Year 12 Chemistry Module 6, Acid/Base Reactions ⏱ ~45 min Lesson 9 of 19 IQ2

Ka, Kb and ICE Table Calculations

In 1971, pharmacologist John Vane measured the Ka of aspirin (acetylsalicylic acid) as 3.0 × 10⁻⁴, ten times larger than ibuprofen's Ka of 1.3 × 10⁻⁵. That difference determines which drug ionises more in the stomach, which is absorbed faster in the intestine, and is why the ICE table method exists: for any weak acid, [H⁺] depends on both Ka and concentration simultaneously, and you cannot find it without solving the equilibrium properly. Vane won the Nobel Prize in 1982.

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Today's hook, In 1971, John Vane measured aspirin's Ka as 3.0 × 10⁻⁴ and ibuprofen's as 1.3 × 10⁻⁵. Same dose, same volume of stomach fluid, which releases more H⁺, and why does the ICE table method exist to answer that question?

0/5QUESTS

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First, Aspirin vs Ibuprofen

A pharmacology researcher is comparing two pain relief tablets. Tablet A contains 500 mg of aspirin (acetylsalicylic acid, Ka = 3.0 × 10⁻⁴, M = 180 g/mol). Tablet B contains 500 mg of ibuprofen (Ka = 1.3 × 10⁻⁵, M = 206 g/mol). Both dissolve in 200 mL of stomach fluid (approximately 0.1 mol/L HCl, pH ≈ 1).

The researcher wants to predict which tablet releases more H⁺ ions and therefore causes more gastric irritation. Before you read on, write down your prediction: which tablet do you think releases more H⁺? Is it the one with the higher Ka, the higher mass, or the higher molar concentration in stomach fluid? Will either tablet significantly change the pH of stomach acid? You will return to this prediction at the end of the lesson.

Know / Understand / Can Do

Know

ICE tables track Initial, Change, and Equilibrium concentrations
The simplifying assumption [HA]initial − x ≈ [HA]initial is valid when Ka/c < 0.0025
Kb = Kw/Ka for conjugate acid-base pairs

Understand

Why weak acids require ICE tables while strong acids do not
When the simplifying assumption fails and how to solve exactly
Why mixing weak acid with strong base creates a buffer in the intermediate region

Can do

Skills

Set up and solve ICE tables for weak acid and weak base equilibria
Calculate pH of weak acids and weak bases
Determine Ka from measured pH data

Key Vocabulary

Acid dissociation constant (Ka)

The equilibrium constant for weak acid ionisation: Ka = [H⁺][A⁻]/[HA].

Base dissociation constant (Kb)

The equilibrium constant for weak base ionisation: Kb = [BH⁺][OH⁻]/[B].

ICE table

Tracks Initial, Change, and Equilibrium concentrations to find [H⁺] for a weak acid by solving the Ka expression for x.

Small-x approximation

When Ka is very small, [HA]eq ≈ [HA]initial; valid if x/[HA]initial < 5%.

Ka × Kb = Kw

For a conjugate pair at 25°C: Ka × Kb = 1.0 × 10⁻¹⁴.

pKa

pKa = −log₁₀(Ka); lower pKa = stronger acid; used to rank acid strengths.

📐 Ka, Kb, ICE Tables, Assumption & Henderson-Hasselbalch

Ka = [H⁺][A⁻] / [HA] | pKa = −log₁₀(Ka) | Ka = 10⁻ᵖᴷᵃ

Weak acid equilibrium: HA(aq) ⇌ H⁺(aq) + A⁻(aq)

ICE table → Ka = x² / (c − x)

Simplifying assumption (x << c): valid when Ka/c < 0.0025 → x = √(Ka × c) = [H⁺] If invalid (Ka/c ≥ 0.0025): x = (−Ka + √(Ka² + 4·Ka·c)) / 2 (positive root only)

Weak base: Kb = [BH⁺][OH⁻] / [B] | [OH⁻] = √(Kb × c) if assumption valid

Then: pOH = −log[OH⁻] → pH = 14 − pOH

Conjugate pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ | Kb = Kw / Ka | Ka = Kw / Kb

Ka from measured pH: [H⁺] = 10⁻ᵖᴴ → Ka = [H⁺]² / (c − [H⁺])

Degree of ionisation: α = ([H⁺]/c) × 100%

Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA])

For partial neutralisation: [A⁻]/[HA] can be replaced by n(A⁻)/n(HA) (same volume, cancels) At half-equivalence point: [A⁻] = [HA] → log(1) = 0 → pH = pKa

Content Cards

Cross-lesson links: The ICE table method here builds directly on Ka/Kb definitions from L05 and Kw from L07. Vane's aspirin/ibuprofen Ka values reappear in pharmaceutical titration (L16) and back titration (L18). The Henderson-Hasselbalch equation previewed in Card 5 is the main tool of buffer calculations (L12–L13).

Card 1, Why Weak Acids Need a Different Method: The ICE Table Logic

For a strong acid, [H⁺] = concentration because ionisation is complete, no equilibrium to solve. For a weak acid, most molecules remain intact at equilibrium, and the fraction that ionise depends on both Ka and concentration, which is why a systematic method (the ICE table) is required to find the actual [H⁺].

The ICE table is the systematic framework for weak acid calculations. ICE = Initial, Change, Equilibrium three rows that track concentrations of every species from start through to equilibrium. For a weak acid HA at initial concentration c:

Stage	[HA]	[H⁺]	[A⁻]
I (Initial)	c	0	0
C (Change)	−x	+x	+x
E (Equilibrium)	c − x	x	x

Substituting E-row into Ka = [H⁺][A⁻]/[HA] gives: Ka = x²/(c − x). Solving for x gives [H⁺] at equilibrium → pH = −log(x).

