# Acid-base equilibria

> Chemistry · CIE A-Level 9701
> Source: https://www.owlsprep.com/study/cie-9701-u18-acid-base-equilibria/

This sub-topic covers core Bronsted-Lowry acid-base theory, pH calculations for strong and weak acids, conjugate acid-base pairs, and the acid dissociation constant Ka, forming the foundation for all further acid-based equilibria topics in CIE A-Level Chemistry.

**Prerequisites:** [Basic chemical equilibrium concepts](https://www.owlsprep.com/study/cie-9701-u17-chemical-equilibrium/); Logarithm arithmetic

## Learning objectives

- Distinguish between Bronsted-Lowry acids and bases
- Identify conjugate acid-base pairs in a reaction
- Calculate pH for strong and weak monoprotic acids
- Interpret the relationship between acid strength, Ka and pKa

## Bronsted-Lowry Acid-Base Theory

**Bronsted-Lowry Acid** — A substance that can donate a proton ($H^+$ ion) to another substance

*Example:* Hydrochloric acid ($HCl$) donates a proton to water in aqueous solution

A Bronsted-Lowry base is defined conversely as a proton acceptor. All bases have at least one lone pair of electrons to form a bond with the donated proton. This theory is more general than the older Arrhenius theory, which only applies to aqueous solutions.

**Worked example:** Identify which species act as the acid and base in the forward reaction: $NH_3(aq) + HCl(g) \rightarrow NH_4Cl(s)$

1. First, track proton movement between the reactants to identify donor and acceptor.
2. $HCl$ loses a proton to form $Cl^-$, so it donates a proton.
3. $NH_3$ gains a proton to form $NH_4^+$, so it accepts a proton.
4. Final answer: $HCl$ = acid, $NH_3$ = base

> **Exam tip:** CIE examiners almost always ask for identification of acid/base species in reactions, always check for proton movement not just charge.

## Conjugate Acid-Base Pairs

**Conjugate Acid-Base Pair** — Two species that differ by exactly one proton, formed when an acid donates a proton or a base accepts a proton. The conjugate base of an acid is the species remaining after proton donation; the conjugate acid of a base forms after proton acceptance.

*Example:* $HCl$ (acid) and $Cl^-$ (conjugate base) form one conjugate pair

Acid strength is inversely related to conjugate base strength: strong acids have very weak conjugate bases, while weak acids have relatively strong conjugate bases. This is because a strong acid fully dissociates, so its conjugate base has almost no tendency to re-accept a proton.

**Worked example:** Write all conjugate acid-base pairs for the equilibrium: $HNO_2(aq) + H_2O(l) \rightleftharpoons NO_2^-(aq) + H_3O^+(aq)$

1. Group species by proton difference across the equilibrium.
2. $HNO_2$ loses one proton to become $NO_2^-$, so this is the first pair: $HNO_2$ (acid) and $NO_2^-$ (conjugate base).
3. $H_2O$ gains one proton to become $H_3O^+$, so this is the second pair: $H_2O$ (base) and $H_3O^+$ (conjugate acid).

## pH Scale and Strong Acid pH Calculations

**pH** — A logarithmic scale measuring hydrogen ion concentration in aqueous solution, ranging from ~0 (strongly acidic) to ~14 (strongly alkaline) at 25°C.

*Notation:* pH = -\log_{10}[H^+_{(aq)}]

Strong acids fully dissociate in aqueous solution, so for monoprotic strong acids (with one acidic proton), $[H^+] = [acid]_{initial}$. For diprotic strong acids like $H_2SO_4$, CIE assumes full dissociation so $[H^+] = 2 \times [acid]_{initial}$.

**Worked example:** Calculate the pH of 0.050 mol dm⁻³ hydrochloric acid at 25°C.

1. HCl is a strong monoprotic acid, so full dissociation occurs:
2. $$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$$
3. Therefore, $[H^+] = [HCl] = 0.050$ mol dm⁻³
4. Substitute into the pH formula:
5. $$pH = -\log_{10}(0.050) = 1.30$$
6. Final pH = 1.30 (2 decimal places)

> **Exam tip:** Always give pH values to 2 decimal places unless the question explicitly states otherwise, this is the CIE marking requirement.

