pH of Weak Acids — Compute pH of a weak acid solution using Ka and ICE tables - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists recognized that some acids were "stronger" than others — vinegar stings less than sulfuric acid, and lemon juice is tart but not dangerous. However, quantifying how much an acid ionizes in water required decades of theoretical development. The journey from qualitative taste-testing to precise pH calculations is a story of equilibrium chemistry coming of age.

1884 • Arrhenius Theory of Acids

Svante Arrhenius proposed that acids are substances that produce H⁺ ions in water. This was the first modern definition of an acid, but it did not distinguish between strong and weak acids quantitatively.

1909 • Sørensen Introduces pH

Søren Sørensen, working at the Carlsberg brewery laboratory, introduced the pH scale as a convenient way to express hydrogen ion concentration. The logarithmic scale made it much easier to compare acidity levels across solutions.

1923 • Brønsted–Lowry Acid–Base Theory

Johannes Brønsted and Thomas Lowry independently defined acids as proton donors and bases as proton acceptors. This broader framework explained why weak acids only partially donate their protons to water.

1930s • Ka Values Tabulated

Chemists systematically measured acid dissociation constants (Ka) for hundreds of weak acids, enabling precise equilibrium calculations. ICE tables became a standard problem-solving tool in chemistry education.

The central question this lesson addresses is straightforward but powerful: if a weak acid only partially ionizes in water, how do we figure out exactly how much H⁺ is produced and what the resulting pH will be? The answer lies in combining the equilibrium constant K_a with a systematic bookkeeping method called the ICE table.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to understand a few foundational ideas that distinguish weak acids from strong acids. A strong acid like HCl ionizes 100% in water — every molecule donates its proton. A weak acid like acetic acid (CH₃COOH) only ionizes partially, establishing a dynamic equilibrium between the intact acid molecules and the ions they produce.

Weak Acid Equilibrium

A weak acid HA dissolves in water and partially ionizes: HA ⇌ H⁺ + A⁻. Both the forward and reverse reactions occur simultaneously, so un-ionized HA molecules coexist with H⁺ and A⁻ ions at equilibrium.

Acid Dissociation Constant (Ka)

K_a measures the strength of a weak acid. It equals [H⁺][A⁻] / [HA] at equilibrium. A larger K_a means greater ionization and a stronger weak acid.

The ICE Table

ICE stands for Initial, Change, Equilibrium. It is an organized table that tracks how concentrations shift as a reaction moves toward equilibrium. Each column represents a species in the reaction.

The pH Scale

pH = −log[H⁺]. Because the log function compresses large ranges, small differences in pH correspond to tenfold differences in H⁺ concentration. A pH below 7 is acidic; above 7 is basic.

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KEY TAKEAWAY

Think of a weak acid like a shy volunteer in class. If you ask 100 students to raise their hands (ionize), a strong acid is like a class where everyone raises their hand immediately. A weak acid is like a class where only a few students raise their hands — the rest keep them down. K_a tells you what fraction of the class actually volunteers.

SECTION 3

Visual Explanation — Weak vs. Strong Acid Ionization

Left: a strong acid like HCl ionizes completely — every molecule separates into H⁺ and Cl⁻. Right: a weak acid like acetic acid barely ionizes — most molecules remain intact as HA (purple), with only a few H⁺ (gold) and A⁻ (green) ions present. This partial ionization is why we need K_a and ICE tables to compute pH.

The diagram above reveals the key difference. In the strong acid beaker, every HCl molecule has donated its proton, so [H⁺] equals the initial acid concentration and pH is easy to compute: pH = −log(0.10) = 1.0. In the weak acid beaker, however, most acetic acid molecules keep their protons. Only a small fraction ionizes, so [H⁺] is much less than the initial concentration. This is why you cannot simply plug the initial concentration into pH = −log[H⁺] for a weak acid — you must first figure out how much ionization actually occurs.

SECTION 4

Mathematical Framework — Ka Expression & ICE Table Setup

The calculation of weak acid pH follows a clear sequence of steps. First, write the ionization equilibrium. Second, set up the K_a expression. Third, use an ICE table to express equilibrium concentrations in terms of a single variable, x. Fourth, solve for x. Finally, take the negative log to find pH.

WEAK ACID IONIZATION

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

HA = the weak acid; H⁺ = hydrogen ion (proton); A⁻ = conjugate base. The double arrow (⇌) indicates that the reaction reaches equilibrium rather than going to completion.

ACID DISSOCIATION CONSTANT

Ka = [H⁺][A⁻] / [HA]

Square brackets denote molar concentrations at equilibrium. K_a is a constant at a given temperature. Water is the solvent and is omitted from the expression.

