Acids and Bases

Section 1 of 29Brønsted–Lowry acids and bases

Theory

• Theory acids and bases are defined by proton transfer between species

Brønsted acid = a species that can donate a proton, such as HCl(aq) → H⁺(aq) + Cl^-(aq)

Brønsted base = a species that can accept a proton, such as OH^-(aq) + H⁺(aq) → H₂O(l)

Water and ammonia

• Water and ammonia water acts as an acid and ammonia as a base

Scope

• Scope the theory applies to the gas phase too, such as HCl donating a proton to ammonia

Proton

• Proton an H⁺ ion has lost its electron from a hydrogen atom, so it is a proton
  • (1 proton, 1 electron, 0 neutrons)

Section 2 of 29The pH scale and pH calculations

Scale

• 0 to 14 , acids below 7, alkalis above 7
  • (below 1 for extremely acidic substances)

Meaning

• Meaning pH measures hydrogen ion concentration, so a lower pH means a higher [H⁺] and a more acidic solution

Definition

• Definition pH = −log[H⁺], with [H⁺] in mol dm^-3
  What the symbols mean: pH has no unit · [H⁺] = concentration of hydrogen ions (mol dm⁻³); each 1 pH unit is a ten-fold change in [H⁺]

Reverse

• [H⁺] = 10^-pH

Logarithmic scale

• Logarithmic scale each pH unit is a factor of 10, so pH 5 is ten times more acidic than pH 6

Precision

• Precision give pH to 2 decimal places

[H+] / mol dm-3	pH
1.0	0
0.1	1
0.01	2
0.001	3
0.0001	4

• Examples
  • [H⁺] = 1.60 × 10^-4 gives pH 3.80, and pH 3.10 gives [H⁺] = 7.94 × 10^-4 mol dm^-3

Section 3 of 29pH of strong acids

Ionisation

• Ionisation strong acids ionise completely, HA(aq) → H⁺(aq) + A^-(aq)

Concentration

• [H⁺] equals the acid concentration, because H⁺ from water is negligible

Example

• 0.01 mol dm^-3 HCl has [H⁺] = 0.01, so pH = 2.00

Section 4 of 29The ionic product of water, K_w

Equilibrium

• H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

K_w = the equilibrium constant for the dissociation of water, 1.00 × 10^-14 mol² dm^-6 at 298 K

Expression

• Expression ionisation is very small, so [H₂O] is constant and K_w = [H⁺][OH^-]

Pure water

• [H⁺] = [OH^-], so K_w = [H⁺]², giving [H⁺] = 1.00 × 10^-7 mol dm^-3 and pH 7

Section 5 of 29Effect of temperature on K_w

Enthalpy

• Enthalpy dissociation of water is endothermic

Shift

• Shift a higher temperature shifts the equilibrium to the right, forming more H⁺ and OH^-, so K_w increases

pH

• pH as K_w increases the pH of pure water decreases

T / °C	Kw / mol2 dm-6	pH
0	0.114 × 10-14	7.47
10	0.293 × 10-14	7.27
20	0.681 × 10-14	7.08
25	1.008 × 10-14	7.00

• Neutrality
  • pure water stays neutral because [H⁺] = [OH^-], even if its pH is not 7

Section 6 of 29pH of strong bases

• Ionisation
  • strong bases ionise completely, BOH(aq) → B⁺(aq) + OH^-(aq), so [OH^-] equals the base concentration

pH of strong bases

Find [OH^-] from the concentration of the base.

Calculate [H⁺] = K_w ÷ [OH^-].

Calculate pH = −log[H⁺].

• Example
  • 0.15 mol dm^-3 NaOH gives [H⁺] = 1 × 10^-14 ÷ 0.15 = 6.66 × 10^-14, so pH 13.17

Section 7 of 29Weak acids and K_a

1

Weak acid = an acid that partially dissociates in aqueous solution, such as ethanoic acid, HCN, H₂S and H₂CO₃

2Equilibrium

• Equilibrium the position lies far to the left, so an equilibrium is established

3Acid dissociation constant, Ka

• K_a = [H⁺][A^-] ÷ [HA], with units mol dm^-3
  What the symbols mean: K_a = acid dissociation constant (mol dm⁻³) · [H⁺], [A⁻], [HA] = equilibrium concentrations (mol dm⁻³); larger K_a = stronger acid

4Size

• Size values are very small, such as 1.74 × 10^-5 mol dm^-3 for ethanoic acid

5Assumption

• H⁺ from the ionisation of water is negligible

6Strength

• Strength the higher the K_a, the more dissociated and stronger the acid

Section 8 of 29Calculating pH from K_a

Simplification

• [H⁺] equals [A^-], so K_a = [H⁺]² ÷ [HA]

