How to Correct for Diluted Aliquots in Back Titration

A powder sample reacts with a known excess of hydrochloric acid. You dilute the entire reaction mixture to 250.0 mL, then take just 25.00 mL for a back titration. That small portion contains 0.00120 mol of leftover acid.

Would you subtract 0.00120 mol from the acid originally added?

Not yet. You’ve measured the acid in only one-tenth of the diluted solution. The original sample was exposed to all the acid, so you must first calculate how much acid remains in the whole flask.

The correction is:

\[ n_{\mathrm{left}}=n_{\mathrm{aliquot}} \times\frac{V_D}{V_P} \]

Here, \(n_{\mathrm{left}}\) is the total amount of leftover reagent in moles, \(n_{\mathrm{aliquot}}\) is the amount measured in the titrated portion, \(V_D\) is the total volume of diluted reaction mixture, and \(V_P\) is the volume pipetted for titration.

In this case, the total leftover acid is:

\[ \begin{aligned} n_{\mathrm{left}} &=0.00120\times\frac{250.0}{25.00}\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

If the sample originally received 0.0240 mol of hydrochloric acid, it consumed \(0.0240-0.0120=0.0120\) mol.

That’s the essential rule: correct the leftover reagent for the aliquot before subtracting it from the reagent originally added. Don’t multiply the amount originally added by the dilution factor.

01Why the experiment uses a diluted aliquot

Imagine trying to measure the magnesium hydroxide in a powdered mixture by adding hydrochloric acid until all the magnesium hydroxide has dissolved.

How would you know the exact moment the reaction had finished?

A solid can dissolve gradually, particularly if its particles are not all the same size. Disappearance of the solid may be difficult to judge when inert material is also present. There isn’t necessarily a sharp, visible endpoint.

A better experimental design is to add a measured excess of hydrochloric acid, allow the reaction to finish, and measure the acid that remains.

This is a back titration.

The magnesium hydroxide consumes hydrochloric acid:

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

Every mole of magnesium hydroxide consumes two moles of HCl.

The leftover hydrochloric acid can then be measured using standard sodium hydroxide solution:

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

This second reaction has a \(1:1\) mole ratio.

The first reaction tells you how much acid the sample consumes. The second tells you how much acid remains.

Follow the liquid through the apparatus

The experiment has several distinct stages.

1. Weigh the powdered sample into a conical flask. Add a precisely measured volume of hydrochloric acid of known concentration.
2. Swirl the flask and allow the reaction to finish. Hydrochloric acid must remain in excess.
3. Transfer the reaction solution quantitatively into a volumetric flask. Rinse the original container and add the rinsings so that the remaining acid is transferred. Remove any insoluble material appropriately without losing acidic solution.
4. Add distilled water to the volumetric flask’s calibration mark. Stopper and invert it repeatedly to mix.
5. Use a volumetric pipette to remove a measured aliquot, such as 25.00 mL, and transfer it to a clean conical flask.
6. Titrate that aliquot with standardised sodium hydroxide from a burette, using a suitable indicator or endpoint method.

Why bother with the volumetric flask and pipette?

Titrating the entire reaction solution could require an inconveniently large volume of sodium hydroxide. Taking a smaller, representative portion gives a manageable titre.

But it introduces a calculation problem: the burette measures the acid in that portion, not the acid remaining from the entire original experiment.

What should happen when the aliquot changes?

Suppose the diluted reaction mixture contains a uniform concentration of leftover HCl.

You titrate 25.00 mL, then separately titrate 50.00 mL using sodium hydroxide of the same concentration.

Would you expect the second titre to double?

Yes. The 50.00 mL portion contains twice as much HCl as the 25.00 mL portion. It therefore needs twice as many moles of NaOH.

Now suppose you take the original 25.00 mL aliquot and add another 75.00 mL of distilled water to its conical flask before titration.

The concentration falls to one-quarter of its previous value, but the amount of HCl doesn’t change. The titre should stay the same, assuming the same endpoint and no contamination.

