Unit 3: CH03 Syllabus Node

Acids, Bases, Buffers & pH Curves

Bronsted-Lowry theory, Kw and water neutrality at elevated temperatures, weak acid Ka calculations, buffer solution action via Henderson-Hasselbalch, and titration curves with indicator selection for OxfordAQA A-Level Chemistry.

1. Bronsted-Lowry Theory, pH & Conjugate Pairs

In quantitative chemistry, acid-base behavior is governed by proton transfer:

Bronsted-Lowry Definitions
  • Acid: A proton (H+) donor.
  • Base: A proton (H+) acceptor.
  • Conjugate Acid-Base Pair: Two species that transform into each other by the gain or loss of a single proton.
    Example: In CH3COOH + H2O <=> CH3COO- + H3O+, CH3COOH is the acid and CH3COO- is its conjugate base; H2O is the base and H3O+ is its conjugate acid.
The pH Scale pH is defined as the negative logarithm to the base 10 of the hydrogen ion concentration:
pH = -log10[H+]
[H+] = 10^(-pH)

2. The Ionic Product of Water (Kw) & High-Temperature Neutrality

Water undergoes a slight self-ionisation (autoprotolysis):

H2O(l) <=> H+(aq) + OH-(aq) [delta H > 0, endothermic]

Because the concentration of undissociated water is immense (~55.5 mol dm-3) and effectively constant, it is incorporated into the equilibrium constant to give the ionic product of water (Kw):

Kw = [H+] [OH-]

At standard temperature (298 K / 25 degrees C):

Kw = 1.00 x 10^-14 mol2 dm-6 => pKw = -log10(Kw) = 14.00

Examiner Distinction: Water Neutrality at Elevated Temperatures

Because the dissociation of water is endothermic (delta H > 0), increasing temperature shifts the equilibrium to the right according to Le Chatelier's principle. Consequently, Kw increases:

  • At 25 degrees C: Kw = 1.00 x 10^-14 mol2 dm-6; [H+] = 1.00 x 10^-7 mol dm-3; pH = 7.00.
  • At 50 degrees C: Kw rises to 5.48 x 10^-14 mol2 dm-6; [H+] = sqrt(Kw) = 2.34 x 10^-7 mol dm-3; pH = 6.63.

Crucial Examination Point: Even though the pH drops below 7 at 50 degrees C, water remains strictly neutral because [H+] is strictly equal to [OH-]. Neutrality does NOT mean pH = 7; neutrality means [H+] = [OH-]!

3. Weak Acids, Ka & pKa Calculations

Strong acids (HCl, HNO3, H2SO4) dissociate completely in aqueous solution, so [H+] = [acid]. Weak acids (CH3COOH, HCOOH, HCN) dissociate only partially:

HA(aq) <=> H+(aq) + A-(aq)

The acid dissociation constant (Ka) is:

Ka = ([H+] [A-]) / [HA] and pKa = -log10(Ka)

A higher Ka (or lower pKa) indicates a stronger weak acid that dissociates to a greater extent.

Standard Approximations for Weak Acid pH Calculations

  1. Equimolar dissociation: We assume all H+ ions come solely from the weak acid (ignoring the negligible contribution from water dissociation): [H+] = [A-].
  2. Negligible dissociation: Because the acid is weak, the amount dissociated is tiny compared to starting concentration: [HA]_eq = [HA]_initial.

Substituting these approximations gives:

Ka = [H+]^2 / [HA] => [H+] = sqrt(Ka * [HA])

4. Buffer Solutions & Mechanism of Action

A buffer solution is a chemical system that resists changes in pH when small amounts of acid (H+) or base (OH-) are added.

Types of Buffers
  • Acidic Buffer (pH < 7): Formed from a weak acid and a salt of its conjugate base (e.g. ethanoic acid, CH3COOH, and sodium ethanoate, CH3COONa).
  • Basic Buffer (pH > 7): Formed from a weak base and a salt of its conjugate acid (e.g. ammonia, NH3, and ammonium chloride, NH4Cl).

Mechanism of Acidic Buffer Action

An acidic buffer contains a large reservoir of undissociated weak acid (HA) and a large reservoir of conjugate base ions (A-):

  1. When a small amount of H+ is added: The added protons react with the large reservoir of conjugate base ions: A- + H+ -> HA. The position of equilibrium shifts to the left, removing added H+ and keeping pH almost constant.
  2. When a small amount of OH- is added: The added hydroxide reacts with the large reservoir of undissociated acid: HA + OH- -> A- + H2O. Protons are donated to neutralise OH-, keeping pH almost constant.

