Corrected exercises: titrations and buffer solutions (202-SN2-RE)

$\mathrm{CH_{3}COOH}+\mathrm{OH^{-}}\rightarrow\mathrm{CH_{3}COO^{-}}+\mathrm{H_{2}O}$. The reaction is quantitative, its constant being $\dfrac{K_{a}}{K_{w}}=1.8\times 10^{9}$. Equivalence: $C_{a}V_{a}=C_{b}V_{b}$, so $V_{\text{eq}}=\dfrac{0.100\times 25.00}{0.100}=25.00$ mL. Equal concentrations make the equivalence volume equal to the initial volume, which is convenient but must not become a reflex.

Acetic acid is weak and ionizes only partially: $[\mathrm{H_{3}O^{+}}]=\sqrt{K_{a}C}=\sqrt{1.8\times 10^{-5}\times 0.100}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}$ M, hence $\mathrm{pH}=2.87$. Checking the approximation: the percent ionization is $\dfrac{1.34\times 10^{-3}}{0.100}=1.3$%, comfortably below the usual 5% limit. A STRONG acid of the same concentration would give $\mathrm{pH}=1.00$: almost two units apart, a factor of seventy-five on the hydronium ion concentration.

At $10.00$ mL, $1.00$ mmol of acetate has formed and $1.50$ mmol of acid remains: this is a buffer, and the Henderson-Hasselbalch equation applies. $\mathrm{pH}=4.74+\log\dfrac{1.00}{1.50}=4.74-0.18=4.57$. At $12.50$ mL, exactly half the equivalence volume, $1.25$ mmol of acetate has formed and $1.25$ mmol of acid remains: the ratio is 1, its logarithm is zero, and $\mathrm{pH}=\mathrm{p}K_{a}=4.74$. This is the second piece of information on the curve, and the most rewarding one: the pKa is READ directly on the vertical axis at the half-equivalence point, with no calculation, without knowing the concentration, and even without the acid being identified. This is how an unknown acid is identified in the laboratory.

At the equivalence point, all the acid has become acetate, at the concentration $C=\dfrac{0.100\times 25.00}{50.00}=0.0500$ M, the dilution by the total volume being mandatory. Acetate is a weak BASE that hydrolyzes: $K_{b}=\dfrac{K_{w}}{K_{a}}=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}$. Then $[\mathrm{OH^{-}}]=\sqrt{5.56\times 10^{-10}\times 0.0500}=5.27\times 10^{-6}$ M, $\mathrm{pOH}=5.28$ and $\mathrm{pH}=8.72$. The pH is not 7 because the product of the titration is not neutral: it is a base. Writing “at the equivalence point the pH equals 7” is the most heavily penalized error of the chapter, and it has an immediate practical consequence: it leads to the wrong choice of indicator.

At $26.00$ mL, there is an excess of $1.00$ mL of strong base: $[\mathrm{OH^{-}}]=\dfrac{0.100\times 1.00}{51.00}=1.96\times 10^{-3}$ M, $\mathrm{pOH}=2.71$ and $\mathrm{pH}=11.29$. The excess strong base completely swamps the hydrolysis of acetate, which is therefore neglected. At $24.00$ mL, we are still in the buffer region: $\mathrm{pH}=4.74+\log\dfrac{24.00}{1.00}=6.12$. The jump between $24.00$ and $26.00$ mL is therefore $11.29-6.12=5.17$ pH units for two millilitres, and it is almost entirely concentrated in the last two drops. This is the fourth piece of information on the curve: a jump of more than two units makes the titration usable, a jump of less than one unit makes it unusable, which happens with very weak or very dilute acids.

$\mathrm{NH_{3}}+\mathrm{H_{3}O^{+}}\rightarrow\mathrm{NH_{4}^{+}}+\mathrm{H_{2}O}$. Equivalence: $n(\mathrm{NH_{3}})=n(\mathrm{HCl})$, that is $0.150\times 20.00=0.100\times V_{\text{eq}}$, so $V_{\text{eq}}=30.00$ mL. Be careful, the concentrations are DIFFERENT here, and the equivalence volume has no reason to equal the initial volume: this is a classic error for anyone who memorized the previous exercise instead of the method.