This is the distinction that controls the whole method: dilution changes concentration, while withdrawing an aliquot selects a fraction of the original substance.

02Keep three reagent amounts separate

Before doing any back-titration calculation, identify these three quantities.

The two amounts being subtracted must refer to the same original sample.

Step 1: Calculate the reagent originally added

Use:

\[ n_A=c_AV_A \]

Here, \(n_A\) is the amount of HCl originally added in mol, \(c_A\) is its concentration in \(\mathrm{mol\,L^{-1}}\), and \(V_A\) is its original measured volume in L.

For example, 40.00 mL of 0.5000 \(\mathrm{mol\,L^{-1}}\) HCl contains:

\[ \begin{aligned} n_A&=0.5000\times0.04000\\ &=0.02000\ \mathrm{mol} \end{aligned} \]

Adding water later does not change these original moles of HCl.

Step 2: Calculate the reagent in the titrated aliquot

If leftover HCl is titrated with NaOH, use the \(1:1\) reaction ratio.

\[ n_T=c_BV_B \]

Here, \(n_T\) is the amount of HCl in the aliquot in mol, \(c_B\) is the NaOH concentration in \(\mathrm{mol\,L^{-1}}\), and \(V_B\) is the NaOH titre in L.

If the titration reaction has a different mole ratio, apply that ratio before identifying the amount of leftover reagent.

Step 3: Scale the leftover reagent to the entire solution

Suppose the finished reaction mixture was diluted to 250.0 mL and you titrated 25.00 mL.

The aliquot represents:

\[ f=\frac{25.00}{250.0}=0.1000 \]

Here, \(f\) is the fraction of the final homogeneous mixture that was titrated.

Only one-tenth was measured, so the whole solution contains ten times the amount found in the aliquot.

\[ n_L=\frac{n_T}{f}=10n_T \]

The symbol \(n_L\) represents the total leftover HCl in mol.

Step 4: Subtract to find the amount consumed

\[ n_C=n_A-n_L \]

Here, \(n_C\) is the amount of HCl consumed by the sample in mol.

The subtraction order matters. The leftover acid is what survived the reaction; the difference is what reacted.

For a valid experiment with HCl genuinely in excess, you should obtain:

\[ 0<n_L<n_A \]

A calculated negative amount of acid consumed is not a strange new chemical phenomenon. It signals inconsistent data, a calculation error, or a failure of the assumptions.

04A contrasting case: Water added after the aliquot is taken

Consider two experiments using the same homogeneous 100.0 mL reaction solution.

In both experiments, a 25.00 mL aliquot is withdrawn.

• In experiment A, the 25.00 mL aliquot is titrated directly.
• In experiment B, the same-sized aliquot is diluted with 75.00 mL of water in its conical flask before titration.

Which experiment needs a larger aliquot correction?

Neither. Both aliquots contain one-quarter of the HCl present in the original 100.0 mL reaction solution.

For both experiments:

\[ \frac{V_D}{V_P}=\frac{100.0}{25.00}=4 \]

The extra 75.00 mL of water in experiment B doesn’t create another sample of the original acid. It only spreads the same amount of acid through a larger volume.

This gives you a useful decision rule:

If water is added before taking the aliquot, use the complete made-up volume when finding the fraction sampled.

If water is added after taking the aliquot, it does not change that fraction.

The real question is always: What fraction of the original reaction mixture reached the titration flask?

06How to choose the correct factor

The fastest reliable method isn’t to memorise one dilution formula. It’s to follow the actual liquid from the original reaction to the burette.

Three further checks are worth making before accepting an answer.

First, all concentration calculations use volumes in litres. Ratios such as 250/25 can use millilitres, provided both volumes use the same unit.

Second, apply the appropriate mole ratio for the titration reaction. The amount of titrant is not necessarily equal to the amount of leftover reagent when the reaction is not \(1:1\).

