.
01What are you actually measuring in a back titration?
Imagine weighing a powder containing sodium carbonate, \(\ce{Na2CO3}\), mixed with an inert substance.
You cannot directly use a simple titration to measure the amount of solid sodium carbonate. Instead, you add a known excess of hydrochloric acid, allow the reaction to finish, and measure how much acid remains.
The two reactions are:
1. Sodium carbonate reacts with hydrochloric acid
\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]
2. Remaining hydrochloric acid reacts with sodium hydroxide
\[ \ce{HCl + NaOH -> NaCl + H2O} \]
The first reaction tells you how much acid the sample can consume. The second allows you to measure how much acid is left.
Notice that the sodium carbonate reaction has a mole ratio of 1:2. One mole of sodium carbonate consumes two moles of hydrochloric acid.
The sodium hydroxide reaction has a mole ratio of 1:1. One mole of sodium hydroxide neutralises one mole of hydrochloric acid.
These ratios are essential when converting your titration results into the mass of the original substance.
Follow the experiment from the sample to the burette
The apparatus and order of measurements matter.
- First, weigh the original dry powder sample. Record its mass before adding anything.
- Transfer the sample into a conical flask and add a measured excess of hydrochloric acid.
- Allow the reaction to finish. If the sample contains carbonate, carbon dioxide bubbles are released.
- Transfer the remaining solution quantitatively into a volumetric flask. Add distilled water to the calibration mark and mix thoroughly.
- Use a pipette to transfer a measured aliquot into a clean conical flask.
- Titrate that aliquot with standardised sodium hydroxide delivered from a burette.
The burette reading tells you how much sodium hydroxide was required to neutralise the acid remaining in the aliquot.
It does not directly tell you how much hydrochloric acid the original sample consumed.
Think about the expected observation.
If two identical samples receive the same amount of hydrochloric acid, but one contains more sodium carbonate, which sample should require more sodium hydroxide in the final titration?
It should require less. The sample containing more sodium carbonate has consumed more of the original hydrochloric acid, leaving less acid for the sodium hydroxide to neutralise.
This is the central relationship behind back titration.
02The three quantities you must keep separate
There are three different amounts of hydrochloric acid in your calculation.
| Quantity | Meaning | How to calculate it |
|---|---|---|
| Acid added | Initial amount of HCl introduced | \(c_{\ce{HCl}}V_{\ce{HCl}}\) |
| Acid left over | Amount of HCl remaining after the sample reacts | Calculated from the NaOH titration |
| Acid consumed | Amount of HCl that reacted with the sample | Acid added − acid left over |
Use the following notation:
- \(n_{\text{added}}\): initial moles of hydrochloric acid, in mol.
- \(n_{\text{left}}\): moles of hydrochloric acid remaining, in mol.
- \(n_{\text{consumed}}\): moles of hydrochloric acid consumed by the sample, in mol.
- \(c\): concentration, in \(\mathrm{mol\,L^{-1}}\).
- \(V\): volume, in L.
The relationship is:
\[ \boxed{ n_{\text{consumed}}= n_{\text{added}}-n_{\text{left}} } \]
All three quantities are treated as positive amounts. The subtraction is always initial acid minus remaining acid.
Step 1: Calculate the acid added
Use the concentration and volume of the original hydrochloric acid.
\[ n_{\text{added}}=c_{\ce{HCl}}V_{\ce{HCl}} \]
Remember to convert millilitres into litres before substituting.
Step 2: Calculate the acid left over
For a complete titration of the remaining acid solution:
\[ n_{\text{left}}=c_{\ce{NaOH}}V_{\ce{NaOH}} \]
This works because hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio.
However, if you titrated only an aliquot, you must account for the portion of the solution that was not titrated.
\[ n_{\text{left}}= c_{\ce{NaOH}}V_{\ce{NaOH}} \times \frac{V_{\text{total}}}{V_{\text{aliquot}}} \]
Here:
- \(V_{\text{total}}\) is the volume of the entire diluted solution.
- \(V_{\text{aliquot}}\) is the volume transferred into the titration flask.
The volume ratio scales the measured moles in the aliquot up to the entire solution.
