Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: titrations and buffer solutions (202-SN2-RE)

Q: Is the pH always 7 at the equivalence point of a titration?

No, almost never. The equivalence point is a condition of stoichiometry, not of neutrality: it is the product formed that sets the pH. Titrating a weak acid leaves its conjugate base in solution, so the pH is basic, often around 8.7. Titrating a weak base leaves its conjugate acid, so the pH is acidic. The pH is 7 only if the acid and the base are both strong.

Q: How do you read the pKa on a titration curve?

You locate the equivalence volume, divide it by two, and read the pH at that volume: it is exactly the pKa. At the half-equivalence point, half of the acid has been converted to its conjugate base, so the two forms are present in equal amounts and the logarithmic term of the Henderson-Hasselbalch equation vanishes. No calculation, no concentration needed.

Q: What happens when you dilute a buffer solution?

Its pH hardly changes, because the Henderson-Hasselbalch equation involves only the ratio of the two forms, and the dilution divides both by the same factor. Its buffer index, however, is divided in the same proportion as the dilution: the pH holds, the robustness does not. Below about one millimole per litre, the equation even stops being reliable.

Q: How do you choose the right acid-base indicator?

You first calculate the pH at the equivalence point, then choose an indicator whose transition range brackets that value and fits entirely inside the pH jump. For a weak acid titrated with a strong base, the equivalence point is basic and phenolphthalein is needed. Using methyl orange in that case throws the titration off by nearly seventy percent.

Q: Does a buffer keep the pH constant whatever amount of acid is added?

No. A buffer has a finite capacity, set by the amount of each of its two forms. When one is used up, the buffer is broken and the pH swings abruptly. It cushions, it does not lock. Its efficiency is greatest when the pH equals the pKa, and it has already lost about two thirds of its value one unit away.

Q: Why do a back titration rather than a direct titration?

Because the direct reaction is sometimes too slow, incomplete, or involves a solid. A known excess of reagent is then added, the reaction is allowed to go to completion, and the excess is titrated. The quantity sought is obtained by difference. It is the standard method for analyzing an antacid tablet, whose carbonate dissolves too slowly.

This sheet does not redo the course on titrations, it lists what loses marks on test day. A single wrong idea explains half of the errors of the chapter, and it is stubborn: believing that the solution is neutral at the equivalence point. It almost never is, and the wrong choice of indicator follows from it, which can throw a titration off by nearly seventy percent.

It is written for CEGEP students in the Natural Science and Health Science programs taking 202-SN2-RE (formerly 202-NYB).

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The thread of the chapter

The equivalence point is a condition of STOICHIOMETRY, never of neutrality: the pH there is 7 only when the acid and the base are both strong, and it is the pH of the PRODUCT formed that decides everything else.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Titration curve
Equivalence volume
Equivalence point
Half-equivalence point
Henderson-Hasselbalch equation
Buffer solution
Buffer index
Hydrolysis of a salt
Titration of a diprotic acid
Acid-base indicator
Transition range
Back titration

The essentials

A titration curve contains four pieces of information

• The EQUIVALENCE VOLUME gives the concentration sought: $C_{a}V_{a}=C_{b}V_{b\text{,eq}}$ for a monoprotic acid and a monobasic base.
• The pH at the HALF-EQUIVALENCE point gives the $\mathrm{p}K_{a}$ directly, with no calculation and without knowing the concentration: that is where $[\text{acid}]=[\text{base}]$ and the logarithm of the Henderson-Hasselbalch equation vanishes.
• The pH at the EQUIVALENCE point tells you which indicator to choose. It is 7 for a strong-strong titration, it is BASIC when a weak acid is titrated, and ACIDIC when a weak base is titrated.
• The HEIGHT OF THE JUMP tells you whether the titration is usable: above two pH units yes, below one unit no, which happens with very weak or very dilute acids.

At the half-equivalence point,

12.50

mL, the pH is exactly the

\mathrm{p}K_{a}

4.74

. At the equivalence point,

25.00

mL, it is

8.72

, clearly basic and not 7.