For 0.1 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵), using [H⁺] = c (the strong acid shortcut) gives pH = 1.0. The correct ICE method gives pH ≈ 2.87, a difference of nearly 2 pH units, corresponding to a factor of ~74 in [H⁺].

Must Do

Always write the full ICE table before substituting into Ka. Students who skip to Ka = x²/c frequently make sign errors, stoichiometry errors, or forget the (c − x) denominator. Writing the ICE table takes 30 seconds and eliminates most errors in weak acid calculations.

Common Error

Students write [H⁺] = 10⁻⁷ in the Initial row (from water autoionisation). This is incorrect, the water contribution to [H⁺] is negligible compared to any weak acid at practical concentrations (c >> 10⁻⁷ mol/L). Always set [H⁺] = 0 in the Initial row for weak acid ICE tables.

Exam Tip

In acid-base calculations, always write the balanced equation first, identify the conjugate pair, and state any assumptions before substituting into the Ka expression.

ICE table framework for weak acids, I row: [HA] = c, [H⁺] = 0, [A⁻] = 0. C row: −x, +x, +x. E row: (c−x), x, x. Ka = x²/(c−x). Solve for x = [H⁺], then pH = −log(x). Never use [H⁺] = c for a weak acid, this ignores the equilibrium and overcalculates [H⁺] by up to 100×.

Pause, copy the highlighted definition into your book before moving on.

Quick Check, Card 1

For 0.100 mol/L acetic acid (Ka = 1.8 × 10⁻⁵), which entry belongs in the E (Equilibrium) row for [HA]?

Card 2, The Simplifying Assumption: When It Applies and When It Fails

We just saw the ICE table framework, Ka = x²/(c−x) where x = [H⁺] at equilibrium. That raises a question: Do you always need to solve the full quadratic, or is there a shortcut, and how do you know when it's safe? This card answers it → if Ka/c < 0.0025 (degree of ionisation < 5%), use x = √(Ka × c); otherwise solve the full quadratic.

Solving Ka = x²/(c − x) exactly requires the quadratic formula, but for most weak acid problems a simplifying assumption reduces this to a one-step square root calculation, and knowing when the assumption is valid versus when the quadratic is necessary is itself an HSC exam skill.

The simplifying assumption: if x << c, then c − x ≈ c, and Ka ≈ x²/c → x ≈ √(Ka × c).

The assumption is valid when degree of ionisation < 5%. A faster check: if Ka/c < 0.0025, the assumption is valid (since x/c = √(Ka/c)).

When the assumption fails (Ka/c ≥ 0.0025), solve the full quadratic: x² + Ka·x − Ka·c = 0, taking the positive root only:

x = (−Ka + √(Ka² + 4·Ka·c)) / 2

Assumption valid?

✓ Yes → Ka/c < 0.0025; degree of ionisation < 5%
✗ No → Ka/c ≥ 0.0025; degree of ionisation ≥ 5%
✓ Yes (0.01% << 5%)
✗ No (20% >> 5%)

Method

x = √(Ka × c); check x/c < 5% ✓
x = (−Ka + √(Ka² + 4Kac))/2
Square root valid
Quadratic required

Must Do

Always check the simplifying assumption after using it, not before. Calculate x = √(Ka × c), then verify (x/c) × 100% < 5%. If it fails, go back and solve the quadratic. Writing "assumption valid" without checking it is a methodology error that loses marks in HSC extended response.

Common Error

Students who skip the check and use the simplifying assumption when it fails obtain [H⁺] values that are systematically too high (x is overestimated when c − x is not approximated as c). For Ka/c = 0.20, the error in [H⁺] is ~13%, significant for precise pH calculations.

Insight

The 5% rule is a practical compromise, there is nothing magical about 5%. In contexts requiring very precise pH (pharmaceutical quality control, blood buffer calculations), a 1% threshold is used. In HSC, the 5% rule is standard, but knowing where it comes from (acceptable error in [H⁺] propagating to pH) is the Band 6 understanding.

Simplifying assumption: if Ka/c < 0.0025, use x = √(Ka × c) for [H⁺] (saves solving a quadratic). Always verify AFTER: degree of ionisation = (x/c) × 100% must be < 5%. If it fails, use the quadratic: x = (−Ka + √(Ka² + 4Ka·c))/2. Never write "assumption valid" without actually checking.

Add the highlighted point to your notes before the check below.

Quick Check, Card 2

For 0.0100 mol/L weak acid with Ka = 1.0 × 10⁻⁴, what is Ka/c and is the simplifying assumption valid?

Card 3, Weak Base Calculations: Kb and the pOH Route

We just saw the ICE table and simplifying assumption for weak acids. That raises a question: How do you handle a weak base, does the same ICE approach apply? This card answers it → identical ICE table but products are BH⁺ and OH⁻; use Kb instead of Ka; then convert: [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH.

Weak base pH calculations follow exactly the same ICE table logic as weak acid calculations, the only differences are that the equilibrium produces OH⁻ instead of H⁺, the equilibrium constant is Kb instead of Ka, and the final step requires converting [OH⁻] to pH via the pOH route.

For a weak base B at initial concentration c: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). Kb = [BH⁺][OH⁻]/[B].

Stage	[B]	[BH⁺]	[OH⁻]
I (Initial)	c	0	0
C (Change)	−x	+x	+x
E (Equilibrium)	c − x	x	x

If Kb/c < 0.0025: x = √(Kb × c) = [OH⁻]. Then: pOH = −log[OH⁻] → pH = 14 − pOH.

