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Summary AQA A-Level Chemistry 1.12 Acids, Bases and Buffers

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These are detailed Revision Notes for Topic 1.12 of AQA A-Level Chemistry (Acids, Bases and Buffers). They were written by me using a combination of the textbook and class notes. I will also be uploading the other topics and creating bundles. Topics Included: - Defining an acid - The pH scale - Weak acids and bases - Acid-base titrations - Choice of indicators for titrations - Buffer solutions

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Summarized whole book?
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Chapter 21
Uploaded on
May 8, 2022
Number of pages
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2021/2022
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Acids, Bases and Buffers
21.1 Defining an Acid:
- A BrØnsted-Lowry acid is a proton donor, and a BrØnsted- 21.1 defining an acid
Lowry base is a proton acceptor. 21.2 the pH scale
- Water has a pH of 7, the [H+] is 1x10-7. The [H+] comes from 21.3 weak acids and bases
the ionisation of water and the [OH-] in pure water is 1x10-7
- Kw = [H+][OH-] 21.4 acid-base titrations
- At 298K Kw is 1x10-14 mol2dm6 21.5 choice of indicators for titrations
- Kw is derived from the equilibrium constant for this dissociation. 21.6 buffer solutions
The value of Kw increases with temperature

21.2 The pH Scale: 21
- The measure of acidity is the pH scale.
- pH = -log10[H+]
- [H+] = 10-pH
- Strong acids completely dissociate in aqueous solution
- For strong monobasic acids the concentration of H+ is equal to the concentration of the acid
- For strong dibasic acids the concentration of H+ is equal to double the concentration of the acid

21.3 Weak Acids and Bases:
- Weak acids only partially dissociate forming an e.g., HCOOH(aq) ⇌HCOO-(aq) + H+(aq)
equilibrium with water. Given the Ka = 1.6x10-4 what is the pH of a
HA ⇌ H+ + A- 0.05moldm-3 solution?
- Ka is the acid dissociation constant, it always have
the units of moldm-3, the higher the Ka the [HA] x Ka = [H+]2
stronger the acid. 0.05 x (1.6x10-4) = [H+]2
[# ! ][%" ]
- 𝐾! = 8x10-6 = [H+]2
[#%}
- pKa = -log10[Ka] [H+] = √8 × 10'(
- In weak acids we assume that [A-] is equal to [H+] [H+} = 2.83x10-3
and that [HA] is equal to the initial concentration pH = -log(2.83x10-3)
[# ! ]# pH = 2.55
- So, 𝐾! = [#%]


21.4 Acid-Base Titrations:
- In a titration, the equivalence point is the point where enough base has been added to just neutralise
the acid.

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