Explain the terms (a) flocculation, (b) pH, (c) acid dissociation constant (Ka), (d) conjugate pair - Leaving Cert Chemistry - Question 8 - 2001

pH is defined as the negative logarithm of the hydrogen ion concentration in a solution:

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

It serves as a measure of the acidity or alkalinity of a solution, with lower values indicating acidic solutions and higher values indicating alkaline solutions.

The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It is represented by the equilibrium expression for the dissociation of the acid in water:

K_a = rac{[H^+][A^-]}{[HA]}

where [H⁺] is the concentration of hydrogen ions, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid.

In the dissociation of chloric(I) acid (HClO), the conjugate pairs are HClO and ClO⁻. When HClO donates a proton (H⁺), it becomes its conjugate base ClO⁻, while the H⁺ represents the conjugate acid.

To raise the pH in water treatment when it's too low, alkaline substances such as lime (calcium hydroxide) may be added to neutralize the acid.

If the pH is too high, acidic substances such as sulfuric acid may be introduced to lower the pH and stabilize the water chemistry.

The conjugate pairs in the dissociation of chloric(I) acid (HClO) are:

• HClO (acid) / ClO⁻ (conjugate base)
• H⁺ (proton) / HClO (conjugate acid)

The expression for the acid dissociation constant (Ka) of chloric(I) acid is:

K_a = rac{[H^+][ClO^-]}{[HClO]}

Choric(I) acid is classified as a weak acid. This is because it does not dissociate completely in solution, meaning that there are undissociated molecules present, which indicates a lower dissociation constant.

To calculate the approximate pH of a 0.5 mol dm⁻³ solution of chloric(I) acid:

1. Use the formula derived from the dissociation equation: If we assume full dissociation,

   $[H^+] ext{ will be } 0.5 ext{ mol dm}^{-3}$

2. Calculate pH:

   $pH = - ext{log}(0.5) ext{ which yields approximately } 0.301$

Therefore, rounding gives a pH of 0.30.

To find the approximate percentage dissociation of chloric(I) acid:

1. Determine the amount dissociated: Let x be the dissociation amount so that

   $[H^+] = x ext{ and } [HClO] = 1.0 - x$

2. If we assume dissociation is low, we find:

   ext{Percentage dissociation} = rac{x}{[HClO]_{initial}} imes 100 ext{%}

Calculating leads to approximately 6.0%.