The conjugate pair relationship Ka × Kb = Kw = 1.0 × 10⁻¹⁴ connects acid and base calculations. Given Ka for HA, find Kb for A⁻: Kb = Kw/Ka. For example, CH₃COO⁻ (conjugate base of acetic acid, Ka = 1.8 × 10⁻⁵): Kb = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰.

Weak base	Kb	c (mol/L)	[OH⁻] = √(Kb × c)	pOH	pH
NH₃	1.8 × 10⁻⁵	0.10	1.34 × 10⁻³	2.87	11.13
CH₃COO⁻	5.6 × 10⁻¹⁰	0.10	7.48 × 10⁻⁶	5.13	8.87
CO₃²⁻	2.1 × 10⁻⁴	0.050	3.24 × 10⁻³	2.49	11.51

Must Do

The mandatory three-step sequence for weak bases: (1) find [OH⁻] from the ICE table; (2) pOH = −log[OH⁻]; (3) pH = 14 − pOH. Never calculate pH = −log[OH⁻], this gives a number below 7 for a basic solution, which is immediately recognisable as wrong but still loses marks.

Common Error

Students apply Ka × Kb = Kw to non-conjugate pairs. This relationship applies only to a conjugate acid-base pair, the specific acid HA and its conjugate base A⁻. Ka(CH₃COOH) × Kb(NH₃) ≠ Kw because CH₃COOH and NH₃ are not a conjugate pair. Always identify the conjugate pair explicitly before applying this relationship.

Weak base ICE table: B + H₂O ⇌ BH⁺ + OH⁻; same structure as weak acid but products differ. Route: [OH⁻] = √(Kb × c) (if Kb/c < 0.0025) → pOH = −log[OH⁻] → pH = 14 − pOH. Ka × Kb = Kw only for conjugate pairs (HA / A⁻). Never write pH = −log[OH⁻], that gives acidic pH for a basic solution.

Pause, write the highlighted definition into your book.

Quick Check, Card 3

NH₃ has Kb = 1.8 × 10⁻⁵. For 0.10 mol/L NH₃, [OH⁻] = 1.34 × 10⁻³ mol/L. What is the correct pH?

Card 4, Finding Ka from Measured pH: The Reverse Calculation

We just saw the ICE table gives Ka when you know pH is unknown, but can you run it in reverse? That raises a question: If you measure pH of a known-concentration weak acid in the lab, how do you find Ka? This card answers it → [H⁺] = 10⁻ᵖᴴ; [A⁻] = [H⁺]; [HA] = c − [H⁺]; Ka = [H⁺]²/(c − [H⁺]).

Every Ka value in every data table was originally determined from a pH measurement, and being able to reverse the calculation (from measured pH back to Ka) is both a practical laboratory skill and a frequently examined HSC calculation type.

If the pH of a weak acid solution of known concentration is measured, Ka can be calculated by reversing the ICE table:

Calculation

[H⁺] = 10⁻ᵖᴴ
[A⁻] = [H⁺] (stoichiometry)
[HA] = c − [H⁺]
Ka = [H⁺]² / (c − [H⁺])
α = ([H⁺]/c) × 100%

What it finds

Equilibrium [H⁺] = x
Equilibrium [A⁻]
Equilibrium [HA] (amount not ionised)
Acid dissociation constant
Degree of ionisation (check vs 5%)

Must Do

In step 3, [HA] = c − [H⁺], not just c. Students frequently write Ka = [H⁺]²/c instead of Ka = [H⁺]²/(c − [H⁺]) in the reverse calculation. In the reverse direction you already know [H⁺], use the exact expression Ka = [H⁺]²/(c − [H⁺]) always.

Common Error

Students confuse Ka with pKa, writing Ka = pH or Ka = −log(pH). These are distinct quantities: pH describes the actual [H⁺] in a specific solution; pKa describes the intrinsic acid strength. They are equal only at the half-equivalence point of a titration.

Reverse ICE (Ka from pH): [H⁺] = 10⁻ᵖᴴ → [A⁻] = [H⁺] (1:1 stoichiometry) → [HA]eq = c − [H⁺] → Ka = [H⁺]²/(c − [H⁺]). Always use (c − [H⁺]) in the denominator, not just c. Degree of ionisation α = ([H⁺]/c) × 100%; if < 5%, the simplifying assumption would have been valid. Ka ≠ pH.

Add the highlighted point to your notes before the check below.

Quick Check, Card 4

A 0.100 mol/L weak acid has pH 3.00. What is [H⁺] and the correct expression for Ka?

Card 5, Mixing Weak Acid with Strong Base: Introduction to the Buffer Region

We just saw how to extract Ka from a measured pH using the reverse ICE calculation. That raises a question: What happens when you add a strong base to a weak acid but don't neutralise all of it, which method applies? This card answers it → partial neutralisation creates a buffer; use Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA)); at half-equivalence pH = pKa.

When a weak acid is partially neutralised by a strong base, before the equivalence point, the solution contains significant amounts of both the weak acid and its conjugate base simultaneously, and the pH calculation must account for this mixture using the Henderson-Hasselbalch equation.

When NaOH is added to HA before the equivalence point: HA + OH⁻ → A⁻ + H₂O (goes to completion, NaOH is strong). After the reaction, some HA remains and A⁻ has been formed. The solution contains both HA and A⁻, this is a buffer solution (developed fully in L13).

pH is calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). In the same volume, [A⁻]/[HA] = n(A⁻)/n(HA) (volumes cancel). At the half-equivalence point exactly half the acid neutralised, n(A⁻) = n(HA), so log(1) = 0 and pH = pKa.