Buffers: what sets the pH and what sets the robustness

• $\mathrm{pH}=\mathrm{p}K_{a}+\log\dfrac{[\text{base}]}{[\text{acid}]}$ , valid only in the range $\mathrm{p}K_{a}\pm 1$ . The pH is set by the RATIO of the two forms.
• The buffer index $\beta=\dfrac{dn_{\text{base}}}{V\,d(\mathrm{pH})}$ is set by the AMOUNT of the two forms. It is maximal at $\mathrm{pH}=\mathrm{p}K_{a}$ , where it equals $0.576\,C_{T}$ .
• Diluting a buffer therefore does NOT change its pH, since the ratio is intact, and divides its buffer index in the same proportion. The pH holds, the robustness does not.
• pH of a weak acid alone: $[\mathrm{H_{3}O^{+}}]=\sqrt{K_{a}C}$ . pH of an amphiprotic species: $\dfrac{\mathrm{p}K_{a1}+\mathrm{p}K_{a2}}{2}$ , independent of the concentration.

The buffer index peaks at the

\mathrm{p}K_{a}

in both cases: dilution does not move the peak, so it does not change the pH, but it makes the peak ten times lower.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. A $25.00$ mL sample of $0.100$ M acetic acid is titrated with $0.100$ M NaOH. What is the pH at the equivalence point?

3 points, plus the wrong indicator that follows from it and spoils the whole titration

What not to write

At the equivalence point all the acid has been neutralized, so the solution is neutral and $\mathrm{pH}=7.00$ .

What to write

At the equivalence point only ACETATE is left, at $0.0500$ M, which is a weak base. $K_{b}=5.56\times 10^{-10}$ , $[\mathrm{OH^{-}}]=5.27\times 10^{-6}$ M and $\mathrm{pH}=8.72$ .

Same starting acid, same equivalence volume: with HCl the equivalence point falls at

7.00

, with acetic acid at

8.72

, because the acetate formed is a base.

Why: The equivalence point is a condition of STOICHIOMETRY: the reactants have been mixed in exact proportions, which says nothing about the pH. It is the PRODUCT formed that decides, and it has no reason to be inert. The word “neutralize” is misleading, it dates from nineteenth-century chemistry and does not mean “make neutral”.

2. The same acetic acid is titrated with methyl orange, which changes around $\mathrm{pH}=4.4$ , rather than with phenolphthalein.

The whole value of the titration, and often 4 points

What not to write

The choice of indicator does not matter, the color change always occurs at the equivalence point since that is where the pH changes abruptly.

What to write

At $\mathrm{pH}=4.4$ , the Henderson-Hasselbalch equation gives a titrated fraction of $0.311$ : the color change occurs at $7.78$ mL instead of $25.00$ mL, an error of $-68.9$ %.

Methyl orange changes at

7.78

mL, well before the equivalence point, whereas phenolphthalein changes at

24.99

mL: the first reading is at

69

% of the right result, the second within

0.035

Why: An indicator changes at ITS OWN transition pH, set by its own $\mathrm{p}K_{i}$ , with no relation to the titration at hand. The worst part is that the color change remains sharp and clear: nothing in the experiment signals the anomaly. The rule is to calculate the pH at the equivalence point BEFORE choosing, then to check that the transition range is contained in the jump.

3. An equimolar acetate buffer of pH $4.74$ is diluted tenfold.

2 points, and a lasting confusion between a buffer and a weak acid

What not to write

The pH rises by one unit, as for an acid diluted tenfold: it goes to $5.74$ .

What to write

The pH stays at $4.74$ . The dilution divides both forms by the same factor, so it leaves the RATIO intact, and the Henderson-Hasselbalch equation depends only on this ratio.

Diluting the buffer a thousand times barely moves its pH: the ratio of the two forms is intact. The red line is the reasoning for a lone acid, wrongly applied.

Why: This is the most counterintuitive behaviour of the chapter, because it contradicts the intuition gained on strong acids. What does change, and strongly, is the maximum theoretical BUFFER INDEX: it goes from $5.76\times 10^{-2}$ to $5.76\times 10^{-3}$ , ten times lower. Remember the sentence: the pH is set by the ratio, the capacity by the amount.

