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Summary 3.9 Acid-Base Equilibria $3.95
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Summary 3.9 Acid-Base Equilibria

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Summary booklet containing the content needed for 3.9 WJEC A level Chemistry. Useful for learning the content and creating flashcards. Includes - Acid and Base definitions - Calculating pH of strong acids - Kw and the ionic product of water - Calculating pH of strong bases - Reactions of st...

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  • April 5, 2023
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3.9 Acids and Bases
Acid Definitions

Acid – A proton donor

Monoprotic Acid – Releases one H+ ion e.g. HCl, HNO3

Diprotic Acid – Releases two H+ ions e.g. H2SO4

Strong Acid – Fully dissociates in solution to release H+ ions e.g. HCl H+ + Cl-

Weak Acid – Partially dissociates in solution to form H+ ions e.g. CH3COOH ⇌ CH3COO- + H+



Calculating pH of strong acids

pH – measure of hydrogen ion concentration of a solution

pH = -log[H+]

[H+] = 10-pH

- For monoprotic acids [acid]=[H+]
- For diprotic acids 2[acid]=[H+]

Dilution of a strong acid

1. Calculate moles in original volume
2. Calculate new conc. with original moles and new volume
3. Put [H+] into pH expression



Kw and Ionic Product of Water

H2O ⇌ H+ + OH-

- When water dissociates it is a reversible reaction and establishes an equilibrium
- There are equal moles of OH- and H+

Able to use equilibrium expression:




Kw = [H+][OH-]

Units – mol2dm-6

Pure water:

[H+] = [OH-]

Kw = [H+]2

, Impure water:

e.g. an alkali




Dissociation of water is endothermic

- As temperature increases, pH decreases
- This means more H+ released and [H+] increases
- Equilibrium shifts to endothermic side as temperature increases to reduce the change
- Shifts to dissociated side so the dissociation of water is endothermic

Does water remain neutral even as pH changes?

- Yes – there is no excess of H+ or OH-
- Neutralisation – H+ + OH- H2O



Base Definitions

Base – Proton acceptor (releases OH- ions in solution)

Monobasic bases – Releases one OH- ion e.g. NaOH

Dibasic bases – Release two OH- ions in solution e.g. Ca(OH)2



Calculating pH of strong bases

- Need to find [H+] but bases only release [OH-] ions in solution
- Use the ionic product of water (Kw)

Kw = [H+][OH-]

Under standard conditions Kw = 1.0 x10-14



1. Calculate [OH-]
Monobasic [base] = [OH-]
Dibasic 2[base] = [OH-]
2. Use Kw to calculate [H+]
3. Put [H+] into pH expression



Dilution of a strong base

1. Calculate moles in original solution
2. Calculate [base] using new volume
3. Use [base] to find [OH-]
4. Use in Kw expression to find [H+]
5. Put [H+] in pH expression

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