Setting Up the ICE Table

An ICE table organizes the concentration changes. Suppose you dissolve a weak acid HA at an initial concentration of C₀ (say 0.10 M). Initially, no ionization has occurred, so [H⁺] = 0 and [A⁻] = 0. As the reaction proceeds toward equilibrium, some amount x of HA ionizes. This means [HA] decreases by x while [H⁺] and [A⁻] each increase by x (because of the 1:1:1 stoichiometry).

	HA	H⁺	A⁻
I (Initial)	C₀	0	0
C (Change)	− x	+ x	+ x
E (Equilibrium)	C₀ − x	x	x

Generic ICE table for weak acid ionization

SUBSTITUTION INTO Ka

Ka = (x)(x) / (C₀ − x) = x² / (C₀ − x)

This equation can be solved exactly using the quadratic formula, or simplified with the 5% approximation: if x is small compared to C₀ (less than 5% of C₀), then C₀ − x ≈ C₀, and the equation simplifies to x ≈ √(K_a × C₀).

FINAL pH CALCULATION

pH = −log[H⁺] = −log(x)

Once you solve for x (which equals [H⁺] at equilibrium), take the negative base-10 logarithm to find pH.

SECTION 5

Detailed Breakdown — The 5% Approximation & When to Use It

Solving x² / (C₀ − x) = K_a exactly requires the quadratic formula, which can be tedious. The 5% approximation offers a shortcut. If the degree of ionization is small — specifically, if x / C₀ < 0.05 (or 5%) — then dropping x from the denominator introduces negligible error. The approximation works well when K_a is small relative to C₀. A common rule of thumb: the approximation is valid when C₀ / K_a ≥ 400.

This decision flowchart shows the two solving paths. After setting up the ICE table, check whether C₀ / K_a ≥ 400. If yes, the approximation x ≈ √(K_a × C₀) works. If not, use the quadratic formula. Either way, both paths converge on pH = −log(x).

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When Does the Approximation Fail?

The approximation breaks down in two common scenarios: (1) when the acid is relatively strong for a "weak" acid (K_a > 10⁻³) or (2) when the solution is very dilute (C₀ < 0.01 M). In either case, x becomes a significant fraction of C₀, and you must use the quadratic formula. Always check your answer: divide x by C₀ and verify it is less than 0.05.

Here is a summary of the complete algorithm. Start by writing the balanced ionization equation and setting up the ICE table. Substitute the equilibrium expressions into the K_a formula. Decide whether the approximation is valid by checking C₀ / K_a. Solve for x, verify the 5% condition, and finally compute pH = −log(x). With practice, this process becomes second nature.

SECTION 6

Worked Example — pH of 0.25 M Acetic Acid

Let's walk through a complete pH calculation for a 0.25 M solution of acetic acid (CH₃COOH), which has K_a = 1.8 × 10⁻⁵.

Find the pH of 0.25 M CH₃COOH (Ka = 1.8 × 10⁻⁵)

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Step 1 — Write the Ionization Equation

Acetic acid donates a proton to water: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq). This is the equilibrium we will analyze.
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Step 2 — Set Up the ICE Table

Initial: [CH₃COOH] = 0.25 M, [H⁺] = 0, [CH₃COO⁻] = 0. Change: [CH₃COOH] decreases by x, [H⁺] increases by x, [CH₃COO⁻] increases by x. Equilibrium: [CH₃COOH] = 0.25 − x, [H⁺] = x, [CH₃COO⁻] = x.
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Step 3 — Substitute into Ka Expression

K_a = x² / (0.25 − x) = 1.8 × 10⁻⁵. Check the approximation: C₀ / K_a = 0.25 / (1.8 × 10⁻⁵) = 13,889. Since 13,889 ≥ 400, the approximation is valid.
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Step 4 — Solve Using the Approximation

Approximate: x² / 0.25 ≈ 1.8 × 10⁻⁵. Therefore x² = (1.8 × 10⁻⁵)(0.25) = 4.5 × 10⁻⁶. Taking the square root: x = √(4.5 × 10⁻⁶) = 2.12 × 10⁻³ M.

[H⁺] = x = 2.12 × 10⁻³ M
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Step 5 — Verify the 5% Approximation

Check: x / C₀ = (2.12 × 10⁻³) / 0.25 = 0.0085, or 0.85%. Since 0.85% < 5%, the approximation is valid.
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Step 6 — Calculate pH

pH = −log(2.12 × 10⁻³) = −log(2.12) − log(10⁻³) = −0.326 + 3 = 2.67.

pH = 2.67

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Sanity Check

The pH of a strong acid at the same concentration would be −log(0.25) = 0.60. Our weak acid gives pH = 2.67, which is higher (less acidic). This makes sense because a weak acid produces fewer H⁺ ions than a strong acid at the same concentration.

