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

Revision sheet: acids and bases (202-SN2-RE)

This whole chapter follows from two constants, $K_{w}$ and $K_{a}$ , and a single question asked before every calculation: what is in the solution at this precise moment? An acid alone, a mixture of an acid and its conjugate base, a conjugate base alone, or a strong base in excess: four situations, four distinct methods.

This sheet gives the four methods in the order in which they appear along a titration, the link between a conjugate pair and the acidic or basic character of a salt, and the list of mistakes on buffers that come back from one exam to the next. It is written for CEGEP students in Chemistry of Solutions, 202-SN2-RE (formerly 202-NYB).

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The thread of the chapter

Before any calculation, decide which REGION you are in: acid alone, buffer mixture, equivalence point, excess base. Each region has its own method, and applying the method of one region in another is the costliest mistake of the chapter.

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

The pH scale
Ion product of water
Conjugate acid-base pair
Acid dissociation constant
Percent ionization
Weak base
Hydrolysis of salts
Buffer solution
Henderson-Hasselbalch equation
Acid-base titration
Equivalence point
Polyprotic acid

The essentials

The constants and the basic relations

• $pH=-\log[H_{3}O^{+}]$ and $pOH=-\log[OH^{-}]$ . At $25$ °C, $K_{w}=[H_{3}O^{+}][OH^{-}]=1.0\times 10^{-14}$ , so $pH+pOH=14$ .
• STRONG acid or base: complete dissociation, so $[H_{3}O^{+}]=C_{0}$ or $[OH^{-}]=C_{0}$ , with no table at all.
• WEAK acid: $K_{a}=\dfrac{[H_{3}O^{+}][A^{-}]}{[HA]}$ , ICE table, approximation tested at $5\%$ . Percent ionization $=\dfrac{x}{C_{0}}\times 100$ .
• Conjugate pair: $K_{a}\times K_{b}=K_{w}$ , so $pK_{a}+pK_{b}=14$ . The stronger an acid, the weaker its conjugate base.
• An acid-base reaction is favoured in the direction that goes from the STRONGER acid to the weaker acid.

The three scales are read together: one pH unit is a factor of

10

[H_{3}O^{+}]

, and the pOH reads backwards, since

pH+pOH=14

The scale is logarithmic: one pH unit represents a factor of $10$ on the concentration. A pH given to two decimal places therefore corresponds to two significant figures on the concentration, no more.

The four regions of a titration

• BEFORE any addition: weak acid alone, ICE table with $K_{a}$ .
• BETWEEN the start and the equivalence point: a mixture of acid and conjugate base, so a BUFFER. $pH=pK_{a}+\log\dfrac{[base]}{[acid]}$ .
• At the HALF-EQUIVALENCE point: the two forms are in equal amounts, so $pH=pK_{a}$ exactly. It is the easiest point on the graph.
• At the EQUIVALENCE point: only the conjugate base remains, so the pH is BASIC, never $7$ . Apply $K_{b}=\dfrac{K_{w}}{K_{a}}$ to the diluted concentration.
• AFTER the equivalence point: strong base in excess; ignore everything else and calculate $[OH^{-}]$ from the amount in excess over the total volume.

The jump does NOT pass through

7

: the equivalence point sits at

8.7

, because only the conjugate base remains at that stage. The half-equivalence point, for its part, gives the

pK_{a}

directly by simple reading.

The equivalence pH is exactly $7$ only for a strong acid titrated with a strong base. With a weak acid it is above $7$ , with a weak base it is below $7$ .

Buffers and salts

• A buffer is a mixture of a weak acid AND its conjugate base, in comparable amounts. It absorbs small additions of acid or base.
• $pH=pK_{a}+\log\dfrac{[base]}{[acid]}$ : the BASE goes in the numerator. The ratio can be written in moles, since the volume cancels.
• A buffer is most effective when $pH=pK_{a}$ , that is, when the two forms are in equal amounts. So choose the acid whose $pK_{a}$ is closest to the target pH.
• Diluting a buffer hardly changes its pH: the ratio of the two forms is preserved.
• Hydrolysis of salts: the cation of a strong base and the anion of a strong acid give a NEUTRAL salt; the anion of a weak acid makes it BASIC; the cation of a weak base makes it ACIDIC.

