Define an acid according to the theory of (i) Arrhenius, (ii) Brønsted and Lowry - Leaving Cert Chemistry - Question 7 - 2021
Question 7
Define an acid according to the theory of (i) Arrhenius, (ii) Brønsted and Lowry.
(iii) State one limitation of Arrhenius' acid-base theory?
(iv) What is a conjugate... show full transcript
Worked Solution & Example Answer:Define an acid according to the theory of (i) Arrhenius, (ii) Brønsted and Lowry - Leaving Cert Chemistry - Question 7 - 2021
Step 1
Define an acid according to the theory of (i) Arrhenius, (ii) Brønsted and Lowry.
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Answer
According to Arrhenius, an acid is a substance that dissociates in water to produce protons (H⁺ ions), whereas Brønsted-Lowry defines an acid as a proton (H⁺) donor, which explains the acid's ability to donate protons to bases.
Step 2
(iii) State one limitation of Arrhenius' acid-base theory.
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Answer
One limitation of Arrhenius' theory is that it does not account for acid-base reactions that occur in non-aqueous solvents, or the behavior of acids that do not produce H⁺ in solution.
Step 3
(iv) What is a conjugate acid-base pair in Bronsted-Lowry theory?
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Answer
A conjugate acid-base pair consists of two substances that differ by the presence of a proton (H⁺). For example, in the pair (HA, A⁻), HA is the acid that can donate a proton to become A⁻, the conjugate base.
Step 4
Distinguish between a strong acid and a weak acid.
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Answer
A strong acid completely dissociates in solution, producing a high concentration of H⁺ ions, while a weak acid only partially dissociates, resulting in a lower concentration of H⁺ ions and an equilibrium between the acid and its ions.
Step 5
Write a balanced equation to show that the conjugate base of H2SO4 acts as a Brønsted-Lowry acid in water.
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Answer
The balanced equation is:
HSO4−+H2O⇌H3O++SO42−
Step 6
(a) Write a balanced equation to show the dissociation into ions in water of a weak monobasic acid represented by HA.
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Answer
HA+H2O⇌H3O++A−
Step 7
(i) find the concentrations of H3O+ ion and A− ion in moles per liter in the solution.
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Answer
Given that HA is 1.5% dissociated in a 0.10 M solution:
Concentration of H₃O⁺:
[H3O+]=[A−]=0.10M×0.015=0.0015M
Step 8
(ii) calculate the pH of the 0.10 M HA solution.
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Answer
Using the formula:
pH=−log[H3O+]=−log(0.0015)≈2.82
Step 9
(iii) hence or otherwise, calculate the value of the acid dissociation constant Ka for HA.
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