## Weak Acids and the Acid Dissociation Constant Ka

**Acid Dissociation Constant (Ka)** — The equilibrium constant for dissociation of a weak acid $HA$ in aqueous solution. $pKa = -\log_{10}Ka$, so lower pKa corresponds to a stronger acid.

*Notation:* K_a = \frac{[H^+][A^-]}{[HA]}

Weak acids only partially dissociate, so we use Ka to quantify their strength. For most weak acids, dissociation is very small, so two simplifying approximations are accepted by CIE: $[HA]_{equilibrium} \approx [HA]_{initial}$ and $[H^+] \approx [A^-]$.

**Worked example:** A 0.10 mol dm⁻³ solution of weak monoprotic acid HA has Ka = 1.8 × 10⁻⁵ mol dm⁻³ at 25°C. Calculate its pH.

1. Write the dissociation equilibrium for HA:
2. $$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$
3. Write the Ka expression:
4. $$K_a = \frac{[H^+][A^-]}{[HA]}$$
5. Substitute the accepted approximations: $[H^+] = [A^-]$, $[HA] \approx 0.10$:
6. $$1.8 \times 10^{-5} = \frac{[H^+]^2}{0.10}$$
7. Rearrange to solve for $[H^+]$:
8. $$[H^+]^2 = 1.8 \times 10^{-6} \rightarrow [H^+] = 1.34 \times 10^{-3} \text{ mol dm}^{-3}$$
9. Calculate pH:
10. $$pH = -\log_{10}(1.34 \times 10^{-3}) = 2.87$$

## Common pitfalls

- **Wrong:** Treating diprotic $H_2SO_4$ as monoprotic for pH calculations
  - Why it fails: CIE assumes full dissociation of sulfuric acid, so $[H^+]$ is twice the acid concentration
  - Correct: Always check the number of acidic protons before calculating $[H^+]$ for strong acids
- **Wrong:** Forgetting the negative sign in the pH formula, leading to negative pH values
  - Why it fails: The negative sign reverses the scale so higher $[H^+]$ gives lower pH, matching standard convention
  - Correct: Always double-check your calculator input to confirm the negative sign is included
- **Wrong:** Using Arrhenius definitions to answer acid-base definition questions
  - Why it fails: CIE expects Bronsted-Lowry definitions for all A-Level acid-base questions, as it is more general
  - Correct: Always define acids as proton donors and bases as proton acceptors
- **Wrong:** Confusing conjugate acid and conjugate base in a pair
  - Why it fails: Students often mix up which species gained or lost the proton
  - Correct: Remember: +1 proton = conjugate acid, -1 proton = conjugate base
- **Wrong:** Not using the weak acid approximation, leading to overly complex quadratic calculations
  - Why it fails: For all weak acids commonly tested in CIE, the approximation is valid and accepted
  - Correct: Use the approximation unless the question explicitly tells you not to, to save exam time

## Cheatsheet

| Concept | Formula / Rule |
| --- | --- |
| Bronsted-Lowry Acid | Proton donor |
| Bronsted-Lowry Base | Proton acceptor |
| pH | $pH = -\log_{10}[H^+]$ |
| Strong monoprotic acid | $[H^+] = [acid]_{initial}$ |
| Strong diprotic acid | $[H^+] = 2[acid]_{initial}$ |
| Ka for weak acid HA | $K_a = \frac{[H^+][A^-]}{[HA]}$ |
| pKa | $pKa = -\log_{10}Ka$ |
| Conjugate acid-base pair | Differs by exactly 1 proton |

## What's next

Acid-base equilibria is the foundation for all further acid-base topics in CIE A-Level Chemistry, including buffer solutions, pH curves, titrations and the ionic product of water. The calculation principles you learned here for Ka and pH will be extended to weak bases and salt hydrolysis in later topics, and underpin practical titration calculations commonly assessed in both Papers 2 and 4. Mastery of core definitions and basic pH calculations here is essential to avoid losing easy marks in extended response questions. Build on this knowledge by exploring the related topics below.

- [Buffer solutions](https://www.owlsprep.com/study/cie-9701-u18-buffer-solutions/)
- [Solubility product](https://www.owlsprep.com/study/cie-9701-u18-solubility-product/)
- [pH titration curves and indicators](https://www.owlsprep.com/study/cie-9701-u18-ph-titration-curves-and-indicators/)

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