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Chapter 18 : Acid-Base Equilibria. 18.1 Acids and Bases in Water 18.3 Proton Transfer and the Brønsted-Lowery Acid-Base Definition 18.9 Electron-Pair Donation and the Lewis Acid- Base Definition 18.2 Autoionization of Water and the pH Scale
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Chapter 18 : Acid-Base Equilibria 18.1 Acids and Bases in Water 18.3 Proton Transfer and the Brønsted-Lowery Acid-Base Definition 18.9 Electron-Pair Donation and the Lewis Acid- Base Definition 18.2 Autoionization of Water and the pH Scale 18.4 Solving Problems Involving Weak-Acid Equilibra 18.5 Weak Bases and Their Relation to Weak Acids 18.7 Acid-Base Properties of Salt Solutions
Acid-Base Equilibria Acid Nomenclature Neutralization reactions Amphoteric
Acid H+ [H3O+] in water Sour taste Turns litmus red Naming Rules Does the anion contain Oxygen? NoYes Hydrocheck ending of anion + element root ite ate + ic acid anion root anion root + ous acid + ic acid HBr HNO2 HNO3 Hydrobromic nitrous nitric acid acid acid Base OH– in water Bitter taste Turns litmus blue slippery Arrhenius Theory Neutralization Reaction: Acid + Base Salt + H2O
In acidic solutions, the protons (H+) that are released into solution will not remain alone due to their large positive charge density and small size. They are attracted to the negatively charged electrons on the oxygen atoms in water, and form hydronium ions. .. .. O H H H+(aq) + H2O(l) = H3O+(l) [H+] = [H3O+]
H atom and H+ ion atomic structure? Use symbols H+ and H3O+ to represent the same thing
Acid H+ donor Base H+ acceptor Bronsted-Lowry Practice worksheet and Unit design Problems • Summary • Acids and Bases can be molecules or ions • Not just defined in aqueous solutions • Conjugate pairs differ by H+ • Some substances are both, AMPHOTERIC • depending on the reaction.
Acid Accepts electron pair BF3 Base Gives electron pair NH3 Lewis Theory Draw Lewis dot structure of BF3 and NH3/ Then show combination according to the Lewis Theory Key point: Lewis Acids unique and rare
Relative Acid Strengths Competition for H+ ion H X bond Strong? Weak? Reactions favor formation of _______________. Strong and Weak Acids
The Extent of Dissociation for Strong and Weak Acids Note reactions. Relate to graphs after dissociation.
[H3O+][A-] Qc = [HA] [H3O+][A-] Qc = [HA] The Acid-Dissociation Constant (Ka) Strong acids dissociate completely into ions in water: HA(g or l) + H2O(l) H3O+(aq) + A-(aq) at equilibrium, Qc = Kc >> 1 Weak acids dissociate very slightly into ions in water: HA(aq) + H2O(aq) H3O+(aq) + A-(aq) at equilibrium, Qc = Kc << 1 Stronger acid higher [H3O+] larger Ka
Acid Base Solutions are aqueous.Need to look atWater Water is amphoteric Self-ionization of water [autoionization] Kc = H2O(l) + H2O(l)
Autoionization of Water The ion-product for water, Kw: Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C) For pure water the concentration of hydroxyl and hydronium ions must be equal: [H3O+] = [OH-] = 1000g 1 mole The molarity of pure water is: = 55.5 M 1 L 18.02 g
[H3O+] > 1 x 10-7 acidic [H3O+] = 1 x 10-7 neutral [H3O+] < 1 x 10-7 basic Examples: 1 x 10 -4 1 x 10-9 Classification of solutions according to [H3O+]
pH • To handle the very large variations in the concentrations of the hydrogen ion in aqueous solutions, a scale called the pH scale is used. • pH is defined as • the - log of the molar hydronium ion concentration: pH = - log[H3O+] Give pH values for [H3O+] 1 x 10 -4 1 x 10-9
Acid and Base Character and the pH Scale pH of a neutral solution = 7.00 pH of an acidic solution < 7.00 pH of a basic solution > 7.00 Summary: pH + pOH = 14 pH pOH pH = - log [H3O+] pOH = - log [OH-] [OH-] [H3O+] Kw = [H+][OH-] = 1 x 10-14 at 25oC
The pH Values of Some Familiar Aqueous Solutions
Relationship of [H3O+], pH, [OH-] and pOH As [H3O+] increases pH ________; [OH-] _______; pOH _________
Calculating [H3O+], pH, [OH-], and pOH Problem: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C. Plan:We know that hydrochloric acid is a strong acid, so it dissociates completely in water; therefore [H3O+] = [HCl]init.. We use the [H3O+] to calculate the [OH-] and pH as well as pOH. Use chart as guide. Solution: Answers: 3; -0 .48; 3.33 x 10-15 ;14.5; and .0024; 2.62; 4.1 x 10-12; 11.4
Experiment 7: pH Study Methods for Measuring the pH of an Aqueous Solution (a) pH paper (b) Electrodes of a pH meter
Chart pH + pOH = 14 pH pOH pH = - log [H3O+] pOH = - log [OH-] [OH-] [H3O+] Kw = [H+][OH-] = 1 x 10-14 at 25oC What type of chemical species yields hydronium ions? How many types are there? How do we know which is which? Next slide Relate hydronium ion concentration to each. What type of chemical species yields hydroxide ions? How many types and how do you know which is which? Relate to hydroxide ion concentrations.