Rearranged

• [H⁺] = √(K_a × [HA]), then pH = −log[H⁺]

Example

• 0.100 mol dm^-3 ethanoic acid with K_a = 1.74 × 10^-5 gives [H⁺] = 1.32 × 10^-3 mol dm^-3 and pH 2.88

Section 9 of 29pK_a

Reason

• K_a values are very small and cover a wide range, so pK_a is easier to use

Definition

• Definition pK_a = −log K_a

Acid	Ka / mol dm-3	pKa
Methanoic acid	1.77 × 10-4	3.75
Ethanoic acid	1.74 × 10-5	4.75
Benzoic acid	6.46 × 10-5	4.18
Carbonic acid	4.30 × 10-7	6.36

Range

• Range most weak acids have pK_a between 3 and 7

Strength

• Strength a lower pK_a means a stronger acid

Section 10 of 29Calculating K_a from pH

1

Write the dissociation equation.

2

Write K_a = [H⁺][A^-] ÷ [HA].

3

Set [A^-] equal to [H⁺] and substitute the values.

4

Work out the units, mol dm^-3.

5

Calculate pK_a = −log K_a.

• Example
  • 0.100 mol dm^-3 ethanoic acid with [H⁺] = 1.32 × 10^-3 mol dm^-3 gives K_a = 1.74 × 10^-5 mol dm^-3 and pK_a about 4.76

Section 11 of 29Titration method

1

Measure a known volume (usually 20 or 25 cm³) of one solution with a pipette into a conical flask.

2

Put the other solution in the burette.

3

Add a few drops of indicator.

4

Add from the burette, portion by portion, until the indicator just changes colour, the end point.

5

Repeat until concordant results are obtained.

• Use
  • titrations find the concentration of an unknown solution and the pH at points during the titration

Section 12 of 29Titration calculations

1Unknown concentration

• 25.00 cm³ of 0.05 mol dm^-3 HCl neutralised by 8.50 cm³ NaOH gives 1.25 × 10^-3 mol, a 1 : 1 ratio, and a concentration of 0.15 mol dm^-3

2pH before the equivalence point

• pH before the equivalence point use the moles of excess acid and total volume to find [H⁺]

3

pH after the equivalence point (excess alkali)

• Find the moles of acid and moles of alkali added.
• Subtract to find the moles of excess alkali.
• Divide by the total volume to find [OH^-].
• Use K_w to find [H⁺] and then the pH.

4Example

• 45 cm³ of 0.10 mol dm^-3 NaOH added to 25.0 cm³ of 0.15 mol dm^-3 HCl leaves 7.5 × 10^-4 mol OH^- in 70 cm³,
• so [OH^-] = 0.0107, [H⁺] = 9.35 × 10^-13 and pH = 12.03

Section 13 of 29Reading pH curves

Shape

• Shape all pH curves are S-shaped

Equivalence point

• Equivalence point the midpoint of the vertical region, where moles of alkali equal moles of acid

Information from a curve

• The starting pH shows the pH of the solution in the flask.
• The pH at the equivalence point.
• The volume at the equivalence point.
• The pH range of the vertical section.

Section 14 of 29Sketching a pH curve

1

Draw volume added (cm³) on the x-axis and pH on the y-axis.

2

Draw a horizontal line at pH 7 to separate the acidic and alkaline regions.

3

Identify the substance in the flask to find the starting pH.

4

Identify whether the acid and alkali are strong or weak.

5

Draw the curve.

In the flask	Starting pH
Strong acid	About 0–2
Weak acid	About 2–4
Strong alkali	About 13–14
Weak alkali	About 11

Section 15 of 29Strong acid with strong alkali

Start

• Start only H⁺ from the strong acid is present, with pH about 1–2

Gradual rise

• Gradual rise the pH rises slowly as H⁺ reacts with OH^- to form water

Equivalence

• Equivalence near the equivalence point the pH surges steeply, and at the equivalence point the pH is 7

End

• End excess OH^- raises the pH to about 13–14

Reverse titration

• Reverse titration adding acid to alkali gives the same shape with initial and final pH swapped, and the equivalence point still 7

Section 16 of 29Strong acid with weak alkali

Start

• Start pH about 1–2 from the strong acid

Equivalence point

• Equivalence point about pH 5.5, not neutral, because all the H⁺ has formed NH₄⁺, a relatively strong acid