Ammonia is a weak base: $[\mathrm{OH^{-}}]=\sqrt{K_{b}C}=\sqrt{1.8\times 10^{-5}\times 0.150}=\sqrt{2.70\times 10^{-6}}=1.64\times 10^{-3}$ M. Hence $\mathrm{pOH}=2.78$ and $\mathrm{pH}=11.22$. A strong base of the same concentration would give $\mathrm{pH}=13.18$: two units higher.

At $15.00$ mL, half of the ammonia has been converted to ammonium ion: $[\mathrm{NH_{3}}]=[\mathrm{NH_{4}^{+}}]$, and $\mathrm{pH}=\mathrm{p}K_{a}(\mathrm{NH_{4}^{+}})$. This pKa equals $14-\mathrm{p}K_{b}=14-4.74=9.26$, so $\mathrm{pH}=9.26$. What is read here is therefore the pKa of the CONJUGATE ACID, and care must be taken not to read the pKb: the pH meter only ever measures pKa values, and the conversion by $\mathrm{p}K_{a}+\mathrm{p}K_{b}=14$ is to be done afterwards.

At the equivalence point, all the ammonia has become ammonium ion, at the concentration $C=\dfrac{0.150\times 20.00}{50.00}=0.0600$ M. Ammonium ion is a weak ACID, with constant $K_{a}=\dfrac{K_{w}}{K_{b}}=5.56\times 10^{-10}$: $[\mathrm{H_{3}O^{+}}]=\sqrt{5.56\times 10^{-10}\times 0.0600}=5.77\times 10^{-6}$ M, that is $\mathrm{pH}=5.24$. The pH is acidic because the species PRESENT at the equivalence point is an acid, and not because the strong acid would be in excess, which it is not. This is the exact mirror image of the previous exercise: titrating a weak acid ends on the basic side, titrating a weak base ends on the acidic side. The titrant has disappeared, only the product counts.

At $35.00$ mL, there is $5.00$ mL of strong acid in excess: $[\mathrm{H_{3}O^{+}}]=\dfrac{0.100\times 5.00}{55.00}=9.09\times 10^{-3}$ M, that is $\mathrm{pH}=2.04$. The jump therefore occurs around $\mathrm{pH}=5.24$, going downward. An indicator is needed whose transition range brackets this value: METHYL RED, which changes between $4.4$ and $6.2$, is the obvious choice. Phenolphthalein must be ruled out absolutely: it changes between $8.2$ and $10.0$, that is well BEFORE the equivalence point, when plenty of ammonia is still left to titrate. The titration would be stopped too early and the concentration seriously underestimated. Bromothymol blue, which changes from $6.0$ to $7.6$, would also be too high, although less seriously.

First reaction: $\mathrm{CO_{3}^{2-}}+\mathrm{H_{3}O^{+}}\rightarrow\mathrm{HCO_{3}^{-}}+\mathrm{H_{2}O}$, the strongest base reacting first. Second: $\mathrm{HCO_{3}^{-}}+\mathrm{H_{3}O^{+}}\rightarrow\mathrm{H_{2}CO_{3}}+\mathrm{H_{2}O}$. Each step consumes one mole of protons per mole of initial carbonate: $n=0.100\times 25.00=2.50$ mmol, so $V_{\text{eq}1}=\dfrac{2.50}{0.100}=25.00$ mL and $V_{\text{eq}2}=50.00$ mL, exactly double. The equality of the two intervals is the signature of a diprotic acid, and it is what lets you tell a diprotic acid from a mixture of two different acids on a curve, where the intervals have no reason to be equal.

At the first equivalence point, the majority species is the hydrogen carbonate ion, which is AMPHIPROTIC: it can give up a proton, with $K_{a2}$, as well as accept one, with $K_{a1}$. Treating these two simultaneous equilibria gives $\mathrm{pH}=\dfrac{\mathrm{p}K_{a1}+\mathrm{p}K_{a2}}{2}=\dfrac{6.37+10.33}{2}=8.35$. The concentration does NOT enter, and this is mathematically remarkable: in the derivation it cancels between the acid reaction and the base reaction, which involve it in the same way. A useful practical consequence: an accidental dilution of the sample does not move this point.

