Ch. 18-3 – Acids & Bases II. pH (p. 644 – 658)

Self-Ionization of Water A. Ionization of Water H2O + H2O H3O+ + OH- Self-Ionization of Water

Kw = [H3O+][OH-] = 1.0  10-14 A. Ionization of Water Ion Product Constant for Water The ion production of water, Kw = [H3O+][OH–] Pure water contains equal concentrations of H+ and OH– ions, so [H3O+] = [OH–] For all aqueous solutions, the product of the hydrogen-ion concentration and the hydroxide-ion concentration equals 1.0 x 10-14

A. Ionization of Water Find the hydroxide ion concentration of 3.0  10-2 M HCl. HCl → H+ + Cl- 3.0  10-2M 3.0  10-2M [H3O+][OH-] = 1.0  10-14 [3.0  10-2][OH-] = 1.0  10-14 [OH-] = 3.3  10-13 M

A. Ionization of Water Find the hydronium ion concentration of 1.4  10-3 M Ca(OH)2. Ca(OH)2 → Ca2+ + 2 OH- 1.4  10-3M 2.8  10-3M [H3O+][OH-] = 1.0  10-14 [H3O+][2.8  10-3] = 1.0  10-14 [H3O+] = 3.6  10-12 M

pouvoir hydrogène (Fr.) B. pH Scale 14 7 INCREASING ACIDITY INCREASING BASICITY NEUTRAL pH = -log[H3O+] pouvoir hydrogène (Fr.) “hydrogen power”

pH of Common Substances B. pH Scale pH of Common Substances

pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14 B. pH Scale pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14

B. pH Scale What is the pH of 0.050 M HNO3? pH = -log[H3O+] Acidic or basic? Acidic

B. pH Scale What is the pH of 0.050 M Ba(OH)2? [OH-] = 0.100 M pOH = -log[OH-] pOH = -log[0.100] pOH = 1.00 pH = 13.00 Acidic or basic? Basic

B. pH Scale What is the molarity of HBr in a solution that has a pOH of 9.60? pH + pOH = 14 pH + 9.60 = 14 pH = 4.40 pH = -log[H3O+] 4.40 = -log[H3O+] -4.40 = log[H3O+] [H3O+] = 4.0  10-5 M HBr Acidic

C. pH Worksheet #6 7 = -log[H+] -7 = log[H+] [H+] = 1 x 10-7 M A swimming pool has a volume of one million liters. How many grams of HCl would need to be added to that swimming pool to bring the pH down from 7.0000 to 4.0000? (Assume the volume of the HCl is negligible) 7 = -log[H+] -7 = log[H+] [H+] = 1 x 10-7 M 1,000,000L Sol’n 1x10-7 mol H+ 1 L soln = 0.1 mol H+ 1,000,000L Sol’n 1x10-4 mol H+ 1 L soln -4 = log[H+] [H+] = 1 x 10-4 M = 100 mol H+

C. pH Worksheet #6 = 3642 g HCl 100 mol H+ – 0.1 mol H+ = 99.9 mol HCl A swimming pool has a volume of one million liters. How many grams of HCl would need to be added to that swimming pool to bring the pH down from 7.0000 to 4.0000? (Assume the volume of the HCl is negligible) 100 mol H+ – 0.1 mol H+ = 99.9 mol HCl 99.9 mol HCl 36.46 g HCl 1 mol HCl = 3642 g HCl

D. pH Sig Figs For the pH, the number of sig figs is shown by the # of decimal places [H+] = 2.26 x 10-4 M => For the molarity from the pH, check decimal places in the pH pH = 4.25 => pH = 3.646 5.6 x 10-5 M