For example, if you titrate 20.00 mL from a total solution volume of 100.0 mL:
\[ \frac{V_{\text{total}}}{V_{\text{aliquot}}} = \frac{100.0}{20.00}=5.000 \]
The whole solution contains five times as much leftover acid as the aliquot.
Step 3: Calculate the acid consumed
Now subtract the total remaining acid from the original acid added.
\[ n_{\text{consumed}}= n_{\text{added}}-n_{\text{left}} \]
This gives the moles of hydrochloric acid that reacted with the sample, assuming no other processes consumed or removed the acid.
Step 4: Convert acid consumed into analyte mass
Return to the balanced chemical equation.
For sodium carbonate:
\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]
The mole ratio is:
\[ 1\ \mathrm{mol}\ \ce{Na2CO3}: 2\ \mathrm{mol}\ \ce{HCl} \]
Therefore:
\[ n_{\ce{Na2CO3}}= \frac{n_{\text{consumed}}}{2} \]
Convert the moles of sodium carbonate into mass:
\[ m_{\ce{Na2CO3}}= n_{\ce{Na2CO3}}M_{\ce{Na2CO3}} \]
The molar mass of sodium carbonate is approximately \(105.99\ \mathrm{g\,mol^{-1}}\).
You now have the calculated mass of sodium carbonate in the original sample.
03How to check the answer against the original sample mass
This is the step that catches physically impossible results.
Suppose your original sample weighed \(0.4000\ \mathrm{g}\), and your calculation gives \(0.2915\ \mathrm{g}\) of sodium carbonate.
Calculate the mass percentage:
\[ \%\ce{Na2CO3}= \frac{0.2915}{0.4000}\times100 \]
\[ =72.88\% \]
This is physically possible. The sodium carbonate accounts for approximately 72.9% of the original sample, leaving approximately 27.1% for other components.
Now suppose your calculation instead gives \(0.4511\ \mathrm{g}\).
\[ \%\ce{Na2CO3}= \frac{0.4511}{0.4000}\times100 \]
\[ =112.8\% \]
That result cannot describe the original sample if sodium carbonate is simply one of its components.
You should not round it down, cap it at 100%, or adjust the answer until it looks reasonable.
Go back through your calculations.
The two essential consistency checks
| Check | What must be true? | What a failure suggests |
|---|---|---|
| Acid balance | \(n_{\text{left}}\leq n_{\text{added}}\) | Incorrect subtraction, concentrations, volumes, or aliquot scaling |
| Sample mass | \(m_{\text{analyte}}\leq m_{\text{sample}}\) | Incorrect mole ratio, molar mass, acid balance, or chemical assumption |
You can also express the sample-mass check as:
\[ 0\leq\%\text{ analyte}\leq100\% \]
There is another useful way to apply the same physical limit.
If the entire original sample were pure sodium carbonate, the maximum amount of hydrochloric acid it could consume would be:
\[ n_{\ce{HCl,max}}= \frac{2m_{\text{sample}}}{M_{\ce{Na2CO3}}} \]
Any calculated acid consumption greater than this maximum is inconsistent with the assumption that sodium carbonate is the only acid-consuming component.
This is the same mass constraint expressed in moles rather than grams.
Important: Passing the mass check does not prove that your chemical identification is correct. If another substance in the sample also reacts with hydrochloric acid, the measured acid consumption may represent several substances combined.
04Worked example: Checking a full back-titration calculation
A \(0.4000\ \mathrm{g}\) powder sample contains sodium carbonate as its only acid-consuming component. The remaining material is inert.
The sample is reacted with \(40.00\ \mathrm{mL}\) of \(0.2500\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
After the reaction is complete, all the remaining acid is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. The titre is \(45.00\ \mathrm{mL}\).
Calculate the mass and mass percentage of sodium carbonate, then check the result against the original sample mass.
Step 1
Convert the volume into litres:
\[ 40.00\ \mathrm{mL}=0.04000\ \mathrm{L} \]
Apply \(n=cV\):
\[ \begin{aligned} n_{\text{added}} &=0.2500\times0.04000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
The original sample received \(0.01000\) mol of hydrochloric acid.
Step 2
Convert the sodium hydroxide titre:
\[ 45.00\ \mathrm{mL}=0.04500\ \mathrm{L} \]
The neutralisation reaction has a 1:1 mole ratio.
\[ \begin{aligned} n_{\text{left}} &=0.1000\times0.04500\\ &=0.004500\ \mathrm{mol} \end{aligned} \]
Therefore, \(0.004500\) mol of hydrochloric acid remained.