At $12.50$ mL, halfway through the first jump, $[\mathrm{CO_{3}^{2-}}]=[\mathrm{HCO_{3}^{-}}]$: this pair is the one of $K_{a2}$, so $\mathrm{pH}=\mathrm{p}K_{a2}=10.33$. At $37.50$ mL, the midpoint between the two equivalence points, $[\mathrm{HCO_{3}^{-}}]=[\mathrm{H_{2}CO_{3}}]$: this pair is the one of $K_{a1}$, so $\mathrm{pH}=\mathrm{p}K_{a1}=6.37$. A single titration curve therefore gives BOTH acid dissociation constants, each at a point that only has to be read. This is the standard method for determining the pKa values of a polyprotic acid, and it explains why these two plateaus are the flattest regions of the whole curve: they are two buffer regions.

At the second equivalence point, everything is in the form of carbonic acid, at the concentration $C=\dfrac{2.50}{75.00}=0.0333$ M, taking into account the dilution by the $50.00$ mL added. It is a weak acid alone: $[\mathrm{H_{3}O^{+}}]=\sqrt{K_{a1}C}=\sqrt{4.3\times 10^{-7}\times 0.0333}=1.20\times 10^{-4}$ M, that is $\mathrm{pH}=3.92$. Only $K_{a1}$ is used, the second ionization being four orders of magnitude weaker and therefore negligible.

First equivalence point, $\mathrm{pH}=8.35$: PHENOLPHTHALEIN, which changes between $8.2$ and $10.0$, is roughly suitable. Second equivalence point, $\mathrm{pH}=3.92$: METHYL ORANGE, which changes between $3.1$ and $4.4$, is the classic choice. The first is much harder to locate, for a quantitative reason visible on the figure: the jump there is small, a fraction of a pH unit, because the two neighbouring constants, $\mathrm{p}K_{a1}=6.37$ and $\mathrm{p}K_{a2}=10.33$, differ by only four units. A sharp jump requires a gap of at least four to five units between two successive pKa values, and here we are at the limit. The second jump, for its part, takes place against a neutral species and is much sharper. In practice it is often preferable to titrate the carbonate in a single run to the methyl orange end point, heating to drive off the dissolved carbon dioxide, which makes the second jump even sharper.

$n_{T}=0.100\times 0.1000=1.00\times 10^{-2}$ mol, split in equal parts: $5.00$ mmol of acetic acid and $5.00$ mmol of acetate ion. This is indeed an equimolar buffer, hence at $\mathrm{pH}=\mathrm{p}K_{a}$.

The strong base attacks the weak acid of the buffer, quantitatively: $\mathrm{CH_{3}COOH}+\mathrm{OH^{-}}\rightarrow\mathrm{CH_{3}COO^{-}}+\mathrm{H_{2}O}$. The acid goes from $5.00$ to $4.00$ mmol and the acetate from $5.00$ to $6.00$ mmol. Henderson-Hasselbalch: $\mathrm{pH}=4.74+\log\dfrac{6.00}{4.00}=4.74+0.18=4.92$. The change is $\Delta\mathrm{pH}=+0.18$, barely noticeable. Note the structure of the calculation: the stoichiometry is done first in MOLES, all the way through, and the Henderson-Hasselbalch equation is applied only afterwards. Mixing the two steps is the main source of error.

In pure water, the sodium hydroxide is alone and fully dissociated: $[\mathrm{OH^{-}}]=\dfrac{1.00\times 10^{-3}}{0.1000}=1.00\times 10^{-2}$ M, so $\mathrm{pOH}=2.00$ and $\mathrm{pH}=12.00$. The change is $+5.00$ units against $+0.18$ in the buffer, a factor of twenty-eight on the pH change and a factor of more than ten thousand on the final hydroxide ion concentration. This is the numerical demonstration of what a buffer is, and it is this comparison, rather than the definition, that you must be able to produce on an exam.

$\beta=\dfrac{\Delta n_{\text{base}}}{V\,\Delta\mathrm{pH}}=\dfrac{1.00\times 10^{-3}}{0.1000\times 0.176}=5.68\times 10^{-2}$ mol/(L·pH). The theoretical maximum is $0.576\times 0.100=5.76\times 10^{-2}$: the agreement is within $1.4$%. The slight gap comes from the fact that $\beta$ is a DERIVATIVE, defined for an infinitesimal addition, whereas we measured a finite change over which $\beta$ has already started to decrease. The top of the curve is indeed at $\mathrm{pH}=\mathrm{p}K_{a}$, which is the reason for choosing a buffer whose pKa is as close as possible to the target pH.