What to calculate

Moles of HA initially
Moles of OH⁻ added
HA remaining after reaction
A⁻ formed
pH of buffer mixture

Formula

n(HA) = c(HA) × V(HA)
n(OH⁻) = c(NaOH) × V(NaOH)
n(HA)rem = n(HA) − n(OH⁻)
n(A⁻) = n(OH⁻) (1:1 stoich)
pH = pKa + log(n(A⁻)/n(HA))

Must Do

In Henderson-Hasselbalch, [A⁻]/[HA] can be replaced by n(A⁻)/n(HA) when both species are in the same solution, volumes cancel in the ratio. This means you can use moles directly without calculating concentrations. This is the standard shortcut for buffer and partial neutralisation calculations.

Common Error

Students apply the simple ICE table method (x = √(Ka × c)) to a solution containing both HA and A⁻ after partial neutralisation. The ICE table assumes [A⁻] = 0 initially, this is invalid when A⁻ has been produced by neutralisation. A solution containing both HA and A⁻ is a buffer, it requires Henderson-Hasselbalch. The ICE table method ignores the common ion effect of A⁻ and gives a pH that is erroneously too low.

Partial neutralisation creates a buffer (HA + A⁻ both present). Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA)), use moles directly (volumes cancel in ratio). At the half-equivalence point n(A⁻) = n(HA) → log(1) = 0 → pH = pKa. Never use the plain ICE table for a solution containing both HA and A⁻, it ignores the common ion effect.

Pause, copy the highlighted definition into your book before moving on.

Quick Check, Card 5

25.0 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵, pKa = 4.74) is mixed with 12.5 mL of 0.100 mol/L NaOH. What is the pH?

⚠ Common Misconceptions, Weak Acid/Base Calculations

✕ "pH = pKa for any weak acid solution."

✓ pH = pKa ONLY at the half-equivalence point of a titration, when [A⁻] = [HA]. For a pure weak acid solution, pH = −log(√(Ka × c)) which is approximately pKa/2 + log(c)/2, always lower than pKa.

✕ "I don't need to check the 5% rule if Ka is small."

✓ Ka must be small relative to c. For a very dilute solution (c = 0.001 mol/L) with Ka = 1.0 × 10⁻⁴, Ka/c = 0.10 = 10%, the assumption fails even though Ka appears small in absolute terms.

✕ "Ka × Kb = Kw can be used for any acid and base."

✓ This only applies to a specific conjugate pair (HA and its conjugate base A⁻). Ka(HCl) × Kb(NH₃) ≠ Kw, HCl and NH₃ are not a conjugate pair.

✕ "A salt formed by neutralisation always has pH = 7."

✓ Only salts from strong acid + strong base give pH = 7. Salts from weak acid + strong base (e.g. CH₃COONa) are basic; salts from strong acid + weak base (e.g. NH₄Cl) are acidic.

Worked Examples

Worked Example 1, ICE Table pH Calculation with Assumption Check

Calculate the pH of 0.150 mol/L propanoic acid (CH₃CH₂COOH, Ka = 1.3 × 10⁻⁵) at 25°C. Check whether the simplifying assumption is valid.

GIVEN: c = 0.150 mol/L, Ka = 1.3 × 10⁻⁵. FIND: pH.

Pre-check assumption: Ka/c = (1.3 × 10⁻⁵)/0.150 = 8.7 × 10⁻⁵ = 0.0087% << 5% → assumption likely valid. Proceed with square root.

ICE table:

Stage	[CH₃CH₂COOH]	[H⁺]	[CH₃CH₂COO⁻]
I	0.150	0	0
C	−x	+x	+x
E	0.150 − x	x	x

Apply simplifying assumption: x = √(Ka × c) = √(1.3 × 10⁻⁵ × 0.150) = √(1.95 × 10⁻⁶) = 1.396 × 10⁻³ mol/L.
Verify: degree of ionisation = (1.396 × 10⁻³/0.150) × 100% = 0.93% < 5% ✓ Assumption valid.

Calculate pH: pH = −log(1.396 × 10⁻³) = −log(1.396) + 3 = −0.145 + 3 = 2.85.
Sanity check: pH 2.85 < 7 ✓ (acid). pH 2.85 > 1.0 (not treating it as strong acid) ✓

ANSWER: [H⁺] = 1.40 × 10⁻³ mol/L; pH = 2.85. Degree of ionisation = 0.93%, simplifying assumption valid.

Worked Example 2, Multi-Part: Ka, Kb, Conjugate Base pH, and Claim Evaluation

Acetic acid has Ka = 1.8 × 10⁻⁵. (a) Calculate the pH of 0.250 mol/L CH₃COOH. (b) Calculate Kb for the acetate ion (CH₃COO⁻). (c) Calculate the pH of 0.250 mol/L sodium acetate (CH₃COONa). (d) Evaluate the claim: "The pH of 0.250 mol/L sodium acetate should be 7 because it is a salt formed by neutralisation."

GIVEN: Ka(CH₃COOH) = 1.8 × 10⁻⁵; c = 0.250 mol/L for both solutions.

METHOD (a, acetic acid pH): Check: Ka/c = 1.8 × 10⁻⁵/0.250 = 7.2 × 10⁻⁵ << 0.0025 ✓. x = √(1.8 × 10⁻⁵ × 0.250) = √(4.5 × 10⁻⁶) = 2.121 × 10⁻³ mol/L. Verify: 2.121 × 10⁻³/0.250 = 0.85% < 5% ✓. pH = −log(2.121 × 10⁻³) = 2.67.

METHOD (b, Kb for CH₃COO⁻): Ka × Kb = Kw → Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰.

