Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.1–14.2 Brønsted–Lowry theory and conjugate pairs-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.1–14.2 Brønsted–Lowry theory and conjugate pairs- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.1–14.2 Brønsted–Lowry theory and conjugate pairs- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
14.1 Brønsted–Lowry Acids and Bases
The Brønsted–Lowry theory defines acids and bases in terms of proton (\( \mathrm{H^+} \)) transfer. This provides a more general definition than earlier theories and applies to a wider range of reactions.
Definitions
- Brønsted–Lowry acid: a substance that donates a proton (\( \mathrm{H^+} \)).
- Brønsted–Lowry base: a substance that accepts a proton (\( \mathrm{H^+} \)).
Key Idea
Acid–base reactions involve the transfer of a proton from an acid to a base.
General Reaction
\( \mathrm{acid + base \rightleftharpoons conjugate\ base + conjugate\ acid} \)
- The acid loses \( \mathrm{H^+} \) and forms its conjugate base.
- The base gains \( \mathrm{H^+} \) and forms its conjugate acid.
Example 1
\( \mathrm{HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)} \)
- \( \mathrm{HCl} \) donates \( \mathrm{H^+} \) → acts as an acid.
- \( \mathrm{H_2O} \) accepts \( \mathrm{H^+} \) → acts as a base.
Example 2
\( \mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)} \)
- \( \mathrm{NH_3} \) accepts \( \mathrm{H^+} \) → base.
- \( \mathrm{H_2O} \) donates \( \mathrm{H^+} \) → acid.
Important Concepts
- Proton transfer is the key feature of all acid–base reactions.
- Acids and bases always occur in pairs.
- The reaction is often reversible.
Key Features
- Acid = proton donor.
- Base = proton acceptor.
- Reactions involve transfer of \( \mathrm{H^+} \).
- Forms conjugate acid–base pairs.
Example 1:
Identify the acid and base in the reaction:
\( \mathrm{HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-} \)
▶️ Answer/Explanation
\( \mathrm{HNO_3} \) donates a proton, so it is the acid.
\( \mathrm{H_2O} \) accepts a proton, so it is the base.
Example 2:
Explain the role of water in the reaction:
\( \mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-} \)
▶️ Answer/Explanation
Water donates a proton to \( \mathrm{NH_3} \), so it acts as a Brønsted–Lowry acid.
This shows that water can act as both an acid and a base depending on the reaction.
14.2 Brønsted–Lowry Conjugate Acid–Base Pairs
In the Brønsted–Lowry theory, acids and bases are related through proton transfer. When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid. Therefore, conjugate acid–base pairs differ by one proton only.
Conjugate Acid–Base Pair
A conjugate acid–base pair consists of two species that differ by one proton, \( \mathrm{H^+} \).
In any Brønsted–Lowry acid–base reaction:
- The acid donates \( \mathrm{H^+} \) and forms its conjugate base.
- The base accepts \( \mathrm{H^+} \) and forms its conjugate acid.
How to Identify Conjugate Pairs
- Look for two species that differ by one \( \mathrm{H^+} \).
- Check whether one species can be converted into the other by gaining or losing a proton.
- One member of the pair must act as an acid and the other as its corresponding base.
General Reaction
\( \mathrm{HA + B \rightleftharpoons A^- + HB^+} \)
Conjugate pairs are:
\( \mathrm{HA / A^-} \) and \( \mathrm{B / HB^+} \)
Example 1
\( \mathrm{HCl + H_2O \rightarrow H_3O^+ + Cl^-} \)
- \( \mathrm{HCl} \) → \( \mathrm{Cl^-} \) (loss of \( \mathrm{H^+} \)) → conjugate pair.
- \( \mathrm{H_2O} \) → \( \mathrm{H_3O^+} \) (gain of \( \mathrm{H^+} \)) → conjugate pair.
Conjugate pairs:
- \( \mathrm{HCl / Cl^-} \)
- \( \mathrm{H_2O / H_3O^+} \)
Example 2
\( \mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-} \)
- \( \mathrm{NH_3} \) → \( \mathrm{NH_4^+} \) → conjugate pair.
- \( \mathrm{H_2O} \) → \( \mathrm{OH^-} \) → conjugate pair.
Conjugate pairs:
- \( \mathrm{NH_3 / NH_4^+} \)
- \( \mathrm{H_2O / OH^-} \)
Important Points
- Conjugate acid–base pairs always differ by exactly one proton.
- They are not just any reactant and product pair.
- Charges usually differ by 1 because of proton transfer.
- A species can act as an acid in one reaction and as a base in another.
Key Features
- Acid → conjugate base after losing \( \mathrm{H^+} \).
- Base → conjugate acid after gaining \( \mathrm{H^+} \).
- Pairs differ by one proton only.
- Acid–base reactions always contain two conjugate pairs.
Example 1:
Identify the conjugate acid–base pairs in the reaction:
\( \mathrm{HNO_2 + H_2O \rightleftharpoons H_3O^+ + NO_2^-} \)
▶️ Answer/Explanation
\( \mathrm{HNO_2} \) donates a proton to form \( \mathrm{NO_2^-} \), so one conjugate pair is \( \mathrm{HNO_2 / NO_2^-} \).
\( \mathrm{H_2O} \) accepts a proton to form \( \mathrm{H_3O^+} \), so the second conjugate pair is \( \mathrm{H_2O / H_3O^+} \).
Example 2:
In the reaction
\( \mathrm{HSO_4^- + NH_3 \rightleftharpoons SO_4^{2-} + NH_4^+} \)
identify the Brønsted–Lowry conjugate acid–base pairs.
▶️ Answer/Explanation
\( \mathrm{HSO_4^-} \) loses a proton to form \( \mathrm{SO_4^{2-}} \), so one conjugate pair is \( \mathrm{HSO_4^- / SO_4^{2-}} \).
\( \mathrm{NH_3} \) gains a proton to form \( \mathrm{NH_4^+} \), so the second conjugate pair is \( \mathrm{NH_3 / NH_4^+} \).