Step 3
\[ \begin{aligned} n_{\text{consumed}} &=n_{\text{added}}-n_{\text{left}}\\ &=0.01000-0.004500\\ &=0.005500\ \mathrm{mol} \end{aligned} \]
The sample consumed \(0.005500\) mol of hydrochloric acid.
Step 4
From the balanced equation, two moles of hydrochloric acid react with one mole of sodium carbonate.
\[ \begin{aligned} n_{\ce{Na2CO3}} &=\frac{0.005500}{2}\\ &=0.002750\ \mathrm{mol} \end{aligned} \]
Step 5
Using \(M_{\ce{Na2CO3}}=105.99\ \mathrm{g\,mol^{-1}}\):
\[ \begin{aligned} m_{\ce{Na2CO3}} &=nM\\ &=0.002750\times105.99\\ &=0.2915\ \mathrm{g} \end{aligned} \]
Step 6
\[ \begin{aligned} \%\ce{Na2CO3} &=\frac{0.2915}{0.4000}\times100\\ &=72.88\% \end{aligned} \]
The calculated sodium carbonate mass is less than the original sample mass.
The result is physically reasonable: approximately 72.9% of the powder is sodium carbonate, with the remaining 27.1% consisting of inert material.
The calculation passes the sample-mass check.
05Worked example: When an incorrect aliquot gives more analyte than the entire sample
This example shows why the original sample mass is such a useful diagnostic tool.
A \(0.3000\ \mathrm{g}\) powder sample contains sodium carbonate as its only acid-consuming component.
It reacts with \(40.00\ \mathrm{mL}\) of \(0.2500\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The remaining solution is quantitatively transferred into a \(100.0\ \mathrm{mL}\) volumetric flask and diluted to the calibration mark.
A \(20.00\ \mathrm{mL}\) aliquot is removed and titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. The titre is \(14.88\ \mathrm{mL}\).
Calculate the sodium carbonate mass and check whether the result is consistent with the original sample mass.
Step 1
\[ \begin{aligned} n_{\text{added}} &=0.2500\times0.04000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
Step 2
\[ \begin{aligned} n_{\ce{NaOH}} &=0.1000\times0.01488\\ &=0.001488\ \mathrm{mol} \end{aligned} \]
Because hydrochloric acid and sodium hydroxide react in a 1:1 ratio, the aliquot contains \(0.001488\) mol of hydrochloric acid.
Step 3
This is the critical step.
The aliquot represents only \(20.00\ \mathrm{mL}\) of the total \(100.0\ \mathrm{mL}\) solution.
\[ \text{Scaling factor}= \frac{100.0}{20.00}=5.000 \]
Therefore:
\[ \begin{aligned} n_{\text{left}} &=0.001488\times5.000\\ &=0.007440\ \mathrm{mol} \end{aligned} \]
The entire solution contains \(0.007440\) mol of hydrochloric acid.
Step 4
\[ \begin{aligned} n_{\text{consumed}} &=0.01000-0.007440\\ &=0.002560\ \mathrm{mol} \end{aligned} \]
Step 5
\[ \begin{aligned} n_{\ce{Na2CO3}} &=\frac{0.002560}{2}\\ &=0.001280\ \mathrm{mol} \end{aligned} \]
Step 6
\[ \begin{aligned} m_{\ce{Na2CO3}} &=0.001280\times105.99\\ &=0.1357\ \mathrm{g} \end{aligned} \]
Step 7
\[ \begin{aligned} \%\ce{Na2CO3} &=\frac{0.1357}{0.3000}\times100\\ &=45.22\% \end{aligned} \]
The result is physically possible. The sample contains approximately \(0.1357\ \mathrm{g}\) of sodium carbonate, representing 45.2% of its original mass.
What happens if you forget to scale up?
Suppose you mistakenly subtract the moles in the 20.00 mL aliquot directly from the initial moles of hydrochloric acid.