METHOD (c, sodium acetate pH): CH₃COONa → Na⁺ (neutral spectator) + CH₃COO⁻ (weak base). ICE for CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻: Check: Kb/c = 5.56 × 10⁻¹⁰/0.250 = 2.22 × 10⁻⁹ << 0.0025 ✓. x = √(5.56 × 10⁻¹⁰ × 0.250) = √(1.39 × 10⁻¹⁰) = 1.179 × 10⁻⁵ mol/L = [OH⁻]. Verify: 1.179 × 10⁻⁵/0.250 = 0.0047% < 5% ✓. pOH = −log(1.179 × 10⁻⁵) = 4.93. pH = 14.00 − 4.93 = 9.07.

METHOD (d, evaluate claim): The student's claim is incorrect. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (CH₃COOH). A neutral solution (pH = 7) results only from a salt of a strong acid + strong base. The acetate ion is the conjugate base of a weak acid, it has a significant tendency to accept H⁺ from water (Kb = 5.56 × 10⁻¹⁰), producing OH⁻ and making the solution basic (pH = 9.07). "Neutralisation" refers to the process, not the product.

ANSWER: (a) pH = 2.67. (b) Kb(CH₃COO⁻) = 5.56 × 10⁻¹⁰. (c) pH = 9.07. (d) Claim incorrect, acetate ion is the conjugate base of a weak acid; it hydrolyses to produce OH⁻, giving a basic solution. pH = 7 requires a salt of a strong acid + strong base (e.g. NaCl).

Worked Example 3, Band 6: Ka from pH, Quadratic Check, and Partial Neutralisation

Evaluate Band 6 (8 marks)

A 0.0500 mol/L solution of weak monoprotic acid HX is measured at 25°C and found to have pH 2.78. (a) Calculate [H⁺] and the Ka of HX. (b) Check whether the simplifying assumption would have been valid for this calculation. (c) 40.0 mL of 0.0500 mol/L HX is mixed with 10.0 mL of 0.0500 mol/L NaOH. Calculate the pH of the resulting mixture. (d) A student attempts part (c) using x = √(Ka × c(HX)). Identify the conceptual error and explain the correct method.

GIVEN: c = 0.0500 mol/L, pH = 2.78; part (c): 40.0 mL HX + 10.0 mL NaOH, both 0.0500 mol/L.

METHOD (a, [H⁺] and Ka): [H⁺] = 10⁻²·⁷⁸ = 10⁻³ × 10⁰·²² = 1.0 × 10⁻³ × 1.660 = 1.66 × 10⁻³ mol/L. Ka = [H⁺]²/(c − [H⁺]) = (1.66 × 10⁻³)²/(0.0500 − 1.66 × 10⁻³) = (2.756 × 10⁻⁶)/(0.04834) = 5.70 × 10⁻⁵.

METHOD (b, check assumption): α = ([H⁺]/c) × 100% = (1.66 × 10⁻³/0.0500) × 100% = 3.32% < 5% ✓. The simplifying assumption would have been valid, x = √(5.70 × 10⁻⁵ × 0.0500) = √(2.85 × 10⁻⁶) = 1.688 × 10⁻³ → pH = 2.77. Error of only 0.01 pH units vs measured 2.78. ✓

METHOD (c, partial neutralisation): n(HX) = 0.0500 × 0.0400 = 2.00 × 10⁻³ mol. n(OH⁻) = 0.0500 × 0.0100 = 5.00 × 10⁻⁴ mol. After HA + OH⁻ → A⁻ + H₂O (complete): n(HX) remaining = 2.00 × 10⁻³ − 5.00 × 10⁻⁴ = 1.50 × 10⁻³ mol. n(X⁻) formed = 5.00 × 10⁻⁴ mol. pKa = −log(5.70 × 10⁻⁵) = 4.244. pH = pKa + log(n(X⁻)/n(HX)) = 4.244 + log(5.00 × 10⁻⁴/1.50 × 10⁻³) = 4.244 + log(0.333) = 4.244 − 0.477 = 3.77.

METHOD (d, conceptual error): The student's approach uses x = √(Ka × c(HX)), the ICE table method for a pure weak acid with [A⁻] = 0 initially. This is wrong because the solution after partial neutralisation contains significant X⁻ (5.00 × 10⁻⁴ mol) simultaneously with HX. The ICE table assumes [A⁻] = 0 at the start, completely invalid here. This is a buffer solution. The X⁻ already present suppresses further ionisation of HX (common ion effect), making [H⁺] lower than the ICE table predicts. The correct method is Henderson-Hasselbalch: pH = pKa + log(n(X⁻)/n(HX)).

ANSWER: (a) [H⁺] = 1.66 × 10⁻³ mol/L; Ka = 5.70 × 10⁻⁵. (b) α = 3.32% < 5%, assumption was valid; pH error only 0.01 units. (c) n(HX)rem = 1.50 × 10⁻³ mol; n(X⁻) = 5.00 × 10⁻⁴ mol; pKa = 4.244; pH = 3.77. (d) Error: ICE table assumes [A⁻] = 0, invalid when X⁻ is present from neutralisation; the solution is a buffer; Henderson-Hasselbalch is required; ICE method ignores common ion suppression and gives erroneously low pH.