Third, check the experimental assumptions. The reaction must be complete before ordinary aliquot sampling, the diluted solution must be homogeneous, and the titration must measure the intended leftover reagent. Dilution cannot correct an incomplete chemical reaction or an interfering substance.

07Questions and solutions

Question 1

A sample containing magnesium oxide as its only acid-reactive substance is treated with 25.00 mL of 0.4000 \(\mathrm{mol\,L^{-1}}\) HCl.

After the reaction finishes, the solution is quantitatively transferred to a 100.0 mL volumetric flask and diluted to the mark.

A 20.00 mL aliquot requires 12.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) NaOH.

Calculate the amount of HCl consumed and the mass of MgO in the sample.

Use \(M(\ce{MgO})=40.30\ \mathrm{g\,mol^{-1}}\).

Solution 1

The sample consumed 0.00400 mol of HCl, corresponding to 0.0806 g of MgO.

The original amount of HCl was:

\[ \begin{aligned} n_A&=0.4000\times0.02500\\ &=0.01000\ \mathrm{mol} \end{aligned} \]

From the \(1:1\) HCl-NaOH reaction, the aliquot contained:

\[ \begin{aligned} n_T&=0.1000\times0.01200\\ &=0.001200\ \mathrm{mol} \end{aligned} \]

The 20.00 mL aliquot represents one-fifth of the original 100.0 mL diluted solution.

\[ \begin{aligned} n_L&=0.001200\times5\\ &=0.006000\ \mathrm{mol} \end{aligned} \]

Therefore:

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

Since \(\ce{MgO}\) and HCl react in a \(1:2\) mole ratio:

\[ \begin{aligned} n(\ce{MgO})&=\frac{0.004000}{2}\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

\[ m(\ce{MgO})=0.002000\times40.30 =0.08060\ \mathrm{g} \]

The key is to multiply the measured leftover acid by five before subtracting it from the original acid. Multiplying the amount of MgO by five at the end would not be equivalent.

Question 2

A 0.3000 g powdered sample contains magnesium hydroxide and inert filler. It reacts completely with 35.00 mL of 0.5000 \(\mathrm{mol\,L^{-1}}\) HCl.

The entire reaction mixture is made up to 200.0 mL.

Student A titrates a 20.00 mL aliquot using 0.1000 \(\mathrm{mol\,L^{-1}}\) NaOH and obtains a titre of 9.00 mL.

Student B takes a separate 40.00 mL aliquot from the same homogeneous solution, adds 160.0 mL of distilled water to the titration flask, and uses 0.1500 \(\mathrm{mol\,L^{-1}}\) NaOH.

Assume both aliquots are representative, all reactions are complete, and no solution is lost.

(a) Predict Student B’s titre.

(b) Calculate the percentage of magnesium hydroxide in the original sample.

(c) Explain why Student B must not introduce an extra dilution correction for the 160.0 mL of added water.

Use \(M(\ce{Mg(OH)2})=58.32\ \mathrm{g\,mol^{-1}}\).

Solution 2

Student B should obtain a titre of 12.00 mL. The original powder contains 82.6% magnesium hydroxide by mass. Adding water after pipetting changes neither the amount of acid selected nor the fraction of the original solution sampled.

Part (a)

Student A’s aliquot contained:

\[ \begin{aligned} n_T&=0.1000\times0.00900\\ &=0.000900\ \mathrm{mol} \end{aligned} \]

Student B sampled twice the volume of Student A, so B’s aliquot contains twice the HCl:

\[ n_T(B)=0.001800\ \mathrm{mol} \]

The NaOH concentration used by B is 0.1500 \(\mathrm{mol\,L^{-1}}\).

Using \(V=n/c\):

\[ \begin{aligned} V_B&=\frac{0.001800}{0.1500}\\ &=0.01200\ \mathrm{L} \end{aligned} \]

Therefore, B’s predicted titre is 12.00 mL.

It is tempting to predict 18.00 mL simply because B takes twice as much solution as A. That ignores the fact that B’s NaOH is 1.5 times as concentrated.