4. An acetate buffer in the ratio $10:1$ has to absorb $1.00$ mmol of sodium hydroxide.

3 points, often on the last question of a problem

What not to write

A buffer keeps the pH constant whatever the amount of base added: the pH will stay at $5.74$ .

What to write

This buffer contains only $0.909$ mmol of acid: the reserve is EXHAUSTED before the end, the buffer is broken, and the excess base sends the pH soaring above 11.

10:1

buffer has a large reserve against an acid,

9.09

mmol, and a very small one against a base,

0.909

mmol:

1.00

mmol of sodium hydroxide uses it up.

Why: A buffer has a FINITE capacity, set by the amount of each of its two forms taken separately. A lopsided buffer resists well in one direction and not at all in the other. This is also why you choose a pair whose $\mathrm{p}K_{a}$ is as close as possible to the target pH: one unit away, about two thirds of the efficiency is already lost.

5. On a titration curve, you read the pH at the half-equivalence point.

2 points, and the missed chance to identify the acid

What not to write

The half-equivalence point gives half of the pH at the equivalence point: $\dfrac{8.72}{2}=4.36$ .

What to write

The half-equivalence point gives the $\mathrm{p}K_{a}$ , here $4.74$ . The word “half” refers to the VOLUME added, never to the pH.

“Half” refers to the VOLUME: at

12.50

mL we read the

\mathrm{p}K_{a}

4.74

, and not half of the equivalence pH.

Why: At half of the equivalence volume, half of the acid has been converted to its conjugate base, so $[\text{acid}]=[\text{base}]$ , the logarithm of the Henderson-Hasselbalch equation vanishes and $\mathrm{pH}=\mathrm{p}K_{a}$ . The numbers happen to make the error believable here, $4.36$ against $4.74$ , and that is exactly why it gets past proofreading.

6. Determining the calcium carbonate in an antacid tablet.

2 points on the procedure question, and a titration that cannot be used in practice

What not to write

The tablet is dissolved in water and titrated directly with hydrochloric acid, following the color change of an indicator.

What to write

A BACK titration is used: a known excess of acid, complete reaction, degassing, then titration of the remaining acid with sodium hydroxide. The quantity sought is obtained by difference.

The back titration in two steps: a KNOWN excess of acid attacks the tablet and drives off the carbon dioxide, then the sodium hydroxide titrates what is left.

Why: Three reasons add up and you must be able to give them. The carbonate is a sparingly soluble SOLID, whose dissolution is too slow for a direct titration. The excipients cloud the solution and make any color change unreadable. Finally the carbon dioxide given off stirs the medium and spoils the reading, which requires degassing before the back titration, failing which the content is underestimated.

Which method to choose

Which pH calculation, according to how far the titration has gone

Locate the volume added relative to the equivalence volume: before, at the half-equivalence point, at the equivalence point, or after.

Each region has its own calculation: before any addition, the buffer region, the half-equivalence point, the equivalence point where the product formed counts, then the excess of strong base.

If BEFORE any addition, the acid or base is alone → Weak acid: $[\mathrm{H_{3}O^{+}}]=\sqrt{K_{a}C}$ . Weak base: $[\mathrm{OH^{-}}]=\sqrt{K_{b}C}$ . Then check that the percent ionization stays below 5%.

Example: $0.100$ M acetic acid: $\mathrm{pH}=2.87$ , against $1.00$ for a strong acid.
If BETWEEN the start and the equivalence point, away from the half-equivalence point → Buffer region, the Henderson-Hasselbalch equation applies: do the stoichiometry IN MOLES first, all the way through, then $\mathrm{pH}=\mathrm{p}K_{a}+\log\dfrac{n_{\text{base}}}{n_{\text{acid}}}$ .

Example: At $10.00$ mL: $1.00$ mmol of acetate against $1.50$ mmol of acid, $\mathrm{pH}=4.57$ .
If Exactly at the HALF-EQUIVALENCE point → $\mathrm{pH}=\mathrm{p}K_{a}$ , with no calculation. It is also the fastest way to identify an unknown acid from a curve.