Classifying the Relative Strengths of Acids and Bases–I Strong acids.There are two types of strong acids: 1. The hydrohalic acids HCl, HBr, and HI 2. Oxoacids in which the number of O atoms exceeds the number of ionizable H atoms by two or more, such as HNO3, H2SO4, HClO4 Strong bases. Soluble compounds containing O2- or OH- ions are strong bases. The cations are usually those of the most active metals: 1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs) 2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba) [MgO and Mg(OH)2 are only slightly soluble, but the soluble portion dissociates completely.]
[H3O+][A-] Ka= [HA] The Dissociation Constant (Ka and Kb) Weak acids dissociate very slightly into ions in water: HA(aq) + H2O(l) H3O+(aq) + A-(aq) Acid dissociation constant Weak bases dissociate very slightly into ions in water MOH(aq) M+(aq) + OH- (aq) Kb = NH3 + H2O(l) Base dissociation constant Large Ka ? [H30+] ? Acid strength Large Kb ? [H30+] ? Base strength
Solving Problems Class examples There are two general types of equilibrium problems involving weak acids and their conjugate bases: 1. Given equilibrium concentrations, find Ka or Kb. 2. Given Ka or Kb and some concentration information, find the other equilibrium concentrations.
Plan: problem-solving approach. ***1. Write the balanced equation and Ka expression; these will help identify what to solve for. 2. Define x as the unknown concentration that changes. 3. Construct a reaction table, I.C.E. table, that incorporates the unknown, x. 4. Make assumptions that simplify the calculation, usually that x is very small relative to the initial concentration. 5. Substitute the values into the Ka expression and solve for x. 6. Check that the assumptions are justified, if not use the quadratic.
The Relation Between Ka and Kb of a Conjugate Acid-Base Pair Acid HA + H2O H3O+ + A- Base A- + H2O HA + OH- 2 H2O H3O+ + OH- [H3O+] [A-] [HA] [OH-] [H3O+] [OH-] = x [HA] [A-] Kw = Ka x Kb Ka = 4.5 x 10-4 Ka x Kb = (4.5 x 10-4)(2.2 x 10-11) = 9.9 x 10-15 For HNO2 or ~ 10 x 10-15 = 1 x 10 -14 = Kw Kb = 2.2 x 10-11
pH of Solutions What type of solutions are each of these? 0.1 M KNO2 pH = 8.2 0.1 M NH4Cl pH = 5.1 0.1 M KCl pH = 7.0 NH4+ makes solution acidic Cl- no effect on pH K+ no effect on pH NO2- makes solution basic K+ no effect on pH Cl- no effect on pH Conclusion:
Salts Solutions / Hydrolysis • Salts: • Definition • Ionization: • Strong electrolytes - 100 % ionization • Form cations and anions • Reaction with water: HYDROLYSIS • Ion interaction with H2O in solution • Use to look at acidic and basic properties of salt solutions
Possible Hydrolysis Reactions • Cation: M+ + H2O MOH + H+ _____?____ solution • Anion: X- + H2O HX + OH- _____?_____ solution Bronsted/Lowry reactions
Hydrolysis and pHStrategy for determining pH of a salt solution 1. Ionize salt 2. Identify ion undergoing hydrolysis (cation or anion or both) 3. Write hydrolysis reaction 4. Identify solution as acidic or basic If needed calculate Kh value.
Hydrolysis Summary 1. Both ions are strong ________ 2. Weak anion / anion hydrolysis X- + H2O HX + OH-________ Kh = Kw / Kaof weak acid formed = Kb 3. Weak Cation / cation hydrolysis M+ + H2O MOH+ H+ ____________ Kh= Kw / Kbof weak base formed = Ka 4. Both ions weak find Kh for each reaction largest value of Kh determines type of solution.
Examples Following the hydrolysis strategy steps determine the type of solution and Kh for each of the following. 1. NaCl 2. NaC2H3O2 3. NH4Cl 4. NH4C2H3O2 Class problems
Summary for Problem Solving 1. Read problem, identify what is asked and what is given. 2. Decide if you are working with Acid, Base or Salt then determine if strong or weak. 3. Write Balanced Chemical Reaction using Bronsted/Lowry theory. Write K expression for reaction 4. Solve problem.