Your incorrect calculation would be:
\[ \begin{aligned} n_{\text{consumed,wrong}} &=0.01000-0.001488\\ &=0.008512\ \mathrm{mol} \end{aligned} \]
Converting this into sodium carbonate gives:
\[ \begin{aligned} m_{\ce{Na2CO3},\text{wrong}} &=\frac{0.008512}{2}\times105.99\\ &=0.4511\ \mathrm{g} \end{aligned} \]
The resulting mass percentage is:
\[ \frac{0.4511}{0.3000}\times100 =150.4\% \]
The result is impossible because the calculated sodium carbonate mass exceeds the entire powder sample.
The error was not in the balanced equation or molar mass. It was in treating an aliquot as though it represented the whole solution.
This is one of the most useful applications of the original sample-mass check: it can expose an error that isn’t immediately obvious from the mole calculations alone.
06What a reasonable mass percentage does not tell you
A result below 100% is necessary, but it is not sufficient evidence that your calculation is correct.
Consider a powder containing two substances that both react with hydrochloric acid.
If you assume that all the acid consumed came from sodium carbonate, you may calculate a sodium carbonate mass that is smaller than the original sample mass.
That result can pass the physical check while still being chemically incorrect.
The back titration measures total acid consumption. It does not automatically identify which substance consumed the acid.
There are several other errors worth recognising:
| Experimental or calculation error | Effect on calculated analyte mass |
|---|---|
| Forgetting the aliquot scaling factor | Overestimates the acid consumed and analyte mass |
| Overshooting the NaOH endpoint | Overestimates leftover acid and underestimates analyte mass |
| Stopping the titration too early | Underestimates leftover acid and overestimates analyte mass |
| Using the wrong mole ratio | Produces an incorrect analyte amount |
| Ignoring another acid-reactive ingredient | May incorrectly attribute its acid consumption to the target analyte |
| Using a different mass instead of the original sample mass | Produces an incorrect mass percentage |
The distinction between a possible answer and a proven answer matters.
If your result exceeds 100%, investigate the calculation or experimental assumptions. If it falls below 100%, continue checking the chemistry.
For a more detailed application of the reaction and subtraction method, see Back Titration of an Antacid: Calculating Calcium Carbonate.
The same principle of using the original sample mass as your denominator also appears in Measuring Mixture Composition by Mass: Method and Error Checks.
07Questions and solutions
Question 1
A \(0.2500\ \mathrm{g}\) powder sample contains sodium hydrogen carbonate, \(\ce{NaHCO3}\), as its only acid-consuming component.
The sample reacts completely with \(30.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The entire remaining acid solution requires \(40.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide for complete neutralisation.
Calculate:
a) The moles of hydrochloric acid initially added.
b) The moles of hydrochloric acid consumed by the sample.
c) The mass and mass percentage of sodium hydrogen carbonate.
d) Whether the result passes the original sample-mass check.
Use \(M_{\ce{NaHCO3}}=84.01\ \mathrm{g\,mol^{-1}}\).
Solution 1
The sample contains \(0.1680\ \mathrm{g}\) of sodium hydrogen carbonate, representing 67.20% of the original sample mass. This passes the physical consistency check.
Step 1: Calculate the initial moles of hydrochloric acid.
\[ \begin{aligned} n_{\text{added}} &=cV\\ &=0.2000\times0.03000\\ &=0.006000\ \mathrm{mol} \end{aligned} \]
Step 2: Calculate the moles of hydrochloric acid remaining.
\[ \begin{aligned} n_{\text{left}} &=0.1000\times0.04000\\ &=0.004000\ \mathrm{mol} \end{aligned} \]
Step 3: Calculate the moles of hydrochloric acid consumed.
\[ \begin{aligned} n_{\text{consumed}} &=0.006000-0.004000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]
Step 4: Calculate the moles of sodium hydrogen carbonate.
The balanced equation is:
\[ \ce{NaHCO3 + HCl -> NaCl + CO2 + H2O} \]
The mole ratio is 1:1.
Therefore:
\[ n_{\ce{NaHCO3}}=0.002000\ \mathrm{mol} \]
Step 5: Calculate the mass.
\[ \begin{aligned} m_{\ce{NaHCO3}} &=nM\\ &=0.002000\times84.01\\ &=0.1680\ \mathrm{g} \end{aligned} \]
Step 6: Calculate the mass percentage.
\[ \begin{aligned} \%\ce{NaHCO3} &=\frac{0.1680}{0.2500}\times100\\ &=67.20\% \end{aligned} \]
The calculated analyte mass is smaller than the original \(0.2500\ \mathrm{g}\) sample mass.