Example: $12.50$ mL out of $25.00$ : $\mathrm{pH}=4.74$ , hence $K_{a}=1.8\times 10^{-5}$ , acetic acid.
If At the EQUIVALENCE point → Identify the species formed and calculate ITS pH, allowing for the dilution by the TOTAL volume. Conjugate base: $\mathrm{pH}>7$ . Conjugate acid: $\mathrm{pH}<7$ . Two inert ions: $\mathrm{pH}=7$ .

Example: Acetate at $0.0500$ M after dilution: $\mathrm{pH}=8.72$ .
If AFTER the equivalence point → The excess of strong reagent swamps everything: consider only it, and neglect the hydrolysis of the salt. Be careful to divide by the TOTAL volume.

Example: At $26.00$ mL: $[\mathrm{OH^{-}}]=\dfrac{0.100\times 1.00}{51.00}$ , hence $\mathrm{pH}=11.29$ .

For a DIPROTIC acid, this scheme is repeated twice, with two equivalence points separated by the same volume. At the first, the species is AMPHIPROTIC and its pH is $\dfrac{\mathrm{p}K_{a1}+\mathrm{p}K_{a2}}{2}$ , independent of the concentration.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating a pH during a titration

When to use it: At every point asked for on a titration curve, whatever pair is involved.

1 Write the titration equation and calculate the equivalence volume, even if the question does not ask for it.
2 Locate the point asked for relative to the equivalence point, and NAME the region: acid alone, buffer region, equivalence point, or excess.
3 Do the material balance IN MOLES, all the way through, before any pH formula.
4 Apply the relation specific to the region, dividing by the TOTAL volume whenever a concentration is needed.
5 Check consistency: the pH must increase steadily as a base is added.

Concluding sentence

“At this stage $1.50$ mmol of acid remains and $1.00$ mmol of conjugate base has formed: the mixture is a buffer, so $\mathrm{pH}=\mathrm{p}K_{a}+\log\dfrac{1.00}{1.50}=4.57$ .”

The trap: Using CONCENTRATIONS in the Henderson-Hasselbalch equation without recalculating them after dilution: since the ratio is the same in moles and in concentrations for a common volume, you can work in moles, and it is safer.

Marking: 1 mark for the equivalence volume, 1 mark for the region named, 2 marks for the material balance, 1 mark for the pH and its consistency.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The pH at the equivalence point is on the right side of 7

Identify the species present at the equivalence point: conjugate base of a weak acid, so basic; conjugate acid of a weak base, so acidic; two inert ions, so neutral.

Acetic acid titrated with sodium hydroxide: $8.72$ , basic. Ammonia titrated with HCl: $5.24$ , acidic. A $7.00$ in either case is wrong.
The pKa read at the half-equivalence point is plausible

Compare the value read with the usual pKa values: acetic acid $4.74$ , ammonium ion $9.26$ , carbonic acid $6.37$ and $10.33$ . A negative pKa or one above 14 signals an error.

A pKa read as $4.36$ betrays the confusion with half of the equivalence pH.
The transition range is INSIDE the jump

Before validating an indicator, check that its transition range brackets the pH at the equivalence point and fits entirely inside the pH jump.

Phenolphthalein, $8.2$ to $10.0$ , for an equivalence point at $8.72$ : good. Methyl orange, $3.1$ to $4.4$ : catastrophic, $-69$ % error.
The total volume is indeed that of the mixture

For every concentration calculated, check that the denominator is the sum of the volumes added, never the initial volume alone.

At the equivalence point, acetate is at $\dfrac{2.50}{50.00}=0.0500$ M and not at $0.100$ M: forgetting the dilution shifts the pH by $0.15$ unit.

The typical problem, taken apart

The titration of acetic acid, point by point

A $25.00$ mL sample of $0.100$ M acetic acid is titrated with $0.100$ M NaOH, with $K_{a}=1.8\times 10^{-5}$ . Calculate the pH initially, at $10.00$ mL, at $12.50$ mL, at the equivalence point and at $26.00$ mL.

The setup of the example: acetic acid in the beaker, sodium hydroxide in the burette, and the pH meter that will follow the five points asked for.