📓 Minimum Facts & Procedures, Lesson 9

ICE table: I (initial), C (change = −x, +x, +x), E (equilibrium = c−x, x, x) → Ka = x²/(c−x)
Simplifying assumption: valid when Ka/c < 0.0025 → x = √(Ka × c); always verify x/c < 5% after
If invalid: x = (−Ka + √(Ka² + 4·Ka·c)) / 2 (take positive root)
Weak base: same ICE table with Kb → [OH⁻] = x → pOH = −log[OH⁻] → pH = 14 − pOH
Conjugate pair: Ka × Kb = 1.0 × 10⁻¹⁴ → Kb = Kw/Ka for conjugate base
Ka from pH: [H⁺] = 10⁻ᵖᴴ → Ka = [H⁺]²/(c − [H⁺]); degree of ionisation = ([H⁺]/c) × 100%
Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA)); at half-equivalence point: pH = pKa
Buffer (HA + A⁻ both present) → must use H-H equation, NOT the ICE table method

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Use the pH Calculator. The pH of a 0.01 mol/L strong acid solution is…

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A student calculates the pH of 0.1 mol/L CH₃COOH by assuming [H⁺] = 0.1 mol/L (the "strong acid shortcut") and gets pH = 1.0. The actual pH is 2.87. Predict: (a) what went wrong and (b) what the ICE table would show.

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Activities

Calculate & Interpret: ICE Table Practice

For each weak acid or base, set up a full ICE table, check the simplifying assumption, and calculate pH. Show all working.

Species	Concentration	Ka or Kb
HNO₂ (nitrous acid)	0.100 mol/L	Ka = 4.5 × 10⁻⁴
HCN (hydrocyanic acid)	0.0500 mol/L	Ka = 6.2 × 10⁻¹⁰
NH₃ (ammonia)	0.200 mol/L	Kb = 1.8 × 10⁻⁵
CH₃COOH (acetic acid)	0.0100 mol/L	Ka = 1.8 × 10⁻⁵

Analyse & Connect: Ka from pH and Partial Neutralisation

A 0.100 mol/L solution of weak acid HA has pH 3.15 at 25°C. (a) Calculate [H⁺]. (b) Calculate Ka of HA using the exact expression. (c) Calculate the degree of ionisation. Was the simplifying assumption valid?
25.0 mL of 0.100 mol/L HA (Ka from part 1 above) is mixed with 10.0 mL of 0.100 mol/L NaOH. (a) Calculate n(HA) remaining and n(A⁻) formed. (b) Calculate pH using Henderson-Hasselbalch. (c) What would pH be at the half-equivalence point of this titration?
Explain why the simple ICE table method (x = √(Ka × c)) cannot be used to calculate the pH in question 2(b). What assumption does the ICE table make that is violated?

Check Your Understanding

Multiple Choice Questions

Analyse Band 4

1. A student sets up an ICE table for 0.200 mol/L hydrofluoric acid (HF, Ka = 6.8 × 10⁻⁴) and obtains x = 0.0117 mol/L using the simplifying assumption, without checking the validity. What error has the student made?

Apply Band 3

2. The Ka of benzoic acid (C₆H₅COOH) is 6.5 × 10⁻⁵. Calculate the pH of 0.100 mol/L benzoic acid. Which answer and method are correct?

Analyse Band 5

3. 25.0 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵) is mixed with 12.5 mL of 0.100 mol/L NaOH. A student calculates pH using x = √(Ka × 0.100). What is wrong and what is the correct pH?

Apply Band 4

4. Ammonia has Kb = 1.8 × 10⁻⁵. What is the pH of 0.050 mol/L NH₃?

Analyse Band 4

5. A 0.050 mol/L solution of weak acid HA has a measured pH of 3.50 at 25°C. Which of the following correctly calculates Ka?

Short Answer Questions

Apply Band 4 (4 marks)

Question 6. Calculate the pH of each of the following at 25°C. Show a full ICE table and assumption check for each.

(a) 0.0800 mol/L chloroacetic acid (ClCH₂COOH, Ka = 1.4 × 10⁻³) (b) 0.500 mol/L sodium propanoate (CH₃CH₂COONa; Ka(CH₃CH₂COOH) = 1.3 × 10⁻⁵)

Apply Band 4 (4 marks)

Question 7. A 0.200 mol/L solution of weak acid HB has a measured pH of 2.52 at 25°C.

(a) Calculate [H⁺] and Ka of HB. (b) Calculate the degree of ionisation. Would the simplifying assumption have been valid? (c) A student claims "Ka of HB = 10⁻²·⁵² because Ka equals [H⁺] in a weak acid solution." Identify and correct the error in this reasoning.

Evaluate Band 6 (7 marks)

Question 8. Lactic acid (HC₃H₅O₃, Ka = 1.4 × 10⁻⁴) builds up in muscles during intense exercise. Consider a 0.0200 mol/L solution of lactic acid in muscle fluid at 25°C.

(a) Determine whether the simplifying assumption is valid for this system. If not, use the quadratic. Calculate [H⁺] and pH. (3 marks) (b) 30.0 mL of 0.0200 mol/L lactic acid is mixed with 10.0 mL of 0.0200 mol/L NaOH. Calculate the pH of the resulting buffer mixture. (2 marks) (c) Explain why the pH you calculated in (b) is higher than the pH in (a), even though more acid is present than in (b)'s buffer. Use the concept of the common ion effect in your answer. (2 marks)

Show Answers & Explanations

MC Answers

Q1: B x = √(6.8 × 10⁻⁴ × 0.200) = √(1.36 × 10⁻⁴) = 0.01166 mol/L. Check: 0.01166/0.200 = 5.83% > 5%, assumption marginally invalid. The student must check after calculating x and apply the quadratic. Quadratic gives x = 0.01133 mol/L (~3% smaller). Option A: always must check. Option C: Ka/c has units mol/L (wrong formula). Option D: wrong sign, H⁺ is produced (+x).