Only the ion that comes from a WEAK species reacts with water:

NH_{4}^{+}

makes the solution acidic, acetate makes it basic, and NaCl, made of two spectator ions, stays at

7

Buffer capacity is finite: as soon as the addition exceeds the amount of one of the two forms, the buffer breaks down and the pH drops abruptly. This is exactly what the jump of a titration shows.

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. Treating a weak acid as a strong acid

the whole question, 6 points

What not to write

“The acetic acid solution is $0.100$ mol/L, so $[H_{3}O^{+}]=0.100$ and $pH=1.00$ .”

What to write

“Acetic acid is WEAK: $\dfrac{x^{2}}{0.100-x}=1.8\times 10^{-5}$ gives $x=1.33\times 10^{-3}$ and $pH=2.87$ .”

The same acetic acid: believing it strong gives

1.00

, treating it with

K_{a}

gives

2.87

. Nearly two units apart, a factor of

75

[H_{3}O^{+}]

Why: Only the six usual strong acids dissociate completely. A weak acid at $0.100$ mol/L is only $1.3\%$ ionized, which changes the pH by almost two units.

2. Setting the equivalence pH of a weak-acid titration at seven

the whole question, 5 points, and the choice of indicator missed

What not to write

“At the equivalence point, all the acid is neutralized, so $pH=7.00$ .”

What to write

“Only the ACETATE ION remains, a weak conjugate base: its hydrolysis gives $pH=8.72$ . The pH equals $7$ only between a strong acid and a strong base.”

The equivalence pH is that of the PRODUCT formed:

7

between two strong species only, above 7 with a weak acid, below 7 with a weak base.

Why: Neutralizing does not mean making neutral: it means consuming the acid. What remains is a base, and its hydrolysis imposes a pH above $7$ .

3. Inverting the ratio in the Henderson-Hasselbalch equation

3 points, and a pH that is the mirror image of the right one about the

pK_{a}

What not to write

“ $pH=pK_{a}+\log\dfrac{[acid]}{[base]}$ .”

What to write

“The BASE goes in the numerator: $pH=pK_{a}+\log\dfrac{[base]}{[acid]}$ . The more base there is, the higher the pH, which is the only consistent reading.”

Adding base must make the pH RISE: only the green curve does. The inverted ratio gives a pH that is the mirror image of the right one about the

pK_{a}

Why: The error is invisible when the two forms are in equal amounts, since the logarithm is then zero. It shows up as soon as the ratio moves away from $1$ , and the check is immediate: adding base must raise the pH.

4. Using the buffer formula outside the buffer region

the whole question, 5 points

What not to write

“After the equivalence point, I still have acetate and a little acid, so I apply Henderson-Hasselbalch.”

What to write

“After the equivalence point there is no weak acid left at all: there is STRONG base in excess, and $[OH^{-}]$ is calculated from the amount in excess divided by the TOTAL volume.”

In the green band, Henderson-Hasselbalch follows the true curve; near the equivalence point it shoots off to infinity, and after it the formula is meaningless: the strong base alone decides.

Why: Henderson-Hasselbalch assumes that both forms of the pair coexist in comparable amounts. A single drop of strong base in excess imposes its own pH and makes the formula inapplicable.

5. Adding the two ionizations of a polyprotic acid

3 points

What not to write

“Phosphoric acid has three ionizations, so $[H_{3}O^{+}]$ is three times what the first gives.”

What to write

“Only the FIRST counts: $K_{a1}=7.1\times 10^{-3}$ against $K_{a2}=6.3\times 10^{-8}$ , a hundred thousand times smaller. The second is negligible next to the first.”

The three constants of phosphoric acid are separated by about five

pK_{a}

units, so by a factor of a hundred thousand. Next to the first, the other two weigh nothing.

Why: Pulling a second proton off a species that is already negatively charged costs much more: successive constants drop by about five orders of magnitude at each step.

6. Believing that every salt gives a neutral solution

4 points

What not to write

“Sodium acetate comes from the neutralization of an acid by a base, so its solution is neutral.”

What to write

“The acetate ion is the conjugate base of a WEAK acid: it hydrolyzes and makes the solution BASIC. Only salts of a strong acid and a strong base are neutral.”

Why: The conjugate base of a strong acid, such as the chloride ion, is so weak that it does not react. That of a weak acid, by contrast, is a measurable base, and its $K_{b}$ equals $\dfrac{K_{w}}{K_{a}}$ .

