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Define an acid according to the theory of (i) Arrhenius, (ii) Brønsted and Lowry - Leaving Cert Chemistry - Question 7 - 2021

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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+H2OH3O++SO42HSO_4^{-} + H_2O \rightleftharpoons H_3O^{+} + SO_4^{2-}

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+H2OH3O++AHA + H_2O \rightleftharpoons H_3O^{+} + 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[H_3O^{+}] = [A^{-}] = 0.10 M \times 0.015 = 0.0015 M

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)2.82pH = -\log[H_3O^{+}] = -\log(0.0015) \approx 2.82

Step 9

(iii) hence or otherwise, calculate the value of the acid dissociation constant Ka for HA.

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Answer

Using the formula for Ka:

\text{After dissociation, } [HA] = 0.10 - 0.0015 \approx 0.0985 M \ K_a = \frac{(0.0015)(0.0015)}{0.0985} \approx 2.25 \times 10^{-5}$$

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