Part (b)

Using Student A’s titration, total leftover HCl is:

\[ \begin{aligned} n_L&=0.000900\times\frac{200.0}{20.00}\\ &=0.009000\ \mathrm{mol} \end{aligned} \]

Originally added HCl:

\[ n_A=0.5000\times0.03500 =0.01750\ \mathrm{mol} \]

Consumed HCl:

\[ n_C=0.01750-0.009000 =0.008500\ \mathrm{mol} \]

From the \(1:2\) magnesium hydroxide-HCl mole ratio:

\[ n(\ce{Mg(OH)2})=\frac{0.008500}{2} =0.004250\ \mathrm{mol} \]

\[ m=0.004250\times58.32 =0.2479\ \mathrm{g} \]

\[ \%\ce{Mg(OH)2} =\frac{0.2479}{0.3000}\times100 \approx82.6\% \]

Part (c)

The additional 160.0 mL of water is added after Student B withdraws the aliquot. It cannot increase or decrease the moles of HCl already selected.

Student B’s aliquot is \(40.00/200.0=1/5\) of the original solution, so the correction factor is five.

The final volume of liquid inside the titration flask is irrelevant to this fraction. It affects concentration, not the amount of acid to be neutralised.

Question 3

An illustrative 0.2500 g sample containing MgO and inert material is treated with 40.00 mL of 0.5000 \(\mathrm{mol\,L^{-1}}\) HCl.

After the reaction finishes, the acidic solution is made up to exactly 100.0 mL and thoroughly mixed.

Unfortunately, 10.00 mL of this homogeneous solution is accidentally discarded.

The remaining 90.00 mL is quantitatively transferred to a 250.0 mL volumetric flask and diluted to the mark.

A 25.00 mL aliquot requires 9.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) NaOH.

(a) Calculate the mass percentage of MgO in the original sample.

(b) Calculate the percentage that would result if the analyst corrected only for the final 250.0 mL to 25.00 mL aliquot.

(c) Explain why knowing the discarded volume allows a correction, but losing an unknown amount of solution would not.

Use \(M(\ce{MgO})=40.30\ \mathrm{g\,mol^{-1}}\).

Assume the discarded liquid was representative, contained no solid, and no further reaction occurred after the first dilution.

Solution 3

The original sample contained 80.6% MgO by mass. Ignoring the discarded fraction would give 88.7%, an overestimate.

Part (a)

The HCl measured in the final aliquot was:

\[ n_T=0.1000\times0.00900 =0.000900\ \mathrm{mol} \]

The entire 250.0 mL diluted solution therefore contains:

\[ \begin{aligned} n_{\mathrm{retained}} &=0.000900\times10\\ &=0.009000\ \mathrm{mol} \end{aligned} \]

But this is only the HCl retained after discarding 10.00 mL from the original 100.0 mL solution.

Since 90% of that homogeneous solution was retained:

\[ \begin{aligned} n_L&=0.009000\times\frac{100.0}{90.00}\\ &=0.01000\ \mathrm{mol} \end{aligned} \]

The original HCl was:

\[ n_A=0.5000\times0.04000 =0.02000\ \mathrm{mol} \]

So the sample consumed:

\[ n_C=0.02000-0.01000 =0.01000\ \mathrm{mol} \]

The MgO mass was:

\[ \begin{aligned} n(\ce{MgO})&=\frac{0.01000}{2}\\ &=0.005000\ \mathrm{mol}\\ m(\ce{MgO})&=0.005000\times40.30\\ &=0.2015\ \mathrm{g} \end{aligned} \]

The percentage is:

\[ \frac{0.2015}{0.2500}\times100=80.6\% \]

Part (b)

Without the correction for discarded solution, the analyst would use \(n_L=0.009000\ \mathrm{mol}\).