7. Believing that diluting a buffer changes its pH

3 points on the reasoning question

What not to write

“I dilute the buffer ten times, so its pH goes up by one unit.”

What to write

“The ratio $\dfrac{[base]}{[acid]}$ is UNCHANGED by dilution, so the pH stays practically the same. It is its CAPACITY that decreases, not its pH.”

The two buffers start at the SAME pH,

4.74

: dilution changed nothing. But

100

mL of the diluted buffer gives way after only

1

mmol of acid, ten times sooner than the original.

Why: Both concentrations are divided by the same factor, and only their ratio enters the formula. A diluted buffer, however, is broken down by an addition ten times smaller.

8. Adding $K_{a}$ and $K_{b}$ instead of multiplying them

3 points

What not to write

“ $K_{a}+K_{b}=K_{w}$ , so $K_{b}=1.0\times 10^{-14}-1.8\times 10^{-5}$ , which is negative.”

What to write

“ $K_{a}\times K_{b}=K_{w}$ , so $K_{b}=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times 10^{-10}$ . It is the $pK$ values that ADD: $pK_{a}+pK_{b}=14$ .”

Why: The relation comes from the sum of the two equilibria, and adding equations MULTIPLIES their constants. The negative result obtained by adding is, moreover, impossible for a constant.

9. Forgetting the dilution caused by the added volume in a titration

2 points, and a shifted equivalence pH

What not to write

“At the equivalence point, the acetate concentration is $0.100$ mol/L, like the starting acid.”

What to write

“The volume has DOUBLED, since $25.0$ mL was added to $25.0$ mL: the acetate concentration is $0.0500$ mol/L.”

The

2.50

mmol of acetate are in

50.0

mL at the equivalence point, not in

25.0

mL: the concentration is halved.

Why: All concentrations in a titration are calculated on the TOTAL volume, burette included. It is one of the few quantities that the buffer lets you forget, since the volume cancels there.

10. Calculating a pH without checking that it is on the right side of seven

3 points, but the error can be spotted in a second

What not to write

“The ammonia solution has a pH of $2.9$ .”

What to write

“Ammonia is a BASE: its pH must exceed $7$ . Calculate $[OH^{-}]$ , then $pOH$ , then $pH=14-pOH=11.1$ .”

A base gives a pH above

7

2.9

is the pOH of the ammonia, and it is

14-2.9=11.1

that must be written.

Why: A weak base is handled with $K_{b}$ and first gives a $pOH$ . Forgetting the last subtraction gives an acidic pH for a base, which plain common sense rejects.

Which method to choose

Which region of the titration are we in

Compare the amount of base added with the starting amount of acid

The count in MOLES designates the region: acid alone, acid and conjugate base together, conjugate base alone, then strong base in excess. Each column has its own formula.

If no base added → weak acid alone: ICE table with $K_{a}$

Example: $0.100$ mol/L acetic acid gives $pH=2.87$
If base added in an amount LESS than the acid → buffer: $pH=pK_{a}+\log\dfrac{n_{base}}{n_{acid}}$ , in moles, the volume cancelling

Example: at the half-equivalence point, $pH=pK_{a}$ exactly
If base added EXACTLY equal to the acid → equivalence point: only the conjugate base remains, apply $K_{b}=\dfrac{K_{w}}{K_{a}}$ to the TOTAL volume

Example: $pH=8.72$ for acetate at $0.0500$ mol/L
If base added in EXCESS → strong base alone: $[OH^{-}]=\dfrac{n_{excess}}{V_{total}}$ , everything else is negligible

Example: $1.0$ mL of excess already gives $pH\approx 11.3$ (0.100 mol/L base, 25.0 mL sample)
If we start from a SALT and not from an acid → identify which of the two ions hydrolyzes, then apply $K_{a}$ or $K_{b}$ to that ion

the ion coming from the STRONG species is a spectator and never takes part

The first line of the answer must always be a count in MOLES, not in concentrations: it decides the region, and it stays valid whatever the volume added.