The remaining \(32.80\%\) can be attributed to the other components of the mixture under the stated assumptions.
The answer is physically consistent.
Question 2
A \(0.3600\ \mathrm{g}\) sample contains calcium carbonate, \(\ce{CaCO3}\), as its only acid-consuming component.
The sample is treated with \(50.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
After the reaction, the remaining solution is transferred into a \(200.0\ \mathrm{mL}\) volumetric flask and diluted to the mark.
A \(25.00\ \mathrm{mL}\) aliquot is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. The titre is \(8.40\ \mathrm{mL}\).
Calculate the mass percentage of calcium carbonate.
A student forgets to account for the aliquot and obtains a calcium carbonate mass greater than the original sample mass.
Identify the student’s error and demonstrate why the original sample-mass check exposes it.
Use \(M_{\ce{CaCO3}}=100.09\ \mathrm{g\,mol^{-1}}\).
Solution 2
The correct calcium carbonate mass is \(0.1641\ \mathrm{g}\), representing 45.6% of the original sample. Forgetting the aliquot correction produces an impossible result of approximately 127.3%.
Step 1: Calculate the initial moles of hydrochloric acid.
\[ \begin{aligned} n_{\text{added}} &=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
Step 2: Calculate the moles of sodium hydroxide used.
\[ \begin{aligned} n_{\ce{NaOH}} &=0.1000\times0.00840\\ &=0.000840\ \mathrm{mol} \end{aligned} \]
The aliquot contains \(0.000840\) mol of hydrochloric acid.
Step 3: Scale up to the entire solution.
The aliquot is one-eighth of the total solution:
\[ \frac{200.0}{25.00}=8.000 \]
Therefore:
\[ \begin{aligned} n_{\text{left}} &=0.000840\times8.000\\ &=0.006720\ \mathrm{mol} \end{aligned} \]
Step 4: Calculate the hydrochloric acid consumed.
\[ \begin{aligned} n_{\text{consumed}} &=0.01000-0.006720\\ &=0.003280\ \mathrm{mol} \end{aligned} \]
Step 5: Calculate the moles of calcium carbonate.
The balanced equation is:
\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]
Therefore:
\[ \begin{aligned} n_{\ce{CaCO3}} &=\frac{0.003280}{2}\\ &=0.001640\ \mathrm{mol} \end{aligned} \]
Step 6: Calculate the mass.
\[ \begin{aligned} m_{\ce{CaCO3}} &=0.001640\times100.09\\ &=0.1641\ \mathrm{g} \end{aligned} \]
Step 7: Calculate the mass percentage.
\[ \begin{aligned} \%\ce{CaCO3} &=\frac{0.1641}{0.3600}\times100\\ &=45.6\% \end{aligned} \]
The result passes the sample-mass check.
Now consider the student’s mistake.
If the student uses the aliquot’s \(0.000840\) mol directly as the total leftover acid:
\[ n_{\text{consumed,wrong}} =0.01000-0.000840 \]
\[ =0.009160\ \mathrm{mol} \]
This gives:
\[ \begin{aligned} m_{\ce{CaCO3},\text{wrong}} &=\frac{0.009160}{2}\times100.09\\ &=0.4584\ \mathrm{g} \end{aligned} \]
The incorrect mass percentage is:
\[ \frac{0.4584}{0.3600}\times100 =127.3\% \]
The calculated calcium carbonate mass is greater than the entire original sample mass.
There is also a useful independent way to recognise the problem. If the whole \(0.3600\ \mathrm{g}\) sample were calcium carbonate, the maximum possible hydrochloric acid consumption would be:
\[ \begin{aligned} n_{\ce{HCl,max}} &=\frac{2\times0.3600}{100.09}\\ &=0.007193\ \mathrm{mol} \end{aligned} \]
The student’s calculated consumption of \(0.009160\) mol exceeds this maximum.
Both checks identify the same problem: the aliquot was treated as the entire solution.
Question 3
A \(0.2500\ \mathrm{g}\) sample contains sodium carbonate as its only acid-consuming component.