This gives:

\[ n_C=0.02000-0.009000 =0.01100\ \mathrm{mol} \]

\[ m(\ce{MgO})=\frac{0.01100}{2}\times40.30 =0.2217\ \mathrm{g} \]

\[ \%\ce{MgO} =\frac{0.2217}{0.2500}\times100 \approx88.7\% \]

The error makes the sample appear to have consumed more acid than it really did.

Part (c)

The known 10.00 mL loss can be corrected because the original solution was homogeneous. Exactly 90% of the leftover acid remained after the loss.

An unknown spill has no known retained fraction. The later titre alone cannot tell us how much HCl disappeared in the spill and how much was consumed by MgO.

Two processes have become indistinguishable: chemical consumption and physical loss. A dilution formula cannot separate them without additional information.

Question 4

A 0.2089 g powder is stated to contain only magnesium oxide, \(\ce{MgO}\), and magnesium carbonate, \(\ce{MgCO3}\).

Both substances react completely with excess HCl:

\[ \ce{MgO + 2HCl -> MgCl2 + H2O} \]

\[ \ce{MgCO3 + 2HCl -> MgCl2 + CO2 + H2O} \]

The powder receives 25.00 mL of 0.5000 \(\mathrm{mol\,L^{-1}}\) HCl.

After the reaction finishes and the carbon dioxide has escaped, the remaining solution is made up to 250.0 mL.

A 50.00 mL aliquot requires 13.00 mL of 0.1000 \(\mathrm{mol\,L^{-1}}\) NaOH.

(a) Calculate the amount of HCl consumed by the sample.

(b) Determine the mass of each compound in the original powder.

(c) Would the answer to part (b) still be unique if an unknown quantity of inert sand were also permitted in the sample? Justify your conclusion.

Assume the endpoint measures only residual HCl.

Use \(M(\ce{MgO})=40.30\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{MgCO3})=84.31\ \mathrm{g\,mol^{-1}}\).

Solution 4

The sample consumed 0.006000 mol of HCl. It contained approximately 0.0403 g of MgO and 0.1686 g of MgCO3. These masses are uniquely determined only because the original powder is specified to contain exactly those two substances.

Part (a)

The titrated aliquot contained:

\[ n_T=0.1000\times0.01300 =0.001300\ \mathrm{mol} \]

The entire solution contained:

\[ \begin{aligned} n_L&=0.001300\times\frac{250.0}{50.00}\\ &=0.006500\ \mathrm{mol} \end{aligned} \]

Initially added HCl:

\[ n_A=0.5000\times0.02500 =0.01250\ \mathrm{mol} \]

Consumed HCl:

\[ n_C=0.01250-0.006500 =0.006000\ \mathrm{mol} \]

Part (b)

Both MgO and MgCO3 consume two moles of HCl per mole of compound.

Let \(x\) be the moles of MgO and \(y\) be the moles of MgCO3.

The acid consumption gives:

\[ 2x+2y=0.006000 \]

Therefore:

\[ x+y=0.003000 \]

The mass of the powder gives a second equation:

\[ 40.30x+84.31y=0.2089 \]

Substituting \(x=0.003000-y\):

\[ \begin{aligned} 40.30(0.003000-y)+84.31y&=0.2089\\ 0.12090+44.01y&=0.2089 \end{aligned} \]

Hence:

\[ y\approx0.002000\ \mathrm{mol} \]

And:

\[ x\approx0.001000\ \mathrm{mol} \]

The corresponding masses are:

\[ m(\ce{MgO})\approx0.0403\ \mathrm{g} \]

\[ m(\ce{MgCO3})\approx0.1686\ \mathrm{g} \]

The small difference from 0.2089 g when using rounded component masses is due to rounding.

Part (c)

No. If inert sand is also allowed, the two measured quantities no longer uniquely determine all the components.

The titration still tells us that \(x+y=0.003000\ \mathrm{mol}\), but now the mass equation becomes:

\[ 40.30x+84.31y+m_s=0.2089 \]

Here, \(m_s\) is the mass of inert sand in g.

There are three unknowns and only two independent equations.