Which formula for which type of solution

The exact nature of the dissolved species

If strong acid → $[H_{3}O^{+}]=C_{0}$ and $pH=-\log C_{0}$ , with no table

Example: $0.0200$ mol/L hydrochloric acid gives $pH=1.70$
If strong base → $[OH^{-}]=C_{0}$ , then $pH=14-pOH$

Example: $0.0100$ mol/L sodium hydroxide gives $pH=12.00$
If weak acid → $K_{a}$ , ICE table, approximation tested at $5\%$

Example: $x=\sqrt{K_{a}C_{0}}$ if the test passes
If weak base, including the conjugate base of a weak acid → $K_{b}=\dfrac{K_{w}}{K_{a}}$ , ICE table, then $pOH$ then $pH$

Example: acetate, ammonia, carbonate ion
If mixture of a weak acid AND its conjugate base → Henderson-Hasselbalch, with the BASE in the numerator

the ratio can be written in moles, which avoids calculating the total volume

An ion like chloride or sodium never does anything: it comes from a strong species, so its conjugate is too weak to react. Spotting it right away often simplifies the question by half.

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.

Preparing a buffer and calculating its pH after an addition

When to use it: The question asks you to make up a buffer at a given pH, or to calculate the effect of adding acid or base to an existing buffer.

1 Calculate $pK_{a}=-\log K_{a}$ and compare it with the target pH: their difference gives the logarithm of the ratio directly.
2 Deduce $\dfrac{n_{base}}{n_{acid}}=10^{pH-pK_{a}}$ , then the two amounts from the imposed total amount.
3 When adding a strong acid, count in MOLES: the added acid converts base into acid, mole for mole. An added base does the opposite.
4 Recalculate the pH with the new amounts, always by Henderson-Hasselbalch.
5 Compare the pH change obtained with what pure water would have undergone, and conclude on the effectiveness of the buffer.

Concluding sentence

“Adding $0.0100$ mol of hydrochloric acid converts as much acetate into acetic acid: the amounts become $0.0543$ and $0.0457$ mol, hence $pH=4.74+\log\dfrac{0.0543}{0.0457}=4.82$ , a drop of only $0.18$ unit.”

The trap: Reasoning in concentrations rather than in moles when adding. If the volume changes, the ratio stays correct in moles but all the concentrations would have to be recalculated: better to spare yourself that.

Marking: Typically 1 point for the $pK_{a}$ , 2 for the ratio and the amounts, 2 for the count after the addition, 2 for the new pH, 1 for the comparison with pure water.

Calculating the pH at the equivalence point of a weak-acid titration

When to use it: The question asks for the pH at the equivalence point, or for the choice of a colour indicator.

1 Calculate the amount of conjugate base formed: it equals the starting amount of acid.
2 Calculate the TOTAL VOLUME, that of the sample plus that added, and deduce the concentration of the conjugate base.
3 Calculate $K_{b}=\dfrac{K_{w}}{K_{a}}$ and write the ICE table for the hydrolysis.
4 Obtain $[OH^{-}]$ , then $pOH$ , then $pH=14-pOH$ , and check that the result exceeds $7$ .
5 Choose an indicator whose transition range brackets this pH, not the one for a strong-strong titration.

Concluding sentence

“At the equivalence point, $0.00250$ mol of acetate is in $50.0$ mL, that is $0.0500$ mol/L; with $K_{b}=5.6\times 10^{-10}$ , we obtain $[OH^{-}]=5.3\times 10^{-6}$ mol/L, so $pOH=5.28$ and $pH=8.72$ .”

The trap: Using the initial concentration of the acid instead of the diluted concentration. The volume has doubled if the two solutions have the same concentration, which shifts the equivalence pH by more than a tenth of a unit.

Check before you hand in

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

The pH on the right side of seven

An acid must give a pH below $7$ , a base a pH above. It is the fastest check and it catches the forgotten subtraction $14-pOH$ .

A pH of $2.9$ reported for ammonia: it is the $pOH$ that was given.
The pH of a buffer close to its $pK_{a}$

A usable buffer has a pH within one unit of its $pK_{a}$ . A gap of three units signals an error in the ratio or a poorly chosen pair.

An acetate buffer reported at $pH=8.5$ when its $pK_{a}$ is $4.74$ : it is no longer a buffer.
The five percent test on the ionization

After the approximation, check that the percent ionization stays below $5\%$ . Otherwise, redo it with the full quadratic equation.

A weak acid diluted to $10^{-4}$ mol/L is ionized by more than $30\%$ : the approximation is invalid.
The total volume in every titration concentration

Go back over each concentration calculated and check that it is divided by the TOTAL volume, burette included.