It is reacted with \(40.00\ \mathrm{mL}\) of \(0.2500\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The correct endpoint for the subsequent titration requires \(65.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.
However, a student overshoots the endpoint by \(2.00\ \mathrm{mL}\) and records a titre of \(67.00\ \mathrm{mL}\).
Calculate:
a) The correct mass of sodium carbonate.
b) The mass reported using the overshot titre.
c) The difference between the two results.
Explain why overshooting the endpoint changes the calculated mass in this direction.
Use \(M_{\ce{Na2CO3}}=105.99\ \mathrm{g\,mol^{-1}}\).
Solution 3
The correct sodium carbonate mass is \(0.1855\ \mathrm{g}\). The overshot titre produces an underestimated mass of \(0.1749\ \mathrm{g}\), a difference of \(0.0106\ \mathrm{g}\).
Step 1: Calculate the initial hydrochloric acid.
\[ \begin{aligned} n_{\text{added}} &=0.2500\times0.04000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
Step 2: Calculate the correct leftover acid.
Using the correct titre:
\[ \begin{aligned} n_{\text{left}} &=0.1000\times0.06500\\ &=0.006500\ \mathrm{mol} \end{aligned} \]
Step 3: Calculate the correct acid consumption.
\[ \begin{aligned} n_{\text{consumed}} &=0.01000-0.006500\\ &=0.003500\ \mathrm{mol} \end{aligned} \]
Step 4: Calculate the correct sodium carbonate mass.
\[ \begin{aligned} n_{\ce{Na2CO3}} &=\frac{0.003500}{2}\\ &=0.001750\ \mathrm{mol} \end{aligned} \]
\[ \begin{aligned} m_{\ce{Na2CO3}} &=0.001750\times105.99\\ &=0.1855\ \mathrm{g} \end{aligned} \]
Step 5: Calculate the mass using the overshot titre.
The student records \(67.00\ \mathrm{mL}\), so:
\[ \begin{aligned} n_{\text{left,wrong}} &=0.1000\times0.06700\\ &=0.006700\ \mathrm{mol} \end{aligned} \]
The calculated acid consumption becomes:
\[ \begin{aligned} n_{\text{consumed,wrong}} &=0.01000-0.006700\\ &=0.003300\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ \begin{aligned} m_{\ce{Na2CO3},\text{wrong}} &=\frac{0.003300}{2}\times105.99\\ &=0.1749\ \mathrm{g} \end{aligned} \]
Step 6: Compare the two results.
\[ \begin{aligned} \Delta m &=0.1855-0.1749\\ &=0.0106\ \mathrm{g} \end{aligned} \]
The overshot titre underestimates the sodium carbonate mass by \(0.0106\ \mathrm{g}\).
Why?
The student added too much sodium hydroxide, so the recorded titre suggests that more hydrochloric acid remained than actually did.
This makes the calculated acid consumption too small. Since sodium carbonate is calculated from the acid consumed, its calculated mass is also too small.
Notice that the incorrect result still passes the sample-mass check:
\[ \frac{0.1749}{0.2500}\times100 =69.95\% \]
This is a good example of why a physically reasonable percentage does not guarantee an accurate experimental result.
Question 4
A \(0.3000\ \mathrm{g}\) powder sample contains sodium carbonate, \(\ce{Na2CO3}\), sodium hydrogen carbonate, \(\ce{NaHCO3}\), and an inert filler.
The sample is reacted with \(50.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The entire remaining acid solution requires \(50.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide for complete neutralisation.
A student assumes that all the hydrochloric acid consumed reacted with sodium carbonate.
a) Calculate the sodium carbonate mass and percentage reported by this student.
b) Explain why passing the original sample-mass check does not prove that the result is correct.
c) Construct two different sample compositions consistent with the experimental data. Use the following molar masses:
- \(\ce{Na2CO3}\): \(105.99\ \mathrm{g\,mol^{-1}}\).
- \(\ce{NaHCO3}\): \(84.01\ \mathrm{g\,mol^{-1}}\).
Solution 4
The student reports \(0.26498\ \mathrm{g}\) of sodium carbonate, or 88.33% of the sample. This result passes the mass check, but it does not establish that the sample actually contains that amount of sodium carbonate.
