LEC 14 CHAP 16 - Lecture 14 •  Strong acids versus...

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Unformatted text preview: Lecture 14 •  Strong acids versus weak acids and strong bases versus weak bases. •  Percent ionization. •  Various ways to describe acid strength. Problems •  Calculating the pH of a strong acid and the pH of a weak acid solution. •  Calculating the pH of a strong base and the pH of a weak base solution. •  Calculating the pH of strong acid mixtures. •  Calculating the pH of strong base mixtures. •  Calculating the pH of strong-acid/weak-acid mixtures. •  Calculating the pH of strong-base/weak-base mixtures. •  Calculating Ka from % dissociation and Ma. •  Calculating Kb from % dissociation and Mb. •  Calculating % dissociation from Ka and Ma. TOUPADAKIS Lecture 14 Chapter 16 Strong Acids and Weak Acids Compared 0.10 M HCl solution has a pH = 1.00 0.10 M CH3CO2H solution has a pH = 2.80 Two aqueous solutions of monoprotic acids with the same molarity (HCl 0.10 M and CH3CO2H 0.10 M), have different pH values. The acid's molarity (0.10 M) simply indicates the amount of the acid (moles of HCl or moles of acetic acid) that was dissolved in water. Many times it is called initial concentration. The pH and thus [H3O+] indicates the total amount of H3O+ (moles of H3O+) per liter of solution. It is the sum of hydronium ions concentration resulting from the reaction of the acid with water, and the hydronium ions concentration resulting from the auto ionization reaction of water. Because the second concentration is practically 0 M, we consider that the total concentration of hydronium ions (i.e. measured) is practically equal to the concentration of the acid that was ionized (i.e. acid that reacted with water). In the case of the strong acid all of the amount dissolved in water and in the case of the weak acid a fraction of the amount dissolved in water. Note that both of the above acids are monoprotic. TOUPADAKIS Lecture 14 Chapter 16 Table 17.3 Ionization Constants of Weak Acids and Bases TOUPADAKIS Lecture 14 Chapter 16 Percent Ionization (definition) HA + H2O H3O+ + A[HA] ionized [HA] originally [HA] ionized [HA] originally Degree of ionization = Percent ionization = x 100% [HA] ionized = [H3O+] from HA = [H3O+] at equilibrium = [H3O+] from pH TOUPADAKIS Lecture 14 Chapter 16 Percent Ionization (explained) HA + H2O H3O+ + A[H3O+][A-] [HA] + - Ka = n H O nA 1 Ka = nHA V 3 Percent ionization = [HA] ionized [HA] originally x 100% TOUPADAKIS Lecture 14 Chapter 16 Various Ways to Describe Acid Strength Property % ionization Ka value Position of the dissociation (ionization) equilibrium Equilibrium concentration of H+ compared with original concentration of HA Strength of conjugate base compared with that of water - Strong Acid % ionization is large Ka is large Far to the right [H+] ~ [HA]0 A much weaker base than H2O - Weak Acid % ionization is small Ka is small Far to the left [H+] << [HA]0 A much stronger base than H2O TOUPADAKIS Lecture 14 Chapter 16 Problems Problems to try before you come to lecture tomorrow. TOUPADAKIS Lecture 14 Chapter 16 Problem 1 Calculating the pH of a strong-base solution. Calculate the pH of a 0.0011 M solution of Ca(OH)2. Answer: pH =11.35 TOUPADAKIS Lecture 14 Chapter 16 Problem 2 pH of a weak-base solution. Calculate the [OH-], pH and percent ionization for a 0.20 M aqueous solution of NH3. Kb (NH3) = 1.8x10-5. Answer: [OH-] = 1.9x10-3, pH =11.28, % ionization = 0.95% TOUPADAKIS Lecture 14 Chapter 16 Problem 3 Initial concentration of a weak base from the pH of its solution. The pH of a household ammonia solution is 11.50. What is its molarity? Kb (NH3) = 1.8x10-5. Answer: [NH3] initial = 0.57 M. TOUPADAKIS Lecture 14 Chapter 16 Problem 4 Calculate the [OH-] for a 0.050 M aqueous solution of HNO3. Answer: [OH-] = 2.0x10-13 M. TOUPADAKIS Lecture 14 Chapter 16 ...
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This note was uploaded on 04/18/2010 for the course CHEM 2B 993029 taught by Professor Toupadakis during the Spring '10 term at UC Davis.

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