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Summary Acid-base Equilibria

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Detailed notes on acid-base equilibria. Notes written using Edexcel Chemistry textbooks, past papers and more. Written by a student with all A*s at GCSE, 3A* predictions at A Level and with an offer for Natural Sciences at Cambridge.

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  • Acid-base equilibria
  • March 30, 2020
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Topic 12: Acid-base Equilibria
1. know that a Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry
base is a proton acceptor
Bronsted-Lowey acid
Bronsted-Lowry acid: a proton donor
Bronsted-Lowry base: a proton acceptor

H+ is never found by itself in water – it combines with H2O to form a hydroxonium ion (H3O+).
HA + H2O  H3O+ + A-

2. know that acid-base reactions involve the transfer of protons
Acid-base reactions involve the transfer of protons.
HCl + H2O ⇌ H H3O+ + Cl-
The proton from HCl is transferred to water in a reversible reaction – it can be transferred back
to the chloride ion.

3. be able to identify Brønsted–Lowry conjugate acid-base pairs
Conjugate pairs
Conjugate pairs are on opposite sides of the reaction equation and are linked by the transfer of a
proton. When acids lose their protons, they donate them to a base (HA + B ⇌ H BH+ + A-).
The acid that has lost a proton is the conjugate base.
The base that has gained a proton is the conjugate acid.
+ -
HCl + H2O ⇌ H HO 3 + Cl
acid base conjugate acid conjugate base



In the forwards reaction, HCl acts as an acid and donates a proton to H 2O. In the backwards
reaction, H3O+ acts as an acid and donates a proton to Cl-

NH4+ + H2O ⇌ H H3O+ + NH 3
acid base conjugate acid conjugate base




4. be able to define the term ‘pH’
pH
pH: a measure of the hydrogen ion concentration

pH = -log10[H+]

[H+] = 10-pH




5. be able to calculate pH from hydrogen ion concentration
Calculating pH
Example: What is the pH of a solution with 0.05moldm-3 of H+
pH = -log10(0.05) = 1.3

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