A concentration at the equivalence point equal to the starting one: the added volume was forgotten.
The direction of change after an addition

Adding an acid must always make the pH FALL, adding a base must make it rise, even in a buffer.

A pH that rises after adding hydrochloric acid: the Henderson ratio was inverted.

The typical problem, taken apart

Preparing an acetate buffer, then pushing it to the breaking point

We want to prepare $500$ mL of a buffer solution at pH $5.00$ from acetic acid and sodium acetate, the total concentration of the two forms being $0.200$ mol/L. Given: $K_{a}=1.8\times 10^{-5}$ .

Determine the amounts of each form, then the pH after adding $0.0100$ mol of hydrochloric acid. Compare with the same addition to $500$ mL of pure water. Finally, determine the amount of hydrochloric acid that would completely break the buffer.

The same addition of

0.0100

mol of hydrochloric acid, in the buffer and in pure water: this is the comparison the example works out.

Step 1

$pK_{a}=-\log(1.8\times 10^{-5})=4.74$ . The target pH being $5.00$ , we have $\log\dfrac{n_{base}}{n_{acid}}=5.00-4.74=0.26$ , so $\dfrac{n_{base}}{n_{acid}}=10^{0.26}=1.80$ .

Why

The gap between the target pH and the $pK_{a}$ gives the logarithm of the ratio directly: that is the whole point of the logarithmic form of Henderson-Hasselbalch.

Step 2

Total amount: $0.200\times 0.500=0.100$ mol. With the ratio $1.80$ : $n_{base}=0.100\times\dfrac{1.80}{2.80}=0.0643$ mol and $n_{acid}=0.0357$ mol.

Why

The total amount is split according to the ratio, which avoids solving a system. The two must add back up to $0.100$ mol, an immediate check.

Step 3

Check: $pH=4.74+\log\dfrac{0.0643}{0.0357}=4.74+0.26=5.00$ , as specified.

Why

Recalculating the pH from the amounts found validates the ratio and the split at once, before moving on to the part that really matters.

Step 4

Addition of $0.0100$ mol of hydrochloric acid: it converts as much acetate into acetic acid. $n_{base}=0.0643-0.0100=0.0543$ mol and $n_{acid}=0.0357+0.0100=0.0457$ mol.

Why

The count is done in MOLES, and the strong acid reacts completely: there is no equilibrium to solve at this step, only bookkeeping.

Step 5

New pH: $4.74+\log\dfrac{0.0543}{0.0457}=4.74+0.075=4.82$ . The drop is only $0.18$ unit.

Why

The volume appears nowhere, since it cancels in the ratio: this is what makes the buffer calculation so quick, and also why dilution does not change its pH.

Step 6

In $500$ mL of pure water: $[H_{3}O^{+}]=\dfrac{0.0100}{0.500}=0.0200$ mol/L, so $pH=1.70$ . The drop is $5.3$ units from $7$ .

Why

The comparison quantifies the effect of the buffer: thirty times less change, which is a factor of more than a hundred thousand on the hydronium ion concentration.

Step 7

Breaking point: the buffer holds as long as acetate remains, that is, up to $0.0643$ mol of hydrochloric acid. With $0.0700$ mol, the excess is $0.0057$ mol, so $[H_{3}O^{+}]=0.0114$ mol/L and $pH=1.94$ .

Why

Buffer capacity is finite and can be calculated: it is the smaller of the two amounts of the pair. Once it is exceeded, we go back to the calculation for a strong acid in excess.

Step 8

Plausibility check: the breaking-point pH, $1.94$ , is indeed close to that of acidified pure water, $1.70$ , which confirms that the buffer no longer plays any role.

Why

The result must rejoin the behaviour of pure water once the capacity is used up. If it stayed far from it, the breaking-point amount would have been miscalculated.

The conclusion, written out

“The buffer is prepared with $0.0643$ mol of acetate and $0.0357$ mol of acetic acid. Adding $0.0100$ mol of hydrochloric acid lowers its pH only from $5.00$ to $4.82$ , against a fall from $7$ to $1.70$ in pure water; it would take $0.0643$ mol to exhaust it completely.”

The classic mistake on this problem: Inverting the ratio and writing $\log\dfrac{n_{acid}}{n_{base}}$ . This gives $4.49$ instead of $5.00$ , a value that is the mirror image of the right result about the $pK_{a}$ : it is this symmetry that makes the error invisible without the direction check.