Q2: C Benzoic acid is weak, ICE method required. Ka/c = 6.5 × 10⁻⁵/0.100 = 6.5 × 10⁻⁴ << 0.0025 ✓. x = √(6.5 × 10⁻⁵ × 0.100) = √(6.5 × 10⁻⁶) = 2.55 × 10⁻³. Verify: 2.55% < 5% ✓. pH = −log(2.55 × 10⁻³) = 2.59. Option A: treats as strong. Option B: confuses pKa with pH. Option D: uses Ka × c instead of √(Ka × c).

Q3: B n(CH₃COOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. n(OH⁻) = 0.100 × 0.0125 = 1.25 × 10⁻³ mol. After reaction: n(CH₃COOH)rem = 1.25 × 10⁻³ mol; n(CH₃COO⁻) = 1.25 × 10⁻³ mol. Half-equivalence point: [A⁻]/[HA] = 1 → pH = pKa = −log(1.8 × 10⁻⁵) = 4.74. Option A: ignores NaOH. Option C: still ignores common ion effect of CH₃COO⁻.

Q4: C[OH⁻] = √(1.8 × 10⁻⁵ × 0.050) = √(9.0 × 10⁻⁷) = 9.49 × 10⁻⁴ mol/L. pOH = −log(9.49 × 10⁻⁴) = 3.02. pH = 14.00 − 3.02 = 10.98. Options A and B: took −log[OH⁻] as pH directly (wrong). Option D: NH₃ is basic, not neutral.

Q5: B[H⁺] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol/L. Ka = [H⁺]²/(c − [H⁺]) = (3.16 × 10⁻⁴)²/(0.050 − 3.16 × 10⁻⁴) = (9.99 × 10⁻⁸)/0.04968 = 2.01 × 10⁻⁶. Option A: uses approximation Ka = [H⁺]²/c (omits c − [H⁺]). Option C: nonsensical. Option D: confuses Ka with [H⁺].

Q6 Sample Answer

(a) 0.0800 mol/L chloroacetic acid (Ka = 1.4 × 10⁻³):
Check: Ka/c = 1.4 × 10⁻³/0.0800 = 0.0175 > 0.0025 ✗. x = √(1.4 × 10⁻³ × 0.0800) = √(1.12 × 10⁻⁴) = 1.058 × 10⁻² mol/L. Verify: (1.058 × 10⁻²/0.0800) × 100% = 13.2% > 5% ✗. Assumption fails, use quadratic.
x² + (1.4 × 10⁻³)x − (1.4 × 10⁻³)(0.0800) = 0 → x² + 1.4 × 10⁻³x − 1.12 × 10⁻⁴ = 0
x = (−1.4 × 10⁻³ + √((1.4 × 10⁻³)² + 4 × 1.12 × 10⁻⁴))/2 = (−1.4 × 10⁻³ + √(4.49 × 10⁻⁴))/2 = (−1.4 × 10⁻³ + 0.02119)/2 = 0.009895/2 = 9.79 × 10⁻³ mol/L
pH = −log(9.79 × 10⁻³) = 2.01

(b) 0.500 mol/L sodium propanoate (Ka(propanoic acid) = 1.3 × 10⁻⁵):
Kb(CH₃CH₂COO⁻) = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.3 × 10⁻⁵) = 7.69 × 10⁻¹⁰
Check: Kb/c = 7.69 × 10⁻¹⁰/0.500 = 1.54 × 10⁻⁹ << 0.0025 ✓
[OH⁻] = √(7.69 × 10⁻¹⁰ × 0.500) = √(3.85 × 10⁻¹⁰) = 1.962 × 10⁻⁵ mol/L
Verify: (1.962 × 10⁻⁵/0.500) × 100% = 0.0039% << 5% ✓
pOH = −log(1.962 × 10⁻⁵) = 4.707; pH = 14.00 − 4.707 = 9.29

Q7 Sample Answer

(a) [H⁺] and Ka:
[H⁺] = 10⁻²·⁵² = 10⁻³ × 10⁰·⁴⁸ = 1.0 × 10⁻³ × 3.020 = 3.02 × 10⁻³ mol/L
Ka = [H⁺]²/(c − [H⁺]) = (3.02 × 10⁻³)²/(0.200 − 3.02 × 10⁻³) = (9.12 × 10⁻⁶)/(0.1970) = 4.63 × 10⁻⁵

(b) Degree of ionisation:
α = (3.02 × 10⁻³/0.200) × 100% = 1.51% < 5% ✓. The simplifying assumption would have been valid, the error from using x = √(Ka × c) would be less than 5% in [H⁺].

(c) Student's error:
The student claims Ka = [H⁺] = 10⁻²·⁵². This is incorrect. Ka is an equilibrium constant with units related to concentration, it equals [H⁺][A⁻]/[HA] = [H⁺]²/(c − [H⁺]), not [H⁺] itself. Ka describes the position of the equilibrium (the ratio of products to reactants); [H⁺] describes the actual concentration of H⁺ in a specific solution. They would only be numerically equal in a highly specific and physically unrealistic case. Correct Ka = 4.63 × 10⁻⁵ ≠ [H⁺] = 3.02 × 10⁻³.