The Henderson-Hasselbalch Equation

pH = pKa + log10([A-] / [HA]) = pKa + log10(moles of salt / moles of acid)

5. pH Titration Curves

Plotting solution pH against the volume of alkali added produces characteristic titration curves:

Comparative pH Titration Curves & Indicator Ranges Volume of 0.10 mol dm-3 NaOH Added / cm3 Solution pH 0 3 7 11 14 25.0 cm3 (Eq) Phenolphthalein (pH 8.3 - 10.0) Methyl Orange (pH 3.1 - 4.4) Strong Acid (HCl) Weak Acid (CH3COOH) Half-Eq (pH = pKa) pH 7 (Strong-Strong) pH 8.9 (Weak-Strong)
Titration Combination Initial pH Vertical Inflection Range pH at Equivalence Suitable Indicator
Strong Acid - Strong Base (HCl + NaOH) ~1 pH 3.0 to 11.0 (long vertical section) 7.0 Both Methyl Orange & Phenolphthalein
Weak Acid - Strong Base (CH3COOH + NaOH) ~3 pH 7.0 to 11.0 (vertical in alkaline region) > 7.0 (~8.9) Phenolphthalein ONLY
Strong Acid - Weak Base (HCl + NH3) ~1 pH 3.0 to 7.0 (vertical in acidic region) < 7.0 (~5.1) Methyl Orange ONLY
Weak Acid - Weak Base (CH3COOH + NH3) ~3 No vertical section (inflection point only) ~7.0 No standard indicator (use pH meter)

6. Indicator Selection & Half-Neutralisation Determination of Ka

Criterion for Indicator Selection An indicator is suitable for a titration if and only if its pH transition range (pKin +/- 1) lies entirely within the vertical section of the pH curve.

Determining Ka from the Half-Neutralisation Point

During a weak acid titration, at the half-neutralisation point (halfway to equivalence volume, e.g. at 12.5 cm3 if equivalence is 25.0 cm3):

[HA] = [A-]

Substituting into Ka = ([H+][A-]) / [HA]:

Ka = [H+] => pH = pKa

Therefore, reading the pH at exactly half the equivalence volume directly gives the pKa of the weak acid!

7. Worked Calculations

Worked Example 1: pH of a Weak Acid Solution
Calculate the pH of a 0.150 mol dm-3 solution of ethanoic acid (CH3COOH) at 298 K, given Ka = 1.74 x 10^-5 mol dm-3.

Step 1: Apply weak acid approximation:

[H+] = sqrt(Ka * [HA])

[H+] = sqrt((1.74 x 10^-5) * 0.150) = sqrt(2.61 x 10^-6) = 1.616 x 10^-3 mol dm-3

Step 2: Calculate pH:

pH = -log10(1.616 x 10^-3) = 2.79

Worked Example 2: pH of an Acidic Buffer Solution
A buffer solution is prepared by mixing 500 cm3 of 0.200 mol dm-3 methanoic acid (HCOOH, Ka = 1.78 x 10^-4 mol dm-3) with 500 cm3 of 0.100 mol dm-3 sodium methanoate (HCOONa).
Calculate the pH of this buffer solution.

Step 1: Calculate moles of acid and salt:

Moles of HCOOH = 0.500 dm3 * 0.200 mol dm-3 = 0.100 mol

Moles of HCOO- = 0.500 dm3 * 0.100 mol dm-3 = 0.050 mol

Step 2: Calculate pKa:

pKa = -log10(1.78 x 10^-4) = 3.75

Step 3: Apply Henderson-Hasselbalch equation:

pH = pKa + log10(moles of salt / moles of acid)

pH = 3.75 + log10(0.050 / 0.100) = 3.75 + log10(0.50) = 3.75 + (-0.301) = 3.45

Exam-Style Practice Questions

Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.

Question 1: At 60 degrees C, Kw is 9.60 x 10^-14 mol2 dm-6. What is the pH of pure water at this temperature, and is the water acidic, basic, or neutral?

Show Answer & Explanation

Correct Answer: C

[H+] = sqrt(Kw) = sqrt(9.60 x 10^-14) = 3.10 x 10^-7 mol dm-3. pH = -log10(3.10 x 10^-7) = 6.51. The water remains neutral because [H+] = [OH-].

Question 2: Which indicator is most suitable for titrating ethanoic acid (weak acid) with sodium hydroxide (strong base)?

Show Answer & Explanation

Correct Answer: C

A weak acid - strong base titration curve has its vertical inflection in the alkaline region (pH 7 to 11). Phenolphthalein's working range (8.3 to 10.0) lies entirely within this vertical section.

Question 3: In a titration of 25.0 cm3 of a weak monoprotic acid with 0.10 M NaOH, equivalence is reached at 30.0 cm3. At what volume of added NaOH does the solution pH equal the pKa of the acid?

Show Answer & Explanation

Correct Answer: C

pH equals pKa at the half-neutralisation point. Halfway to the 30.0 cm3 equivalence point is 15.0 cm3.

Question 4: What happens to the pH of an acidic buffer when a small volume of hydrochloric acid is added?

Show Answer & Explanation

Correct Answer: C

The large reservoir of conjugate base ions (A-) reacts with added protons (A- + H+ -> HA), preventing any major change in [H+] and keeping pH virtually constant.