Salt Hydrolysis and Solution pH
Neutralisation is the exothermic reaction between an acid and a base forming a salt and water. The pH of the resulting salt solution depends on whether the conjugate ions undergo hydrolysis with water.
Neutralisation Reactions
The general equation for neutralisation:
Acid + Base → Salt + Water
The net ionic equation (for strong acid + strong base):
H+(aq) + OH-(aq) → H2O(l)
Summary of Salt Solution Types
- Strong Acid + Strong Base → Neutral Salt (pH = 7): E.g. \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\) (neither \(\text{Na}^+\) nor \(\text{Cl}^-\) hydrolyses).
- Strong Acid + Weak Base → Acidic Salt (pH < 7): E.g. \(\text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl}\) (\(\text{NH}_4^+\) releases \(\text{H}^+\)).
- Weak Acid + Strong Base → Basic Salt (pH > 7): E.g. \(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\) (\(\text{CH}_3\text{COO}^-\) produces \(\text{OH}^-\)).
Salt Hydrolysis
When a salt dissolves, its ions may react with water (hydrolyse). The pH of the solution depends on the parent acid and base:
| Salt Type | Made From | Solution pH | Example |
|---|---|---|---|
| Neutral | SA + SB | = 7 | NaCl, KNO3 |
| Basic | WA + SB | > 7 | CH3COONa, Na2CO3 |
| Acidic | SA + WB | < 7 | NH4Cl, FeCl3 |
Why Do Some Salt Solutions Have a Non-Neutral pH?
Acidic Salt Hydrolysis (NH₄⁺)
The ammonium ion (\(\text{NH}_4^+\)) is the conjugate acid of the weak base \(\text{NH}_3\). It donates a proton to water:
\[\text{NH}_4^+(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_3(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \implies \text{pH} < 7\]Basic Salt Hydrolysis (CH₃COO⁻)
The ethanoate ion (\(\text{CH}_3\text{COO}^-\)) is the conjugate base of weak ethanoic acid. It accepts a proton from water:
\[\text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) \implies \text{pH} > 7\]Predicting pH of Sodium Carbonate
Problem: Predict whether aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) is acidic, basic, or neutral.
Solution: \(\text{Na}_2\text{CO}_3\) forms \(\text{Na}^+\) (spectator from strong base \(\text{NaOH}\)) and \(\text{CO}_3^{2-}\) (conjugate base of weak acid \(\text{HCO}_3^-\)).
\[\text{CO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{HCO}_3^-(\text{aq}) + \text{OH}^-(\text{aq}) \implies \textbf{Basic (pH > 7)}\]Identifying Spectator vs Hydrolysing Ions
Spectator ions from strong acids (\(\text{Cl}^-, \text{Br}^-, \text{I}^-, \text{NO}_3^-, \text{SO}_4^{2-}\)) and strong bases (\(\text{Group 1 cations}, \text{Ba}^{2+}\)) do not undergo hydrolysis. Only conjugate species of weak parents hydrolyse!
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