In the $10:1$ ratio, the $10.0$ mmol in total split into $9.09$ mmol of acetate and $0.909$ mmol of acetic acid. Adding $1.00$ mmol of base would require consuming $1.00$ mmol of acid, whereas only $0.909$ mmol is there: the reserve is EXHAUSTED before the end, the buffer is broken, and the excess base is left free in solution, which sends the pH soaring above 11. With only $0.500$ mmol: the acid falls to $0.409$ mmol and the acetate rises to $9.59$ mmol, so $\mathrm{pH}=4.74+\log\dfrac{9.59}{0.409}=4.74+1.37=6.11$. The change is $6.11-5.74=0.37$ unit for HALF of the base, that is $\beta=\dfrac{5.00\times 10^{-4}}{0.1000\times 0.37}=1.35\times 10^{-2}$, four times lower than at the top. The lesson is twofold: a buffer one unit away from its pKa has already lost about two thirds of its efficiency, and above all its BUFFERING CAPACITY, that is the amount of base it can take before breaking down, is limited by the amount of each of its two forms taken separately. A very lopsided buffer resists well in one direction and not at all in the other.

The pH stays at $4.74$. The Henderson-Hasselbalch equation, $\mathrm{pH}=\mathrm{p}K_{a}+\log\dfrac{[\text{base}]}{[\text{acid}]}$, involves only the RATIO of the two concentrations. A dilution divides both by the same factor, so it leaves the ratio strictly unchanged, and the logarithm is still zero. This result is disconcerting because it contradicts the intuition gained on strong acids, whose pH rises by one unit with every tenfold dilution. A buffer, for its part, is insensitive to dilution, and that is precisely one of its qualities: a poorly rinsed pipette does not spoil the measurement.

Concentrated sample, $5.00$ mmol of each form. After the addition: $5.10$ mmol of acetate and $4.90$ mmol of acid, hence $\mathrm{pH}=4.74+\log\dfrac{5.10}{4.90}=4.74+0.0174=4.76$, that is $\Delta\mathrm{pH}=0.017$. Diluted sample, $0.500$ mmol of each form. After the addition: $0.600$ mmol of acetate and $0.400$ mmol of acid, hence $\mathrm{pH}=4.74+\log\dfrac{0.600}{0.400}=4.74+0.176=4.92$, that is $\Delta\mathrm{pH}=0.18$. The same addition produces ten times more drift in the diluted buffer.

$\beta_{\text{concentrated}}=\dfrac{1.00\times 10^{-4}}{0.1000\times 0.0174}=5.76\times 10^{-2}$ and $\beta_{\text{diluted}}=\dfrac{1.00\times 10^{-4}}{0.1000\times 0.176}=5.68\times 10^{-3}$. The ratio is $10.1$, in agreement with the prediction $\beta_{\max}=0.576\,C_{T}$, which is proportional to the total concentration. The buffer index therefore follows the dilution exactly, whereas the pH is not sensitive to it at all.

The pH of a buffer is set by the RATIO of the two forms, its capacity is set by their AMOUNT: diluting changes the second without touching the first. This is the sentence to remember, and it has a practical consequence in the laboratory: you choose the pair from the target pH, looking for a nearby pKa, and you choose the concentration from the amount of acid or base the buffer will have to absorb. The two decisions are independent.

No, and the reasoning is the same as for a very dilute strong acid. At $1.00\times 10^{-7}$ M, the concentrations of the two forms of the buffer are of the same order as those produced by the AUTOIONIZATION of water, which on its own supplies $10^{-7}$ M of hydronium and hydroxide ions. The implicit assumption of the Henderson-Hasselbalch equation, that the amounts of acid and base introduced remain large compared with what water produces, collapses. The full balance would then have to be written, with conservation of matter, electrical neutrality and the ion product of water. The result, physically obvious, is that the pH drifts back toward $7.00$: almost pure water is water, and a buffer cannot be made out of nothing. The practical rule is that the Henderson-Hasselbalch equation requires concentrations of at least $10^{-3}$ M to remain reliable.

