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Summary 12 - Acid-base Equilibria - Notes

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Detailed and comprehensive notes for Topic 12 - Acid-Base Equilibria for A-Level Edexcel Chemistry. Based directly off the spec points, so you know you're not missing any content.

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Topic 12 – Acid-Base Equilibria

In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include a series of pH titrations to determine titration curves, calculating an ionisation constant for a pH indicator, making up and
investigating the properties of buffer solutions and using a pH meter. Mathematical skills that could be developed in this topic include the use of
logarithms and exponentials for converting from concentration to pH and vice versa, rearranging K a expressions into expressions suitable for
calculating pH of a buffer solution, plotting and interpreting titration curves. Within this topic, students can consider how the historical development
of theories explaining acid and base behaviour show that scientific ideas change as a result of new evidence and fresh thinking. They can also relate
their study of buffer solutions to a range of applications in living cells, medicines, foods and the natural environment.


1. know that a Brønsted–Lowry acid is [REDACTED] and a Brønsted–Lowry base is [REDACTED]

State the definitions of Brønsted–Lowry acids/bases.

A BL acid is a proton donor. A BL base is a proton acceptor.


2. know that acid-base reactions involve the transfer of protons

What does an acid-base reaction involve? Give an example.

An acid-base reaction involves the transfer of protons from an acid to a base. For example, HCl + H 2O ---> H3O+ + Cl-.


Complete the balanced symbol equation for NH3 + ___ -> NH4Cl.

NH3(aq) + HCl(aq) -> NH4Cl(aq)


Write the balanced symbol equation for sodium carbonate + nitric acid -> ??

Na2CO3 (s) + 2HNO3 (aq) -> 2NaNO3 (aq) + H2O (l) + CO2 (g)


Give the ionic equation for the reaction between sulfuric acid and potassium hydroxide (aq).

H+ (aq) + OH-(aq) ----> H2O (l)


3. be able to identify Brønsted–Lowry conjugate acid-base pairs

Clearly label the conjugate acid-base pairs in the following equations.

, 4. be able to define the term ‘pH’

What does pH stand for and what is its definition?

pH means ‘potential of hydrogen’ and can be defined as pH = – log10[H+]


5. be able to calculate pH from hydrogen ion concentration

What three factors affect pH?

Acid strength (degree of dissociation), concentration, and temperature.


6. be able to calculate the concentration of hydrogen ions, in mol dm -3, in a solution from its pH

Calculate the concentration of hydrogen ions in a solution of HCl with pH 1.70.

[H+] = 10-1.7 = 0.02moldm-3


7. understand the difference between a strong acid and a weak acid in terms of degree of dissociation

State the difference between a weak and strong acid.

Strong acids FULLY dissociate into H+ ions in solution. A weak acid is only slightly dissociated in solution.


8. be able to calculate the pH of a strong acid

Calculate the pH of 0.04moldm-3 H2SO4.

[H+] = 2(0.04) = 0.08 pH = - log(0.08) = 1.10


Calculate the pH of 50gdm-3 HNO3.

mol = mass/Mr = 50/63 [H+] = (50/63)moldm-3 pH = - log(50/63) = 0.10


9. deduce the expression for the acid dissociation constant (K a) for a weak acid + carry out calculations

What is the general expression for Ka?

K a =¿ ¿

The pH of a weak acid, HA, of conc 0.01M was found to be 4.0. Calculate the value of pK a for the acid.

[H+] = 10-4 Ka = (10-4)2/0.01 = 1x10-6 moldm-3 pKa = -log(1x10-6) = 6.0


10. be able to calculate the pH of a weak acid making relevant assumptions

Calculate the pH of 0.2moldm-3 hydrofluoric acid (Ka = 5.62x10-4 moldm-3).

Assume [F-] = [H+] Assume ionisation of [HF] is negligible Assume dissociation of H2O is negligible

5.62x10-4 = [H+].2 [H+] = 0.0106moldm-3 pH = -log0.0106 = 1.97


Calculate the concentration of chloroethanoic acid with a pH of 1.59 (K a = 1.3x10-3 moldm-3).

Assume that [CH2COO-] = [H+], that ionisation of [CH2ClCOOH] is negligible, and that dissociation of H2O is negligible

[H+] = 10-1.59 = 0.0257 1.3x10-3 = (0.0257)2 / [CH2ClCOOH] [CH2ClCOOH] = 0.508moldm-3




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