Step 1

$\mathrm{CH_{3}COOH}+\mathrm{OH^{-}}\rightarrow\mathrm{CH_{3}COO^{-}}+\mathrm{H_{2}O}$ , and $V_{\text{eq}}=\dfrac{0.100\times 25.00}{0.100}=25.00$ mL

Why

The equivalence volume is calculated FIRST, always, even when the question does not ask for it: it is what then lets you name the region of each point, and so choose the right formula.

Step 2

Acid alone: $[\mathrm{H_{3}O^{+}}]=\sqrt{K_{a}C}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}$ M, so $\mathrm{pH}=2.87$

Why

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 ions. The percent ionization is $1.3$ %, which validates the approximation.

Step 3

At $10.00$ mL: $1.00$ mmol of acetate formed, $1.50$ mmol of acid left, so $\mathrm{pH}=4.74+\log\dfrac{1.00}{1.50}=4.57$

Why

You do the stoichiometry IN MOLES all the way through, and only then the Henderson-Hasselbalch equation. Mixing the two steps is the main source of error, and since the ratio has no units, you can stay in millimoles and never divide by the volume.

Step 4

At $12.50$ mL, exactly half: $1.25$ mmol of each, the ratio is 1, so $\mathrm{pH}=\mathrm{p}K_{a}=4.74$

Why

This is the most rewarding reading of the whole curve: the pKa is obtained 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.

Step 5

At the equivalence point: acetate at $\dfrac{2.50}{50.00}=0.0500$ M, $K_{b}=\dfrac{10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}$ , so $[\mathrm{OH^{-}}]=5.27\times 10^{-6}$ M and $\mathrm{pH}=8.72$

Why

Two mandatory moves here: divide by the TOTAL volume of $50.00$ mL, and recognize that the species formed is a BASE. The pH is not 7, and it is this value of $8.72$ that will call for phenolphthalein.

Step 6

At $26.00$ mL: excess of $1.00$ mL of strong base, $[\mathrm{OH^{-}}]=\dfrac{0.100\times 1.00}{51.00}=1.96\times 10^{-3}$ M, so $\mathrm{pH}=11.29$

Why

After the equivalence point, the excess strong base completely swamps the hydrolysis of acetate, which is neglected. The jump between $24.00$ and $26.00$ mL is $5.17$ units, plenty to make the titration usable.

The conclusion, written out

$2.87$ , then $4.57$ , $4.74$ at the half-equivalence point, $8.72$ at the equivalence point and $11.29$ afterwards. The pKa is read at $12.50$ mL, and the pH at the equivalence point calls for phenolphthalein.

The classic mistake on this problem: Stating $7.00$ at the equivalence point, or forgetting to divide by the total volume of $50.00$ mL when calculating the acetate concentration: the first error spoils the choice of indicator, the second shifts the pH by $0.15$ unit.

Learn by heart

• $C_{a}V_{a}=C_{b}V_{b\text{,eq}}$ for a monoprotic acid and a monobasic base. For a diprotic acid, two equivalence points separated by the same volume.
• At the half-equivalence point, $\mathrm{pH}=\mathrm{p}K_{a}$ . The word “half” refers to the VOLUME, never to the pH.
• At the equivalence point: $7.00$ if strong-strong, BASIC if a weak acid is titrated, ACIDIC if a weak base is titrated.
• Henderson-Hasselbalch: $\mathrm{pH}=\mathrm{p}K_{a}+\log\dfrac{[\text{base}]}{[\text{acid}]}$ , valid only between $\mathrm{p}K_{a}-1$ and $\mathrm{p}K_{a}+1$ .
• Diluting a buffer does not change its pH and divides its buffer index by the same factor. $\beta_{\max}=0.576\,C_{T}$ at $\mathrm{pH}=\mathrm{p}K_{a}$ .
• Amphiprotic species, such as $\mathrm{HCO_{3}^{-}}$ : $\mathrm{pH}=\dfrac{\mathrm{p}K_{a1}+\mathrm{p}K_{a2}}{2}$ , independent of the concentration.
• Transition ranges: methyl orange $3.1$ to $4.4$ , methyl red $4.4$ to $6.2$ , bromothymol blue $6.0$ to $7.6$ , phenolphthalein $8.2$ to $10.0$ .