End

• End the pH rises above 7 but stays below that of a strong alkali because NH₃ is weak

Reverse titration

• Reverse titration the shape is the same with initial and final pH swapped, and the equivalence point still about 5.5

Section 17 of 29Weak acid with strong alkali

Start

• Start pH about 2–3 for a weak acid such as ethanoic acid

Equivalence point

• Equivalence point about pH 9, slightly basic, because CH₃COO^- is a relatively strong base

End

• End the pH rises to about 13–14

Sketch of a worked case

• 10.0 cm³ of 0.150 mol dm^-3 weak acid starting at pH 5.3 with 0.100 mol dm^-3 NaOH has its vertical section at 15.0 cm³, with the equivalence point above pH 7 and the curve ending at pH 12.5 at 20 cm³

Reverse titration

• Reverse titration the same shape with initial and final pH swapped, and equivalence still about 9

Section 18 of 29Weak acid with weak alkali and indicators

1Weak acid with weak alkali

• Weak acid with weak alkali there is no vertical region, only a point of inflexion at the equivalence point,
  • and the curve gives little other information

2Indicators

• Indicators substances that change colour in acidic or alkaline solution, chosen using the pH of the equivalence point

Indicator	pH range	Colour change
Methyl orange	3.1–4.4	Red to yellow
Phenolphthalein	8.3–10.0	Colourless to pink

Section 19 of 29Choosing an indicator

• Rule
  • the indicator must change colour within the vertical region of the pH curve, close to the equivalence point

Titration	Suitable indicator
Strong acid and strong alkali	Either methyl orange or phenolphthalein, since both ranges lie in the vertical region
Strong acid and weak alkali	Methyl orange only, since only it changes near the equivalence point
Weak acid and strong alkali	Phenolphthalein only, since the methyl orange range falls below the curve
Weak acid and weak alkali	Neither, so a different method is needed

Section 20 of 29Required Practical 6: aim and calibration

Aim

• Aim to investigate how the pH of ethanoic acid changes as sodium hydroxide is added

Apparatus

• Apparatus two 50 cm³ burettes, two funnels, 100 cm³ beaker, wash bottle of deionised water, stand and clamp, pH meter or probe, 0.100 mol dm^-3 NaOH, 0.100 mol dm^-3 ethanoic acid, standard buffers at pH 4.00, 7.00 and 9.20, stirring rod and graph paper

1

Rinse the pH probe with deionised water and shake off excess.

2

Place it in the pH 7.00 buffer with the bulb fully immersed and record the reading.

3

Repeat with the pH 4.00 and 9.20 buffers, rinsing the probe before each reading.

4

Plot the recorded pH (x-axis) against the buffer pH (y-axis) as a calibration graph.

Section 21 of 29Required Practical 6: measuring pH during the titration

1

Use one burette to put exactly 20.0 cm³ of ethanoic acid into a clean 100 cm³ beaker.

2

Rinse and fill the second burette with NaOH solution.

3

Clamp the pH probe with the bulb fully immersed in the acid.

4

Stir gently and record the starting pH.

5

Add 2.0 cm³ of NaOH, stir and record the pH.

6

Continue in 2.0 cm³ portions up to 18 cm³.

7

Near the end point add 0.20 cm³ portions until 22.0 cm³.

8

Then add 2.0 cm³ portions up to 40 cm³.

• Practical tips
  • Wait a few seconds for the pH to stabilise before reading.
  • Do not stir too fast, because bubbles make the reading unsteady.

Section 22 of 29Required Practical 6: results and evaluation

• Results table
  • volume of NaOH, measured pH and corrected pH

1

Use the calibration graph to correct each pH reading.

2

Enter the corrected values in a new column.

3

Plot corrected pH (y-axis) against volume of NaOH (x-axis).

4

Join the points sensibly and comment on the shape of the curve.

Interpretation

• Interpretation the equivalence point is where neutralisation occurs, such as 25 cm³ of base in the example curve

Indicator

• Indicator phenolphthalein suits this weak acid and strong base titration because it changes colour near the equivalence point

Section 23 of 29Buffer solutions

Buffer = a solution that resists changes in pH when small amounts of acid or alkali are added

Mechanism

• Mechanism it keeps the concentrations of H⁺ and OH^- almost constant

Limit

• Limit buffers cannot cope with large additions of acid or alkali, because the pH then changes significantly

Section 24 of 29Acidic buffers

Composition

• Composition a weak acid and a soluble salt of that acid, such as ethanoic acid with sodium ethanoate

Species present

• Ethanoic acid partially ionises, CH₃COOH ⇌ H⁺ + CH₃COO^-, so it is mostly molecules.
• Sodium ethanoate fully ionises, giving a high concentration of CH₃COO^-.
• The mixture therefore holds high concentrations of both CH₃COOH and CH₃COO^-.