SECTION 7

Strengths & Limitations of the ICE-Table Method

Aspect	Strength	Limitation
Organization	ICE tables provide a clear, systematic framework that prevents errors in tracking concentration changes.	For polyprotic acids (like H₂SO₃), multiple ICE tables are needed — one for each ionization step.
5% Approximation	Simplifies a quadratic equation to a straightforward square root calculation, saving significant time on tests.	Fails for dilute solutions or moderately strong weak acids, requiring you to fall back to the quadratic formula.
Accuracy	Gives precise results for monoprotic weak acids in pure water at reasonable concentrations.	Ignores autoionization of water (Kw), which matters only at extremely low acid concentrations (below ~10⁻⁶ M).
Versatility	The same ICE framework applies to weak bases (using Kb), buffer systems, and solubility equilibria.	Does not account for activity coefficients in concentrated or ionic solutions, which affect real-world accuracy.

Strengths and limitations of the Ka / ICE table approach

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KEY TAKEAWAY

The ICE table is like a spreadsheet for chemical reactions: it tracks every deposit and withdrawal of concentration, so your books always balance. The 5% approximation is a shortcut that works when the numbers are lopsided enough — like rounding $999,999 to a million. But when the shortcut introduces too much rounding error, you need the exact quadratic solution to stay accurate.

SECTION 8

Connection to Advanced Theory — Buffers & Polyprotic Acids

The skills you build here with weak acid pH calculations directly extend to more complex equilibrium problems. In AP Chemistry and college courses, you will encounter buffer solutions (mixtures of a weak acid and its conjugate base) and polyprotic acids (acids that can donate more than one proton). Both rely on the same ICE-table logic, just applied in layers.

Feature	Weak Acid pH (This Lesson)	Buffer pH (Next Topic)
Species Present	Only HA dissolved in pure water.	HA plus its conjugate base A⁻ (e.g., from a salt like NaCH₃COO).
Initial [A⁻]	Starts at 0 — builds up only through ionization.	Starts at a significant non-zero value from the added salt.
Equation Used	Ka = x² / (C₀ − x), solved by square root or quadratic.	Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Key Advantage	Predicts pH from a single known quantity (initial [HA]).	Resists pH change when small amounts of acid or base are added.

Weak acid pH vs. buffer pH — building on the same equilibrium foundation

Understanding weak acid equilibria also connects to biology and environmental science. Your blood maintains a pH of about 7.4 using a carbonic acid/bicarbonate buffer system. Ocean acidification — caused by dissolved CO₂ forming carbonic acid — is fundamentally a weak acid equilibrium problem. Mastering K_a and ICE tables gives you the tools to analyze these real-world systems quantitatively.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A 0.10 M solution of hydrofluoric acid (HF, K_a = 6.8 × 10⁻⁴) has a pH of about 2.08. Explain why this pH is higher than that of a 0.10 M solution of HCl (pH = 1.0), even though both are at the same concentration.

Show answer

HCl is a strong acid that ionizes 100%, so all 0.10 mol/L of HCl becomes H⁺, giving [H⁺] = 0.10 M and pH = 1.0. HF is a weak acid that only partially ionizes — the equilibrium lies to the left, meaning most HF molecules remain intact. Fewer H⁺ ions are produced (about 7.8 × 10⁻³ M), so the pH is higher (less acidic). The K_a value quantifies this: a K_a much less than 1 tells us that the undissociated form dominates at equilibrium.

Question 2: BASIC CALCULATION

Calculate the pH of a 0.50 M solution of nitrous acid (HNO₂), given K_a = 4.5 × 10⁻⁴.

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Set up the ICE table: HNO₂ ⇌ H⁺ + NO₂⁻. Initial: [HNO₂] = 0.50, [H⁺] = 0, [NO₂⁻] = 0. Change: −x, +x, +x. Equilibrium: 0.50 − x, x, x. Check approximation: C₀/K_a = 0.50 / 4.5 × 10⁻⁴ = 1,111 ≥ 400, so the approximation is valid. Solve: x² = K_a × C₀ = (4.5 × 10⁻⁴)(0.50) = 2.25 × 10⁻⁴. x = √(2.25 × 10⁻⁴) = 0.0150 M. pH = −log(0.0150) = 1.82. Verification: 0.0150/0.50 = 3.0%, which is less than 5%.