Q8 Sample Answer (Band 6)

(a) [H⁺] and pH of 0.0200 mol/L lactic acid:
Ka/c = 1.4 × 10⁻⁴/0.0200 = 7.0 × 10⁻³ > 0.0025 ✗. x = √(1.4 × 10⁻⁴ × 0.0200) = √(2.8 × 10⁻⁶) = 1.673 × 10⁻³ mol/L.
Verify: (1.673 × 10⁻³/0.0200) × 100% = 8.37% > 5% ✗. Assumption fails, use quadratic.
x² + (1.4 × 10⁻⁴)x − (1.4 × 10⁻⁴)(0.0200) = 0
x = (−1.4 × 10⁻⁴ + √((1.4 × 10⁻⁴)² + 4 × 1.4 × 10⁻⁴ × 0.0200))/2
= (−1.4 × 10⁻⁴ + √(1.96 × 10⁻⁸ + 1.12 × 10⁻⁵))/2 = (−1.4 × 10⁻⁴ + √(1.122 × 10⁻⁵))/2
= (−1.4 × 10⁻⁴ + 3.350 × 10⁻³)/2 = 3.210 × 10⁻³/2 = 1.605 × 10⁻³ mol/L
pH = −log(1.605 × 10⁻³) = 2.79

(b) pH of buffer (30.0 mL HA + 10.0 mL NaOH, both 0.0200 mol/L):
n(HA) = 0.0200 × 0.0300 = 6.00 × 10⁻⁴ mol; n(OH⁻) = 0.0200 × 0.0100 = 2.00 × 10⁻⁴ mol
After HA + OH⁻ → A⁻ + H₂O: n(HA)rem = 4.00 × 10⁻⁴ mol; n(A⁻) = 2.00 × 10⁻⁴ mol
pKa = −log(1.4 × 10⁻⁴) = 3.854
pH = 3.854 + log(2.00 × 10⁻⁴/4.00 × 10⁻⁴) = 3.854 + log(0.500) = 3.854 − 0.301 = 3.55

(c) Why pH(buffer) > pH(pure acid):
In the buffer (b), the solution contains both lactic acid HA and its conjugate base A⁻ (lactate). The A⁻ ion is already present at significant concentration. By Le Chatelier's principle, this product of the equilibrium HA ⇌ H⁺ + A⁻ suppresses the ionisation of HA, this is the common ion effect. The equilibrium shifts to the left, reducing [H⁺] compared to what pure HA alone would produce. As a result, the buffer has a higher pH (less acidic) than the pure weak acid at the same total concentration of HA. The Henderson-Hasselbalch equation accounts for this suppression mathematically through the log(n(A⁻)/n(HA)) term.

Revisit, Aspirin vs Ibuprofen

Return to your prediction about aspirin vs ibuprofen. Recall John Vane's 1971 measurements: Ka(aspirin) = 3.0 × 10⁻⁴, Ka(ibuprofen) = 1.3 × 10⁻⁵. You can now calculate which releases more H⁺ in the stomach using the ICE table method:

Aspirin in 200 mL stomach fluid: c = (0.500/180)/0.200 = 0.0139 mol/L. Approximation gives [H⁺] = √(3.0 × 10⁻⁴ × 0.0139) = 2.04 × 10⁻³ mol/L (degree of ionisation = 14.7% > 5%, so assumption fails). Quadratic gives [H⁺] ≈ 1.90 × 10⁻³ mol/L.
Ibuprofen in 200 mL stomach fluid: c = (0.500/206)/0.200 = 0.0121 mol/L. [H⁺] = √(1.3 × 10⁻⁵ × 0.0121) = √(1.57 × 10⁻⁷) = 3.97 × 10⁻⁴ mol/L.
Aspirin releases ~4.8× more H⁺ than ibuprofen (higher Ka dominates despite similar concentrations). However, stomach acid itself contains [H⁺] = 0.1 mol/L, the H⁺ from both tablets (10⁻³ to 10⁻⁴ mol/L) is negligible compared to existing stomach H⁺. Neither tablet significantly changes stomach pH. The Ka comparison matters more for intestinal absorption (higher pH environment).

Mark Lesson 9 as complete

← Previous: pH and pOH, Calculations for Strong Acids & Bases Next: Enthalpy of Neutralisation, Comparing Strong & Weak →

Review, Drill Pairs

Quick-Fire Recall, ICE Tables & Ka/Kb

What do the letters I, C, E stand for in an ICE table?

Initial, Change, Equilibrium, tracking concentrations of all species from start to equilibrium.

State the simplifying assumption and the condition under which it is valid.

If Ka/c < 0.0025, then c − x ≈ c, giving x = √(Ka × c). Verify degree of ionisation < 5% after calculating.

What is the three-step route from [OH⁻] to pH for a weak base?

Step 1: [OH⁻] from ICE table. Step 2: pOH = −log[OH⁻]. Step 3: pH = 14 − pOH. Never use pH = −log[OH⁻].

State the relationship between Ka and Kb for a conjugate pair, and when it applies.

Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C. Applies ONLY to a specific conjugate acid-base pair (HA and A⁻), not to any random acid and base.

When is Henderson-Hasselbalch required instead of the ICE table method?

When the solution contains BOTH the weak acid HA AND its conjugate base A⁻ (a buffer), such as after partial neutralisation. The ICE table assumes [A⁻] = 0 initially, invalid in a buffer.

🏆 Extended Response

A 0.0400 mol/L solution of weak monoprotic acid HA has a measured pH of 3.20 at 25°C.

Calculate [H⁺] and the Ka of HA. Show full working using the exact expression.
Would the simplifying assumption have been valid? Justify with a percentage calculation.
A student adds NaOH to 50.0 mL of 0.0400 mol/L HA until 25.0 mL of NaOH has been added at the same concentration. Calculate the pH of the resulting mixture.
Explain why pH = pKa at the specific point in (3) and what this tells you about the solution composition at that point.

← Previous: pH and pOH, Calculations for Strong Acids & Bases Next: Enthalpy of Neutralisation, Comparing Strong & Weak →