HCl with NaOH: the product is NaCl, the salt of a strong acid and a strong base, which does not hydrolyze, so $\mathrm{pH}_{\text{eq}}=7.00$. The jump there is enormous, of the order of six units between $24.9$ and $25.1$ mL, and it encloses both transition ranges: BOTH indicators are suitable, which is why strong-strong titrations are so convenient. Acetic acid with NaOH: $\mathrm{pH}_{\text{eq}}=8.72$, calculated in exercise 1. Only phenolphthalein is suitable, its range of $8.2$ to $10.0$ containing this value; methyl orange, which changes four units lower, is ruled out.

Beyond the equivalence point, the pH is set by the excess of strong base. $\mathrm{pH}=8.2$ gives $\mathrm{pOH}=5.8$, that is $[\mathrm{OH^{-}}]=1.58\times 10^{-6}$ M. This excess occupies about $50$ mL, so $n_{\text{excess}}=1.58\times 10^{-6}\times 0.0500=7.9\times 10^{-8}$ mol, which corresponds to $V=\dfrac{7.9\times 10^{-8}}{0.100}=7.9\times 10^{-7}$ L, that is $7.9\times 10^{-4}$ mL. The relative error is $\dfrac{7.9\times 10^{-4}}{25.00}=3.2\times 10^{-5}$, that is $0.0032$%. In other words less than one hundredth of a drop: completely undetectable, and negligible next to the reading uncertainty of the burette, which is of the order of $0.05$ mL, that is $0.2$%.

At $\mathrm{pH}=4.4$, we are still squarely in the buffer region, so the Henderson-Hasselbalch equation applies: $4.4=4.74+\log\dfrac{x}{1-x}$, where $x$ is the fraction of acid already converted to acetate. Hence $\log\dfrac{x}{1-x}=-0.34$, that is $\dfrac{x}{1-x}=0.457$ and $x=0.311$. Volume read: $0.311\times 25.00=7.78$ mL. Relative error: $\dfrac{7.78-25.00}{25.00}=-68.9$%. The disaster is total: you would conclude that the solution is three times less concentrated than it really is. And the worst of it is that nothing in the experiment signals the anomaly, since the color change occurs sharply, with a clear change of color.

$8.2=4.74+\log\dfrac{V}{25.00-V}$ gives $\dfrac{V}{25.00-V}=10^{3.455}=2851$, hence $V=\dfrac{25.00\times 2851}{2852}=24.991$ mL. Relative error: $\dfrac{24.991-25.00}{25.00}=-3.5\times 10^{-4}$, that is $-0.035$%. This is two thousand times better than methyl orange and, again, far below the precision of the burette. The right indicator therefore brings the systematic error well below the reading error, which is exactly the selection criterion.

Operating rule: the transition range of the indicator must be contained in the pH JUMP of the titration, and ideally it brackets the pH at the equivalence point. Since this pH is almost never 7, it must be calculated BEFORE choosing the indicator, not the other way round. A titration followed with a pH meter avoids the problem completely because it relies on no arbitrary threshold: you plot the whole curve and look for the inflection point, that is the maximum of the derivative, which is the equivalence point by mathematical definition. The indicator is then no more than a visual convenience. This is also why the pH meter is indispensable whenever the jump is small, with very weak or very dilute acids, where no indicator would change color sharply.

Interval $[24.80, 24.90]$: $\dfrac{7.76-7.46}{0.10}=3.0$, plotted at $V=24.85$. Interval $[24.90, 25.00]$: $\dfrac{8.72-7.76}{0.10}=9.6$, at $V=24.95$. Interval $[25.00, 25.10]$: $\dfrac{9.68-8.72}{0.10}=9.6$, at $V=25.05$. Interval $[25.10, 25.20]$: $\dfrac{9.98-9.68}{0.10}=3.0$, at $V=25.15$. The point of method that costs marks on an exam is the placement on the HORIZONTAL axis: a derivative computed between two volumes is plotted at the MIDPOINT of the interval, never at one of its endpoints.