Learn by heart

• $pH+pOH=14$ at $25$ °C, and $K_{a}\times K_{b}=K_{w}$ , so $pK_{a}+pK_{b}=14$ .
• STRONG acid: $[H_{3}O^{+}]=C_{0}$ . WEAK acid: ICE table with $K_{a}$ , $5\%$ test.
• $pH=pK_{a}+\log\dfrac{[base]}{[acid]}$ : the BASE goes in the numerator.
• Half-equivalence point: $pH=pK_{a}$ exactly. Equivalence point of a weak acid: $pH>7$ .
• After the equivalence point, ignore the pair: it is the strong base in excess that sets the pH.
• Diluting a buffer does not change its pH, only its CAPACITY.
• Polyprotic acid: only the FIRST ionization counts, the following ones are a hundred thousand times weaker.
• A salt is neutral only if both ions come from STRONG species; otherwise it hydrolyzes.

Frequently asked questions

Why is the pH at the equivalence point of a weak acid not seven?

Because at the equivalence point all the acid has been turned into its conjugate base, which is a weak base. This base hydrolyzes and releases hydroxide ions, which makes the solution basic. The equivalence pH is seven only when a strong acid is titrated with a strong base, because the ions left over are then inert.

How do I choose the right pair to prepare a buffer?

Choose the weak acid whose pKa is closest to the desired pH, ideally within one unit. The buffer is at its best when the pH equals the pKa, because the two forms are then present in equal amounts and it can absorb as much acid as base. Beyond one unit of difference, the capacity collapses on one side.

Does the pH of a buffer change when it is diluted?

Practically not. The Henderson equation involves only the ratio of the two forms, and dilution divides both by the same factor. What decreases is the capacity: a buffer diluted ten times is broken down by an addition ten times smaller, even though its starting pH is identical.

Why do we count only the first ionization of a polyprotic acid?

Because the successive constants drop by about five orders of magnitude at each step. Pulling a second proton off a species that is already negatively charged is much harder. The contribution of the second ionization to the hydronium ions is therefore negligible next to that of the first, unless the question explicitly asks about the minor species.

Is a salt solution always neutral?

No. It is neutral only if both of its ions come from strong species, like sodium chloride. If the anion is the conjugate base of a weak acid, it hydrolyzes and the solution becomes basic. If the cation is the conjugate acid of a weak base, the solution becomes acidic. Ammonium chloride, for example, gives an acidic solution.

How do I know which method to use during a titration?

Compare, in moles, the base added with the starting acid. Before any addition, it is a weak acid alone. Between the start and the equivalence point, it is a buffer. At the exact equivalence point, it is the conjugate base alone. After that, it is the strong base in excess that sets everything. Four regions, four methods, and the count in moles tells them apart.

Practise it

Corrected exercises: Acids and bases, 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

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn2-acids-and-bases. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: acids and bases (202-SN2-RE)

The essentials

The constants and the basic relations

The four regions of a titration

Buffers and salts

The mistakes that cost marks

1. Treating a weak acid as a strong acid

2. Setting the equivalence pH of a weak-acid titration at seven

3. Inverting the ratio in the Henderson-Hasselbalch equation

4. Using the buffer formula outside the buffer region

5. Adding the two ionizations of a polyprotic acid

6. Believing that every salt gives a neutral solution

7. Believing that diluting a buffer changes its pH

8. Adding KaK_{a}Ka​ and KbK_{b}Kb​ instead of multiplying them

9. Forgetting the dilution caused by the added volume in a titration

10. Calculating a pH without checking that it is on the right side of seven

Which method to choose

Which region of the titration are we in

Which formula for which type of solution

How the answer is expected to be written

Preparing a buffer and calculating its pH after an addition

Calculating the pH at the equivalence point of a weak-acid titration

Check before you hand in

The typical problem, taken apart

Preparing an acetate buffer, then pushing it to the breaking point

Learn by heart

Frequently asked questions

Why is the pH at the equivalence point of a weak acid not seven?

How do I choose the right pair to prepare a buffer?

Does the pH of a buffer change when it is diluted?

Why do we count only the first ionization of a polyprotic acid?

Is a salt solution always neutral?

How do I know which method to use during a titration?

Corrected exercises: Acids and bases, 202-SN2-RE

See also

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

8. Adding $K_{a}$ and $K_{b}$ instead of multiplying them