Define pH. Write an expression for K_w, the ionic product (the dissociation constant) of water - Leaving Cert Chemistry - Question 9 - 2022

The ionic product of water, $K_w$, is expressed as:

$K_w = [H^+][OH^-]$

at 25°C.

The pH scale is only effective for dilute solutions and is not reliable in concentrated solutions.

Given the pH of the mouthwash is 5.65:

1. Calculate [H⁺]: $[H^+] = 10^{-5.65} = 2.2 imes 10^{-6} ext{ moles per litre}$.

2. Since HX is a strong acid, it completely dissociates so: $[HX] = [H^+] = 2.2 imes 10^{-6} ext{ moles per litre}$.

Using the water dissociation expression:

1. Calculate [OH⁻] using: $K_w = [H^+][OH^-]$ Given that $K_w = 1.0 imes 10^{-14}$ at 25°C: $[OH^-] = \frac{K_w}{[H^+]}$ $[OH^-] = \frac{1.0 imes 10^{-14}}{2.2 imes 10^{-6}} = 4.5 imes 10^{-9} ext{ moles per litre}$.

20.0 cm³ corresponds to the volume of NaOH solution required to neutralize solution A.

(i) From the graph, it can be observed that 20.0 cm³ of 1.0 M NaOH neutralizes solution A at a pH change.

(ii) To deduce the molar concentrations of solutions A, B, C, and D:

• Solution A: 1.0 M
• Solution B: 0.5 M
• Solution C: 1.0 M
• Solution D: 0.1 M

(iii) Solution C has the greater acid dissociation constant (K_a) value, which is determined by the steeper slope in the graph indicating a quicker change in pH, thus greater strength in weak acids.

(iv) D is $CH_3COOH$, and its conjugate base is $CH_3COO^-$. The answer to part (iii) further justifies the choice.