The equivalence point is the inflection point of $\mathrm{pH}=f(V)$, hence the maximum of the derivative. Here the two central values are exactly equal, $9.6$ at $24.95$ and $9.6$ at $25.05$: the derivative is symmetric about the midpoint of these two abscissas, so its maximum lies at $V_{\text{eq}}=\dfrac{24.95+25.05}{2}=25.00$ mL. The symmetry argument is what lets you go beyond the resolution of the measurements: the pH was never measured exactly at the equivalence point, and yet it is located to the hundredth of a millilitre.

$C_{a}=\dfrac{C_{b}V_{\text{eq}}}{V_{a}}=\dfrac{0.1000\times 25.00}{20.00}=0.1250$ M. The order-of-magnitude check is immediate: more base than acid was needed by volume although the base is less concentrated, so the acid must be more concentrated than $0.1000$ M, and it is.

The acid is WEAK: a strong acid titrated with a strong base would give $\mathrm{pH}_{\text{eq}}=7.00$, and we observe $8.72$, clearly basic, which signals the hydrolysis of the conjugate base formed. The pKa is read at the half-equivalence point, that is at $\dfrac{25.00}{2}=12.50$ mL, where the pH meter read $4.74$: so $\mathrm{p}K_{a}=4.74$, that is $K_{a}=1.8\times 10^{-5}$. This is the value for acetic acid, which strongly suggests the identification. Two readings on a single curve thus give the concentration AND the identity of the acid.

Two objective advantages. First OBJECTIVITY: two experimenters drawing a tangent freehand do not obtain the same point, whereas the derivative is a reproducible calculation that depends on no visual judgment. Second RESOLUTION: as in b), the symmetry of the surrounding points makes it possible to interpolate between two measurements and to locate the equivalence point more finely than the step of the burette. You can even go further with the SECOND derivative, which vanishes exactly at the equivalence point and provides an even more convenient linear interpolation. The limitation is NOISE: differentiating amplifies measurement fluctuations, since small differences are divided by small intervals. An unstable pH meter or additions that are too small produce a sawtooth derivative in which the maximum becomes impossible to pin down, and the data must then either be smoothed or measured with a larger step. The $0.10$ mL step used here is a good compromise for a jump of this size.

False, and this is the costliest error of the chapter. At the equivalence point, the reactants have been mixed in stoichiometric proportions, which says nothing about the pH: everything depends on the PRODUCT formed. Strong-strong titration, the salt is inert, $\mathrm{pH}=7.00$. Titration of a weak acid, its conjugate base is formed and hydrolyzes, $\mathrm{pH}>7$, here $8.72$. Titration of a weak base, its conjugate acid is formed, $\mathrm{pH}<7$, here $5.24$. Correct statement: the equivalence point is a condition of STOICHIOMETRY, not of neutrality, and the pH of the product must be calculated to know it. It is from this error that the wrong choice of indicator in exercise 6 follows, with its $69$% deviation.

False twice over. First, a pH cannot be divided, it is a logarithm: halving a concentration lowers the pH by $\log 2=0.30$ unit, not by half. Then, and above all, a dilution does NOT change the pH of a buffer, since the Henderson-Hasselbalch equation involves only the ratio of the two forms, which the dilution leaves intact. Correct statement: diluting a buffer leaves its pH practically unchanged and divides its buffer index in the same proportion as the dilution. The numerical demonstration is the one of exercise 5, where the pH stays at $4.74$ while $\beta$ goes from $5.76\times 10^{-2}$ to $5.68\times 10^{-3}$.

False by excess. A buffer has a finite CAPACITY, set by the amount of its two forms: when one of them is used up, it is broken and the pH swings abruptly. Exercise 4 gives the example: a buffer in the ratio $10:1$ containing $0.909$ mmol of acid cannot absorb $1.00$ mmol of base. Correct statement: a buffer LIMITS the change in pH as long as you stay within its capacity, roughly $\mathrm{p}K_{a}\pm 1$, and it stops working altogether beyond it. It cushions, it does not lock.