Frequently asked questions

Is the pH always 7 at the equivalence point of a titration?

No, almost never. The equivalence point is a condition of stoichiometry, not of neutrality: it is the product formed that sets the pH. Titrating a weak acid leaves its conjugate base in solution, so the pH is basic, often around 8.7. Titrating a weak base leaves its conjugate acid, so the pH is acidic. The pH is 7 only if the acid and the base are both strong.

How do you read the pKa on a titration curve?

You locate the equivalence volume, divide it by two, and read the pH at that volume: it is exactly the pKa. At the half-equivalence point, half of the acid has been converted to its conjugate base, so the two forms are present in equal amounts and the logarithmic term of the Henderson-Hasselbalch equation vanishes. No calculation, no concentration needed.

What happens when you dilute a buffer solution?

Its pH hardly changes, because the Henderson-Hasselbalch equation involves only the ratio of the two forms, and the dilution divides both by the same factor. Its buffer index, however, is divided in the same proportion as the dilution: the pH holds, the robustness does not. Below about one millimole per litre, the equation even stops being reliable.

How do you choose the right acid-base indicator?

You first calculate the pH at the equivalence point, then choose an indicator whose transition range brackets that value and fits entirely inside the pH jump. For a weak acid titrated with a strong base, the equivalence point is basic and phenolphthalein is needed. Using methyl orange in that case throws the titration off by nearly seventy percent.

Does a buffer keep the pH constant whatever amount of acid is added?

No. A buffer has a finite capacity, set by the amount of each of its two forms. When one is used up, the buffer is broken and the pH swings abruptly. It cushions, it does not lock. Its efficiency is greatest when the pH equals the pKa, and it has already lost about two thirds of its value one unit away.

Why do a back titration rather than a direct titration?

Because the direct reaction is sometimes too slow, incomplete, or involves a solid. A known excess of reagent is then added, the reaction is allowed to go to completion, and the excess is titrated. The quantity sought is obtained by difference. It is the standard method for analyzing an antacid tablet, whose carbonate dissolves too slowly.

Practise it

Corrected exercises: Titrations and buffer solutions, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

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Revision sheet: titrations and buffer solutions (202-SN2-RE)

The essentials

A titration curve contains four pieces of information

Buffers: what sets the pH and what sets the robustness

The mistakes that cost marks

1. A 25.0025.0025.00 mL sample of 0.1000.1000.100 M acetic acid is titrated with 0.1000.1000.100 M NaOH. What is the pH at the equivalence point?

2. The same acetic acid is titrated with methyl orange, which changes around pH=4.4\mathrm{pH}=4.4pH=4.4, rather than with phenolphthalein.

3. An equimolar acetate buffer of pH 4.744.744.74 is diluted tenfold.

4. An acetate buffer in the ratio 10:110:110:1 has to absorb 1.001.001.00 mmol of sodium hydroxide.

5. On a titration curve, you read the pH at the half-equivalence point.

6. Determining the calcium carbonate in an antacid tablet.

Which method to choose

Which pH calculation, according to how far the titration has gone

How the answer is expected to be written

Calculating a pH during a titration

Check before you hand in

The typical problem, taken apart

The titration of acetic acid, point by point

Learn by heart

Frequently asked questions

Is the pH always 7 at the equivalence point of a titration?

How do you read the pKa on a titration curve?

What happens when you dilute a buffer solution?

How do you choose the right acid-base indicator?

Does a buffer keep the pH constant whatever amount of acid is added?

Why do a back titration rather than a direct titration?

Corrected exercises: Titrations and buffer solutions, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

1. A $25.00$ mL sample of $0.100$ M acetic acid is titrated with $0.100$ M NaOH. What is the pH at the equivalence point?

2. The same acetic acid is titrated with methyl orange, which changes around $\mathrm{pH}=4.4$ , rather than with phenolphthalein.

3. An equimolar acetate buffer of pH $4.74$ is diluted tenfold.

4. An acetate buffer in the ratio $10:1$ has to absorb $1.00$ mmol of sodium hydroxide.