For example, a sample containing 0.002000 mol MgO and 0.001000 mol MgCO3 would have a reactive mass of:

\[ 0.08060+0.08431=0.16491\ \mathrm{g} \]

Adding 0.04399 g of inert sand gives the same total sample mass of 0.2089 g.

This alternative mixture would consume exactly the same 0.006000 mol of HCl, yet its composition would be different.

The tempting conclusion is that a precise back titration identifies the individual substances responsible for acid consumption. It doesn’t. It measures the combined acid consumption. The known two-component composition and original sample mass provide the extra constraint needed to separate them.

Question 5

A 0.4000 g powder contains magnesium oxide and inert material. It is treated with 30.00 mL of 1.000 \(\mathrm{mol\,L^{-1}}\) HCl.

An analyst makes a serious procedural mistake: before the MgO reaction has finished, the entire mixture is transferred to a 250.0 mL volumetric flask and made up to the mark.

Some undissolved MgO is still present.

The acidic liquid is homogeneous, although the solid remains at the bottom. The analyst withdraws 25.00 mL of clear liquid, taking no solid, and titrates it immediately with 0.1000 \(\mathrm{mol\,L^{-1}}\) NaOH.

The first titre is 20.00 mL.

The remaining 225.0 mL of reaction mixture is left undisturbed until the MgO finishes reacting completely. No further solution is added or removed, and the liquid volume is assumed unchanged.

A second 25.00 mL aliquot of the now-completed mixture requires 12.00 mL of the same NaOH solution.

(a) Calculate the total amount of HCl consumed by MgO across the experiment.

(b) Calculate the mass and mass percentage of MgO in the original sample.

(c) Explain why multiplying the second aliquot’s HCl by \(250/25=10\) gives an incorrect result.

(d) Suppose the analyst had instead allowed the entire original mixture to finish reacting before removing either aliquot. Assuming the same amount of MgO reacted, what titre would each 25.00 mL aliquot require?

Assume there are no other acid-consuming substances, no evaporation or acid loss, complete mixing of the acidic liquid at each sampling time, and negligible liquid-volume changes due to reaction.

Use \(M(\ce{MgO})=40.30\ \mathrm{g\,mol^{-1}}\).

Solution 5

The MgO consumed 0.01720 mol of HCl. The original sample contained 0.3466 g of MgO, or 86.6% by mass. If the reaction had finished before either aliquot was removed, both aliquots would instead require 12.80 mL of NaOH.

The crucial complication is that the first aliquot was removed while the reaction was still happening. That acid was no longer available to react with the MgO remaining in the main flask.

Part (a): Track the acid through both stages

Originally, the analyst added:

\[ \begin{aligned} n_A&=1.000\times0.03000\\ &=0.03000\ \mathrm{mol} \end{aligned} \]

The first 25.00 mL aliquot contained:

\[ \begin{aligned} n_1&=0.1000\times0.02000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

This 0.002000 mol of HCl left the original reaction flask permanently.

The undissolved MgO continued reacting with the acid in the remaining 225.0 mL.

At the end, the second 25.00 mL aliquot contained:

\[ \begin{aligned} n_2&=0.1000\times0.01200\\ &=0.001200\ \mathrm{mol} \end{aligned} \]

At the moment the second aliquot was taken, it represented \(25.00/225.0\) of the remaining reaction solution, not \(25.00/250.0\).

Therefore, the acid still in that remaining solution was:

\[ \begin{aligned} n_{\mathrm{remaining}} &=0.001200\times\frac{225.0}{25.00}\\ &=0.01080\ \mathrm{mol} \end{aligned} \]

But the first aliquot also contained acid that was never consumed by MgO.

Total acid not consumed by the sample:

\[ \begin{aligned} n_L&=0.002000+0.01080\\ &=0.01280\ \mathrm{mol} \end{aligned} \]

Thus:

\[ \begin{aligned} n_C&=0.03000-0.01280\\ &=0.01720\ \mathrm{mol} \end{aligned} \]

There is a useful second way to see the same result.