False, and the confusion comes from a vocabulary analogy, the word “half” applying to the VOLUME and not to the pH. At the half-equivalence point, half of the equivalence volume has been added, so half of the acid has been converted, hence $[\text{acid}]=[\text{base}]$ and the logarithm of the Henderson-Hasselbalch equation vanishes. Correct statement: at the half-equivalence point, $\mathrm{pH}=\mathrm{p}K_{a}$. The numerical counterexample is immediate: acetic acid gives $4.74$ at the half-equivalence point and $8.72$ at the equivalence point, and $4.74$ is not half of $8.72$.

False. An indicator changes color at ITS OWN transition pH, which depends only on its pKi, and absolutely not on where the equivalence point of the titration at hand lies: the two have no reason to coincide. The calculation of exercise 6 puts a number on it: titrating acetic acid with methyl orange makes the color change at $7.78$ mL instead of $25.00$ mL, an error of $-69$%, and the color change is perfectly sharp, which makes the error invisible. Correct statement: the indicator must be chosen so that its transition range is contained in the jump of the titration, which means calculating the pH at the equivalence point BEFORE choosing it. Only in the strong-strong case, where the jump spans six units, does the choice become immaterial.

Attack of the tablet: $\mathrm{CaCO_{3}}+2\,\mathrm{H_{3}O^{+}}\rightarrow\mathrm{Ca^{2+}}+\mathrm{CO_{2}}+3\,\mathrm{H_{2}O}$, with release of carbon dioxide. Titration of the excess: $\mathrm{H_{3}O^{+}}+\mathrm{OH^{-}}\rightarrow 2\,\mathrm{H_{2}O}$. A back titration is used for three reasons that add up. Calcium carbonate is a sparingly soluble SOLID, whose direct dissolution in a flask during a titration would be far too slow, the end of the reaction never being sharp. The tablet also contains excipients, starch, cellulose, flavourings, which cloud the solution and make any color change unreadable. Finally the evolution of gas stirs the medium and prevents any stable reading. By drowning the solid in an excess of acid, you give it all the time it needs to react completely, and then you titrate a clear, homogeneous liquid.

Acid introduced: $n=0.05000\times 0.500=2.500\times 10^{-2}$ mol, that is $25.00$ mmol. Acid remaining, equal to the amount of sodium hydroxide added: $n=0.02240\times 0.250=5.600\times 10^{-3}$ mol, that is $5.600$ mmol. Acid consumed by the carbonate: $25.00-5.600=19.40$ mmol. This is the entire structure of a back titration, a simple subtraction, and it requires the first volume to be measured with care as well: a volumetric pipette, never a graduated cylinder.

The stoichiometry is TWO protons per carbonate: $n(\mathrm{CaCO_{3}})=\dfrac{19.40}{2}=9.700$ mmol. Forgetting this factor of 2 is the most common error of the exercise, and it doubles the result without anything else looking wrong. Mass: $m=9.700\times 10^{-3}\times 100.09=0.9709$ g. Mass percentage: $\dfrac{0.9709}{1.250}=0.7767$, that is $77.7$%. A tablet with nearly eighty percent active ingredient is plausible, the rest being excipients and binder.

The carbon dioxide formed dissolves partially and forms carbonic acid, which is a WEAK acid but a very real one. Without degassing, the sodium hydroxide would titrate some of it in addition to the remaining hydrochloric acid, and the equivalence volume would therefore be too LARGE. The remaining acid would then be overestimated, so the acid consumed would be underestimated, so the carbonate content would be underestimated. The error systematically acts toward a content that is too low, and that is why the procedure requires gentle heating or prolonged stirring before titrating.

An uncertainty of $\pm 0.05$ mL on $22.40$ mL represents $0.22$% on the volume of sodium hydroxide, hence on the $5.600$ mmol remaining, that is $\pm 0.0125$ mmol. But this uncertainty applies to a quantity that is SUBTRACTED from $25.00$ mmol: relative to the $19.40$ mmol consumed, it weighs only $\dfrac{0.0125}{19.40}=0.064$%. The back titration is therefore particularly favourable here, because the titrated excess is small compared with the quantity sought: the error is diluted in the subtraction. The situation would be reversed if the excess were very large, say if $22.0$ mmol remained to be titrated for $3.0$ mmol consumed, in which case the same absolute uncertainty would weigh $0.4$% on the result. The practical rule is therefore to choose a modest excess, typically twenty to thirty percent beyond what is needed, and certainly not ten times too much.