Chapter 15 Acids and Bases 15.2 the Acids and Bases properties of water 15.3 PH- a measure of acidity Dr Laila Al-Harbi

Acids and Bases Acid: Substance that produces hydrogen ions in water solution. HCl (aq) → H+(aq) + Cl‐(aq) Base: Substance that produces hydroxide ions in water solution. NaOH (aq) → Na+(aq) + OH‐(aq) An acid neutralizes a base H+(aq) + OH‐(aq) → H2O(ℓ) Dr Laila Al-Harbi

15.2 the Acids and Bases properties of water water is unique solvent , it can act as acid or base. In pure water, a few molecules act as bases and a few act as acids. acid(1) + base (1) ⇄ acid(1) + base (1) This is referred to as autoionization of water The equilibrium expression for this process is Kc = [H3O+] [OH−] This special equilibrium constant is referred to as the ion-product constant for water, Kw. At 25°C, Kw = 1.0  10−14 H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Dr Laila Al-Harbi

Because in pure water [H3O+] = [OH−], Kw = [H3O+] [OH−] = 1.0  10−14 Because in pure water [H3O+] = [OH−], [H3O+] = (1.0  10−14)1/2 = 1.0  10−7 In acidic solution [H3O+] > [OH−] In basic solution [H3O+] < [OH−] Dr Laila Al-Harbi

Example 15.2 Calculate the [OH-] ions in a 1.3 M HCl. Kw = [H3O+][OH−]= 1.0  10−14 [OH−] = 1.0  10−14/ [H3O+] [OH−] = 1.0  10−14/ 1.3 [OH−] = 7.7  10−15 M Calculate the [H+] ions in ammonia , [OH-] =0.0025 M Kw = [H3O+] [OH−]= 1.0  10−14 [H3O+] = 1.0  10−14/ [OH−] [H3O+] = 1.0  10−14/ 0.0025 [H3O+] = 4.0  10−12 M Dr Laila Al-Harbi

15-3 pH - A Measure of Acidity pH is defined as the negative base-10 logarithm of the hydronium ion concentration. In the same manner In pure water, pH + pOH = 14 pH = pOH = 7 pH = −log [H3O+] …. [H3O+] = 10-pH pOH = −log [OH-] ….. [OH-] =10-pOH pKw = −log [14×10-14 ] = 14 Dr Laila Al-Harbi

pH Range [H+]>[OH-] [H+] = [OH-] [OH-]>[H+] Acidic Basic 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Neutral [H+]>[OH-] [H+] = [OH-] [OH-]>[H+] Acidic Basic Increase the acidity Increase the basisty pH = pOH = 7 pH >7 pH < 7 Dr Laila Al-Harbi

Example 15.3 The [H+]=3.2 x 10–4 M. The [H+]=1.0 x 10–3 M. What is the pH in the two occasions. pH = −log [H3O+] pH = −log 3.2 x 10–4 = 3.49 pH = −log 1.0 x 10–3 = 3.00 [H3O+] increase ,pH decrease >>> more acidic The [H+]=0.76 M, nitric acid solution ,What is the pH . pH = −log [H3O+] pH = −log 0.76 = 0.12 Dr Laila Al-Harbi

Example 15.4 The pH = 4.82 , What is the [H+] of the rain water . [H3O+] = 10-pH [H3O+] = 10-4.82 [H3O+] = 1.5 × 10–5M The pH = 3.33 , What is the [H+] of orange juice [H3O+] = 10-pH [H3O+] = 10-3.33 [H3O+] = 4.7 × 10–4M Dr Laila Al-Harbi

Example 15.5 The [OH-]=2.9 x 10–4 M. What is the pH of the NaOH solution pOH = −log [OH-] pOH = −log 2.9 x 10–4 = 3.54 pH + pOH = 14 pH = 14 – pOH = 14-3.54=10.46 The [OH-]=2.5 x 10–7 M. What is the pH of solution the blood? pOH = −log [OH-] pOH = −log 2.5 x 10–7 = pH + pOH = 14 pH = 14 – pOH = 14-3.54=7.4 Dr Laila Al-Harbi

[H3O+] [OH-] 14 pOH pH 1.0  10−14 Acidic solution basic solution Dr Laila Al-Harbi