(a) Define (i) acid, (ii) conjugate pair, according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 7 - 2007

A conjugate pair consists of an acid and a base that differ by a proton (hydrogen ion, H⁺). This implies that when an acid donates a proton, the species that remains is called its conjugate base. Conversely, the species formed when a base accepts a proton is called its conjugate acid.

In the dissociation of nitrous acid (HNO₂):

HNO₂ + H₂O ⇌ NO₂⁻ + H₃O⁺,

• The conjugate acid-base pair for HNO₂ is HNO₂ (acid) and NO₂⁻ (conjugate base).
• The conjugate acid-base pair for H₂O is H₂O (base) and H₃O⁺ (conjugate acid).

A strong acid is characterized by its ability to completely dissociate into ions in a dilute aqueous solution, leading to a high concentration of H⁺ ions. In contrast, a weak acid only partially dissociates in a dilute solution, resulting in a lower concentration of H⁺ ions. Thus, a strong acid is a good proton donor, whereas a weak acid is a poor proton donor.

To calculate the pH of a 0.1 M solution of nitrous acid (HNO₂), we use the formula:

$ext{pH} = - ext{log} [ ext{H}^+]$

First, we need to determine [H⁺]. The dissociation can be expressed as:

$ext{HNO}_2 ⇌ ext{H}^+ + ext{NO}_2^-$

Using the acid dissociation constant (Kₐ):

K_a = rac{[ ext{H}^+][ ext{NO}_2^-]}{[ ext{HNO}_2]}

Substituting in for the concentrations at equilibrium:

Given Kₐ = 5.0 × 10⁻⁴ and initial concentration [HNO₂] = 0.1 M: Assuming x = [H⁺] = [NO₂⁻]:

K_a = rac{x^2}{0.1 - x} o 5.0 × 10^{-4} = rac{x^2}{0.1}

Solving for x, we find: $x^2 = 5.0 × 10^{-5}$ $x = ext{[H}^+] = ext{0.00707 M}$

Then we can calculate pH:

$ext{pH} = - ext{log}(0.00707) ext{ which gives approximately } 2.15.$ Therefore, the pH of the 0.1 M nitrous acid solution is 2.15.

Nitric acid (HNO₃) is a strong acid, which means it completely dissociates in solution. Therefore, for a 0.1 M HNO₃ solution,

$ext{[H}^+] = 0.1 ext{ M}$

Using the pH formula:

$ext{pH} = - ext{log}(0.1) = 1.$ Thus, the pH of a nitric acid solution of the same concentration (0.1 M) is 1.