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

One limitation of Arrhenius' acid-base theory is that it does not account for acid-base reactions that occur in non-aqueous solvents. Additionally, it fails to explain the behavior of acids and bases in reactions that do not produce hydroxide (OH⁻) ions.

A conjugate acid-base pair consists of two species that differ by the presence or absence of a proton (H⁺). For example, in the acid-base reaction HA (acid) donating a proton to form A⁻ (conjugate base), HA and A⁻ are the conjugate pair.

A strong acid is one that completely dissociates in solution, producing a high concentration of H⁺ ions (e.g., hydrochloric acid, HCl). In contrast, a weak acid only partially dissociates, resulting in a lower concentration of H⁺ ions (e.g., acetic acid, CH₃COOH).

The balanced equation illustrating this is:

$HSO_4^- + H_2O \rightleftharpoons H_3O^+ + SO_4^{2-}$

In this reaction, the bisulfate ion (HSO₄⁻) donates a proton to water, forming hydronium ions (H₃O⁺) and sulfate ions (SO₄²⁻).

The balanced equation for the dissociation of the weak monobasic acid HA in water is:

$HA + H_2O \rightleftharpoons H_3O^+ + A^-$

Here, the acid HA donates a proton to water, producing hydronium ions and its conjugate base A⁻.

If HA is 1.5% dissociated in a 0.10 M solution, the concentration of H⁺ ions (H₃O⁺) is calculated as follows:

1.5% of 0.10 M = 0.0015 M, Thus,

$[H_3O^+] = [A^-] = 0.0015 M$.

Using the formula:

$pH = -\log[H_3O^+] = -\log(0.0015) \approx 2.82$.

The acid dissociation constant $K_a$ is calculated using the expression:

$K_a = \frac{[H_3O^+][A^-]}{[HA]}$, Substituting values into the equation gives:

$K_a = \frac{(0.0015)(0.0015)}{0.1} = 2.25 \times 10^{-5}$.