Step 1: Calculate the initial hydrochloric acid.
\[ \begin{aligned} n_{\text{added}} &=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
Step 2: Calculate the leftover hydrochloric acid.
\[ \begin{aligned} n_{\text{left}} &=0.1000\times0.05000\\ &=0.005000\ \mathrm{mol} \end{aligned} \]
Step 3: Calculate the acid consumed.
\[ \begin{aligned} n_{\text{consumed}} &=0.01000-0.005000\\ &=0.005000\ \mathrm{mol} \end{aligned} \]
Step 4: Calculate the mass under the student’s assumption.
The student assumes that only sodium carbonate consumed hydrochloric acid.
The reaction is:
\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]
Therefore:
\[ n_{\ce{Na2CO3}}= \frac{0.005000}{2} =0.002500\ \mathrm{mol} \]
The calculated mass is:
\[ \begin{aligned} m_{\ce{Na2CO3}} &=0.002500\times105.99\\ &=0.26498\ \mathrm{g} \end{aligned} \]
The calculated percentage is:
\[ \begin{aligned} \%\ce{Na2CO3} &=\frac{0.26498}{0.3000}\times100\\ &=88.33\% \end{aligned} \]
This result is below 100%, so it passes the basic mass check.
However, the student’s assumption has not been established.
Step 5: Test a second possible composition.
Sodium hydrogen carbonate also reacts with hydrochloric acid:
\[ \ce{NaHCO3 + HCl -> NaCl + CO2 + H2O} \]
Unlike sodium carbonate, it consumes only one mole of hydrochloric acid per mole of sodium hydrogen carbonate.
Consider a mixture containing:
- \(0.002000\) mol of sodium carbonate.
- \(0.001000\) mol of sodium hydrogen carbonate.
The acid consumed by sodium carbonate is:
\[ 2\times0.002000=0.004000\ \mathrm{mol} \]
The acid consumed by sodium hydrogen carbonate is:
\[ 1\times0.001000=0.001000\ \mathrm{mol} \]
The total acid consumption is:
\[ 0.004000+0.001000 =0.005000\ \mathrm{mol} \]
This exactly matches the experimental result.
Now calculate the masses.
Sodium carbonate:
\[ m=0.002000\times105.99 =0.21198\ \mathrm{g} \]
Sodium hydrogen carbonate:
\[ m=0.001000\times84.01 =0.08401\ \mathrm{g} \]
The total mass of these two substances is:
\[ 0.21198+0.08401=0.29599\ \mathrm{g} \]
The remaining inert filler mass is:
\[ 0.3000-0.29599=0.00401\ \mathrm{g} \]
Both compositions are physically possible and produce the same measured acid consumption.
The experiment therefore cannot uniquely establish the sodium carbonate mass.
The important distinction is that a back titration measures total acid consumption. Converting that result into the mass of one particular substance requires a justified assumption about which substances reacted.
Question 5
A \(0.3000\ \mathrm{g}\) powder sample is known to contain exactly two substances: sodium carbonate, \(\ce{Na2CO3}\), and sodium hydrogen carbonate, \(\ce{NaHCO3}\).
There is no inert filler, moisture, or other component.
The sample reacts completely with \(50.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The entire remaining acid solution requires \(55.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide for complete neutralisation.
A student assumes that all the acid consumed reacted with sodium carbonate. They calculate a sodium carbonate mass below the original sample mass and conclude that their answer is physically reasonable.
However, the sample is known to contain both substances.
Determine the actual mass of each substance in the sample.
Explain why the original sample mass, which was initially just a consistency check, allows you to obtain more information about the composition in this particular situation.
Use:
- \(M_{\ce{Na2CO3}}=105.99\ \mathrm{g\,mol^{-1}}\).
- \(M_{\ce{NaHCO3}}=84.01\ \mathrm{g\,mol^{-1}}\).
Solution 5
The sample contains approximately \(0.1334\ \mathrm{g}\) of sodium carbonate and \(0.1666\ \mathrm{g}\) of sodium hydrogen carbonate.
The key is that the sample mass provides a second independent equation. Because the sample contains exactly two known substances and nothing else, the two equations can determine both amounts.