Adding acid (H⁺)

• The equilibrium shifts left, as H⁺ reacts with CH₃COO^- to form CH₃COOH.
• Large reserves of both species mean their concentrations barely change.
• The pH stays almost constant.

Adding alkali (OH^-)

• OH^- reacts with H⁺ to form water, H⁺(aq) + OH^-(aq) → H₂O(l).
• The equilibrium shifts right, as more CH₃COOH ionises to replace the H⁺.
• The pH stays almost constant.

Section 25 of 29Basic buffers

Role

• Role they keep the pH above 7

Composition

• Composition a weak base and its salt, such as aqueous ammonia with ammonium chloride

Equilibrium

• NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) lies to the left, and NH₄Cl fully dissociates, giving high concentrations of NH₃ and NH₄⁺

Adding acid

• NH₃(aq) + H⁺(aq) ⇌ NH₄⁺(aq), which removes the added H⁺

Adding alkali

• NH₄⁺(aq) + OH^-(aq) ⇌ NH₃(aq) + H₂O(l), which removes the added OH^-

Section 26 of 29Applications of buffers

Blood

• HCO₃^- ions buffer blood pH between 7.35 and 7.45
• Cells produce CO₂ in aerobic respiration, and CO₂(g) + H₂O(l) ⇌ H⁺(aq) + HCO₃^-(aq).
• More H⁺ shifts the equilibrium left and less H⁺ shifts it right, keeping pH constant.
• Without control the blood pH would fall, causing acidosis, malfunction and eventually coma.

Everyday products

• Everyday products buffers are used in shampoos and detergents

Section 27 of 29Buffer pH calculations

1Method

• Method use K_a and the equilibrium concentrations of the weak acid and its conjugate base (salt)

2Rearranged

• [H⁺] = K_a × [acid] ÷ [salt]

3Henderson–Hasselbalch equation

• Henderson–Hasselbalch equation pH = pK_a + log([A^-] ÷ [HA])

4Example

• 0.305 mol dm^-3 ethanoic acid and 0.520 mol dm^-3 sodium ethanoate with K_a = 1.74 × 10^-5 gives [H⁺] = 1.02 × 10^-5 mol dm^-3 and pH 4.99

Section 28 of 29Key terms

Term	Meaning
Brønsted–Lowry acid	A proton donor.
Brønsted–Lowry base	A proton acceptor.
pH	The negative logarithm to base 10 of the hydrogen ion concentration.
Ionic product of water	Kw = [H+][OH-].
Weak acid	An acid that dissociates only partially in water.
Acid dissociation constant	Ka, the equilibrium constant for dissociation of a weak acid.

Term	Meaning
pKa	The negative logarithm of Ka.
Equivalence point	The point at which the moles of acid equal the moles of alkali.
End point	The point at which the indicator changes colour.
Indicator	A substance that changes colour over a specific pH range.
Titration curve	A graph of pH against volume of titrant added.

Term	Meaning
Buffer solution	A solution that resists changes in pH when small amounts of acid or alkali are added.
Acidic buffer	A weak acid and a soluble salt of that acid.
Basic buffer	A weak base and its salt.
Calibration	Checking a pH probe against standard buffers and correcting its readings.
Henderson–Hasselbalch equation	pH = pKa + log([salt] ÷ [acid]).

Section 29 of 29Exam technique

1

Saying pH is not logarithmic — each pH unit is a tenfold change.

2

Using [H⁺] = [acid] for a weak acid — only strong acids ionise completely.

3

Quoting pH to one decimal place — give two decimal places.

4

Saying pure water at a high temperature is acidic — it is still neutral, but its pH is lower than 7.

1

Saying the equivalence point is always pH 7 — it depends on the strength of the acid and alkali.

2

Using the original volume instead of the total volume for excess concentration — use the total volume.

3

Drawing a vertical section for weak acid with weak alkali — it has only an inflexion.

4

Starting a weak acid at pH 0 — a weak acid starts at about 2–4.

1

Using methyl orange for weak acid with strong alkali — its range is below the vertical region.

2

Saying buffers work with any amount of acid — they resist only small additions.

3

Forgetting to calibrate the pH probe — raw readings must be corrected.

4

Using the weak acid concentration alone in a buffer calculation — the salt concentration is also needed.