Question 3: INTERMEDIATE

A 0.020 M solution of chloroacetic acid (ClCH₂COOH) has K_a = 1.4 × 10⁻³. Determine whether the 5% approximation is valid, and then calculate the pH using the appropriate method.

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First check: C₀/K_a = 0.020 / 1.4 × 10⁻³ = 14.3, which is far below 400. The approximation is NOT valid; you must use the quadratic formula. The equilibrium expression gives: x² / (0.020 − x) = 1.4 × 10⁻³. Rearranging: x² + 1.4 × 10⁻³ x − 2.8 × 10⁻⁵ = 0. Using the quadratic formula: x = [−1.4 × 10⁻³ + √((1.4 × 10⁻³)² + 4(2.8 × 10⁻⁵))] / 2 = [−0.0014 + √(0.00000196 + 0.000112)] / 2 = [−0.0014 + 0.01068] / 2 = 0.00464 M. Check: 0.00464 / 0.020 = 23.2% — well over 5%, confirming the approximation would have failed. pH = −log(0.00464) = 2.33.

Question 4: APPLIED

Vitamin C (ascorbic acid, HC₆H₇O₆) has K_a = 8.0 × 10⁻⁵. A student dissolves a 500 mg vitamin C tablet (molar mass = 176.12 g/mol) in enough water to make 250 mL of solution. What is the pH of this solution?

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First, find the initial concentration. Moles of vitamin C = 0.500 g / 176.12 g/mol = 2.84 × 10⁻³ mol. Molarity = 2.84 × 10⁻³ mol / 0.250 L = 0.01136 M. Check approximation: C₀/K_a = 0.01136 / 8.0 × 10⁻⁵ = 142, which is less than 400, so use the quadratic. x² + 8.0 × 10⁻⁵ x − 9.09 × 10⁻⁷ = 0. Solving: x = [−8.0 × 10⁻⁵ + √(6.4 × 10⁻⁹ + 3.636 × 10⁻⁶)] / 2 = [−0.00008 + 0.001908] / 2 = 9.14 × 10⁻⁴ M. pH = −log(9.14 × 10⁻⁴) = 3.04. This slightly acidic pH is consistent with the tart taste of vitamin C tablets dissolved in water.

Question 5: CRITICAL THINKING

Two weak acids, HA and HB, both have the same initial concentration of 0.10 M. Acid HA has K_a = 1.0 × 10⁻⁴ and acid HB has K_a = 1.0 × 10⁻⁸. Without doing a full calculation, predict which acid will have a pH closer to 7.0 and explain why. Then calculate both pH values to verify your prediction and determine the actual difference in pH units.

Show answer

Prediction: HB (K_a = 1.0 × 10⁻⁸) will have a pH closer to 7.0 because its smaller K_a means it ionizes far less, producing fewer H⁺ ions. Verification — For HA: x = √(1.0 × 10⁻⁴ × 0.10) = √(1.0 × 10⁻⁵) = 3.16 × 10⁻³ M. Check: 3.16 × 10⁻³ / 0.10 = 3.16% < 5% ✓. pH = −log(3.16 × 10⁻³) = 2.50. For HB: x = √(1.0 × 10⁻⁸ × 0.10) = √(1.0 × 10⁻⁹) = 3.16 × 10⁻⁵ M. Check: 3.16 × 10⁻⁵ / 0.10 = 0.032% < 5% ✓. pH = −log(3.16 × 10⁻⁵) = 4.50. The difference is 4.50 − 2.50 = 2.00 pH units. This result reveals an important pattern: when K_a changes by a factor of 10⁴ (at the same concentration), the pH changes by exactly 2 units, because pH depends on the square root of K_a.

SUMMARY

Lesson Summary

Summary

A weak acid only partially ionizes in water, establishing an equilibrium HA ⇌ H⁺ + A⁻. The extent of ionization is quantified by the acid dissociation constant Ka, defined as [H⁺][A⁻] / [HA]. To compute pH, you set up an ICE table (Initial, Change, Equilibrium) that tracks each species' concentration as the reaction reaches equilibrium. Substituting the equilibrium row into the K_a expression yields the equation x² / (C₀ − x) = K_a, where x = [H⁺].

When C₀ / K_a ≥ 400, the 5% approximation lets you simplify to x ≈ √(K_a × C₀). Otherwise, use the quadratic formula. After finding x, compute pH = −log(x). Always verify your approximation by checking that x / C₀ < 0.05. This framework extends to weak bases, buffers, and polyprotic acids — making it one of the most versatile tools in equilibrium chemistry.