Step 1: Calculate the initial hydrochloric acid.
\[ \begin{aligned} n_{\text{added}} &=0.2000\times0.05000\\ &=0.01000\ \mathrm{mol} \end{aligned} \]
Step 2: Calculate the leftover hydrochloric acid.
\[ \begin{aligned} n_{\text{left}} &=0.1000\times0.05500\\ &=0.005500\ \mathrm{mol} \end{aligned} \]
Step 3: Calculate the total acid consumed.
\[ \begin{aligned} n_{\text{consumed}} &=0.01000-0.005500\\ &=0.004500\ \mathrm{mol} \end{aligned} \]
This is the total amount of hydrochloric acid consumed by both substances.
Step 4: Define the two unknown amounts.
Let:
- \(x\) = moles of sodium carbonate.
- \(y\) = moles of sodium hydrogen carbonate.
The balanced equations are:
\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]
\[ \ce{NaHCO3 + HCl -> NaCl + CO2 + H2O} \]
One mole of sodium carbonate consumes two moles of hydrochloric acid.
One mole of sodium hydrogen carbonate consumes one mole of hydrochloric acid.
Therefore, the total acid consumption gives:
\[ 2x+y=0.004500 \]
This is our first equation.
Step 5: Use the original sample mass to construct a second equation.
The original sample contains only sodium carbonate and sodium hydrogen carbonate.
Therefore, their masses must add to \(0.3000\ \mathrm{g}\).
Using mass = moles × molar mass:
\[ 105.99x+84.01y=0.3000 \]
This is our second equation.
We now have:
\[ \begin{aligned} 2x+y&=0.004500\\ 105.99x+84.01y&=0.3000 \end{aligned} \]
Step 6: Solve the simultaneous equations.
Rearrange the first equation:
\[ y=0.004500-2x \]
Substitute into the mass equation:
\[ 105.99x+84.01(0.004500-2x)=0.3000 \]
Expand:
\[ 105.99x+0.378045-168.02x=0.3000 \]
Collect like terms:
\[ -62.03x=-0.078045 \]
Therefore:
\[ x=0.001258\ \mathrm{mol} \]
Substitute this into the first equation:
\[ \begin{aligned} y&=0.004500-2(0.001258)\\ &=0.001984\ \mathrm{mol} \end{aligned} \]
We have now determined the moles of both substances.
Step 7: Convert both amounts into masses.
For sodium carbonate:
\[ \begin{aligned} m_{\ce{Na2CO3}} &=0.001258\times105.99\\ &=0.1334\ \mathrm{g} \end{aligned} \]
For sodium hydrogen carbonate:
\[ \begin{aligned} m_{\ce{NaHCO3}} &=0.001984\times84.01\\ &=0.1666\ \mathrm{g} \end{aligned} \]
Step 8: Check against the original sample mass.
\[ 0.1334+0.1666=0.3000\ \mathrm{g} \]
The two calculated masses add to the original sample mass.
Their mass percentages are:
\[ \%\ce{Na2CO3} = \frac{0.1334}{0.3000}\times100 =44.47\% \]
\[ \%\ce{NaHCO3} = \frac{0.1666}{0.3000}\times100 =55.53\% \]
The composition is consistent with both the acid consumption and the original sample mass.
Why was the student’s first answer insufficient?
If the student assumes that all \(0.004500\) mol of hydrochloric acid reacted with sodium carbonate, they calculate:
\[ n_{\ce{Na2CO3}}= \frac{0.004500}{2} =0.002250\ \mathrm{mol} \]
This gives:
\[ m_{\ce{Na2CO3}}= 0.002250\times105.99 =0.2385\ \mathrm{g} \]
The result represents 79.50% of the original sample.
It passes the simple mass check because \(0.2385\ \mathrm{g}\) is less than \(0.3000\ \mathrm{g}\).
But it is not the actual sodium carbonate mass. It incorrectly attributes the hydrochloric acid consumed by sodium hydrogen carbonate to sodium carbonate.
The deeper point is that the original sample mass can do more than reject an impossible answer.
When the sample contains exactly two known substances and no other components, the total mass provides a second equation that can help determine their individual amounts.
If an unknown inert filler were also present, that extra unknown would make the system underdetermined again. You would need another independent measurement.
That is the distinction to carry into more complex HSC stoichiometry problems: a back titration gives you the total amount of reagent consumed, while the original sample mass constrains what that consumption could represent. Before accepting an answer, check both the chemistry and the physical limits of the sample.