At the first sampling time, 0.002000 mol in 25.00 mL means there was 0.02000 mol HCl in the complete 250.0 mL mixture.

So MgO had already consumed 0.01000 mol HCl.

After withdrawing the first aliquot, 0.01800 mol HCl remained in the reactor. By the time the reaction finished, only 0.01080 mol remained there.

The additional HCl consumed was:

\[ 0.01800-0.01080=0.007200\ \mathrm{mol} \]

Adding the earlier consumption gives \(0.01000+0.007200=0.01720\ \mathrm{mol}\), confirming the result.

Part (b): Convert acid consumption to MgO mass

Magnesium oxide consumes HCl in a \(1:2\) mole ratio.

\[ \begin{aligned} n(\ce{MgO})&=\frac{0.01720}{2}\\ &=0.008600\ \mathrm{mol} \end{aligned} \]

\[ \begin{aligned} m(\ce{MgO})&=0.008600\times40.30\\ &=0.3466\ \mathrm{g} \end{aligned} \]

\[ \%\ce{MgO} =\frac{0.3466}{0.4000}\times100 \approx86.6\% \]

The calculated MgO mass is below the original sample mass, as required.

Part (c): Why the usual factor of ten fails

Ordinarily, a 25.00 mL aliquot from 250.0 mL represents one-tenth of a complete, homogeneous reaction mixture.

But after the first withdrawal, the remaining reaction flask contained only 225.0 mL.

More importantly, the acid concentration continued changing because MgO was still reacting.

The second aliquot represents the final acid concentration in those remaining 225.0 mL. It cannot be scaled to 250.0 mL as if that volume still existed at the same final concentration.

The 25.00 mL removed earlier contained a different amount of HCl, measured before the reaction finished. It must be accounted for separately.

Part (d): What if the reaction had finished before sampling?

The same quantity of MgO would still consume 0.01720 mol HCl.

With no early aliquot removed, the remaining HCl would be:

\[ 0.03000-0.01720=0.01280\ \mathrm{mol} \]

All of this leftover acid would be uniformly distributed in the 250.0 mL final solution.

Each 25.00 mL aliquot would contain:

\[ \begin{aligned} n_T&=0.01280\times\frac{25.00}{250.0}\\ &=0.001280\ \mathrm{mol} \end{aligned} \]

The expected NaOH volume would be:

\[ \begin{aligned} V_B&=\frac{0.001280}{0.1000}\\ &=0.01280\ \mathrm{L} \end{aligned} \]

So both titres would be 12.80 mL.

Why isn’t the answer 12.00 mL, matching the original second titre?

Because the revised procedure keeps the first aliquot’s acid in the main flask while the reaction finishes. Once the MgO has completely reacted, the leftover HCl is distributed across the full 250.0 mL instead of the reduced 225.0 mL.

This is why ordinary back-titration aliquot calculations require the reaction to be complete before representative aliquots are taken. Without that condition, you must track the actual movement and reaction of the reagent, not simply apply a memorised volume factor.

08The next check: Does the result fit the chemistry?

The aliquot correction is only one part of the calculation. Once you’ve found the reagent consumed, you still need to apply the correct reaction mole ratio and ask whether the inferred sample composition is possible.

For instance, an analyte mass greater than the original powder mass is impossible. But an apparently reasonable mass can still be wrong if some leftover reagent was lost, another substance consumed reagent, or the reaction had not finished when the aliquot was taken.

For a closer look at that final numerical check, see How to Check a Back-Titration Answer Against Sample Mass.

If you need to revisit the underlying acid-added-minus-acid-leftover method before dealing with multiple dilution stages, the back titration of an antacid provides a simpler starting case.

The next useful skill is deciding whether a titre genuinely represents all the leftover reagent. Once that is established, the mole calculation is straightforward: measure the aliquot, restore the correct fraction of the original reaction mixture, subtract from the amount added, and use the balanced equation to find what reacted.