Presentation on theme: "Chemistry: The Molecular Science"— Presentation transcript:
1Chemistry: The Molecular Science Moore, Stanitski and JursChapter 16: Acids and Bases
2Arrhenius DefinitionArrhenius: any substance which ionizes in water to produce…protons (H+) is an acid.hydroxide ions (OH-) is a base.Better version of the Arrhenius definition:Acid: hydronium ions (H3O+) in water are acidic.Base: hydroxide ions (OH-) in water are basic.
3Acid-Base ReactionsA strong acid is an acid that ionizes completely in water (product favored).There are six strong acids:HCl - hydrochloric acidHBr – hydrobromic acidHI – hydroiodic acidH2SO4 - sulfuric acidHNO3 – nitric acidHClO4 – perchloric acid
4Acid-Base Reactions Strong Acids and Bases A strong base is a base that is present entirely as ions (product favored).The hydroxides of Group IA, IIA (except Be and Mg hydroxides) are strong bases.So, why is NH3 (aq) basic?
5Brønsted-Lowry Concept An alternative definition:acid = proton (H+) donorbase = proton (H+) acceptorWorks for non-aqueous solutions and explains why NH3 is basic:NH3(g) + H2O(l) NH4+(aq) + OH-(aq)indirectly producedBase:H+ acceptorAcid:H+ donor
6Brønsted-Lowry Concept Strong acids and bases almost completely ionize (product favored).100%HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)Weak acids and bases do not fully ionize (reactant favored).3%HF(aq) + H2O(l) H3O+(aq) + F-(aq)Acid:H+ donorBase:H+ acceptorAcid:H+ donorBase:H+ acceptorNote: the products are a new acid and base pair.LeChatelier – a more dilute solution (more water), will ionize more.
7Water’s Role as Acid or Base Water acts as a base when an acid dissolves in water: HBr(aq) + H2O(l) H3O+(aq) + Br-(aq)acid base acid baseBut water acts as an acid for some bases:H2O(l) + NH3(aq) NH4+(aq) + OH-(aq)acid base acid baseWater is amphiprotic - it can donate or accept a proton (act as acid or base).
8PracticeComplete the equations: (acid or base?, mass/charge balance?, single or double arrow?)HClO4 + H2OCN- + H2OH2S + H2ONO H2OCa(OH)2 + H2OClO H3O+HCN + HO-HS- + H3O+HNO2 + HO-Ca OH-
9Conjugate Acid-Base Pairs Molecules or ions related by the loss/gain of one H+.Conjugate Acid Conjugate BaseH3O+ H2OCH3COOH CH3COO-NH4+ NH3H2SO4 HSO4-HSO4- SO42-HCl Cl-donate H+accept H+NH4+ and NH2- are not conjugate(conversion requires 2 H+)
10Conjugate Acid-Base Pairs Identify the base conjugate to HC2H3O2(aq) and the acid conjugate to HCO3-(aq)HC2H3O H2O H3O+ +C2H3O2 -ConjugateBaseAcid:H+ donorHCO H2O OH- +H2CO3Base:H+ acceptorConjugate Acid160 of 200Show that HCO3- is amphoteric.
11Practice: Conjugate Acid-Base Pairs Write the equation for the acid or base in water and identify the acid-base pairs.HBrHSO3- (as an acid)HSO3- (as a base)PH4+
12Relative Strength of Acids & Bases Strong acids are better H+ donors than weak acids.Strong bases are better H+ acceptors than weak bases.Strong acids have weak conjugate bases.Weak acids have strong conjugate bases.Strong acid + H2O H3O+ + weak conjugate baseFully ionized, reverse reaction essentially does not occur.The conjugate base is weak.Weak acid + H2O H3O+ + strong conjugate baseWeakly ionized, reverse reaction readily occurs.The conjugate base is strong.
14Relative Strength of Acids & Bases ProblemIs the following aqueous reaction product or reactant favored?Acid strength increasingBase strength increasingConj acid. Conj. baseH2SO4 HSO4-HBr Br-HCl Cl-HF F-NH4+ NH3OH- O2-H2 H-CH4 CH3-NH4+ + F NH3 + HFHF is a stronger acid than NH4+.(NH3 is a stronger base than F-)HF has greater tendency to ionize than NH4+.(NH3 more readily accepts H+ than F-)Reactant Favored
17Autoionization of Water Pure water conducts a very small electrical current.Autoionization occurs:H2O(l) + H2O(l) H3O+(aq) + OH-(aq)BaseAcidAcidBaseKw = [ H3O+ ] [ OH- ]Kw = 1.0 x (at 25°C)Kw = ionization constant for waterHeavily reactant favored.
18Ionization Constant for Water Kw is a special equilibrium constant for the autoionization of water. It is T-dependent. T = 25°C (77°F) is usually used as the standard T.T (°C) Kwx 10-14x 10-14x 10-14x 10-14x 10-14x 10-14Kw increases with temperature. Is it endo or exothermic?
19Autoionization of Water Hydronium and hydroxide ions are produced in equal numbers in pure water,2H2O(l) H3O+(aq) + OH-(aq)Kw = [ H3O+ ] [ OH- ]Thus, the concentrations of H3O+ and OH- in pure water are both 1.0 x 10-7 M.2
20Autoionization of Water H3O+ and OH- are present in all aqueous solutions. Neutral solution. Pure water 25°C): [H3O+] = 10-7 M = [OH-] Acidic solution If acid is added to water:[ H3O+ ] is increased, disturbing the equilibrium:2 H2O H3O+ + OH-LeChatelier: equilibrium shifts to the left. [OH-]Equilibrium is reestablished: [ H3O+ ] > 10-7 M > [OH-]Product of ion concentrations is the same[ H3O+ ][OH-] = Kw = 1.0 x 10-14
21Autoionization of Water Basic solution.If base is added to water.LeChatelier: equilibrium shifts to the left, [H3O+]New equilibrium: [H3O+] < 10-7 M < [OH-][ H3O+ ][OH-] = Kw = 1.0 x at 25°C2 H2O H3O+ + OH-ExampleCalculate the hydronium and hydroxide ion concentrations at 25°C in a 6.0 M aqueous sodium hydroxide solution.Kw = [H3O+][OH-] = 1.0 x 10-14NaOH (aq) is strong (100% ionized) so [OH-] = 6.0 M.[H3O+](6.0) = 1.0 x 10-14[H3O+] = 1.7 x M [ OH- ] = 6.0 M
22Autoionization of Water The relative concentrations of H3O+ and OH- indicate acidic, neutral or basic character of a solution:In a neutral solution, the concentrations of H3O+ (aq) and OH-(aq) are equal.In an acidic solution, the concentration of H3O+ (aq) is greater than that of OH-(aq).In a basic solution, the concentration of OH-(aq) is greater than that of H3O+ (aq).2
23Acid or Base At 25 oC, we observe the following conditions. In an acidic solution, [H3O+ ] > 1.0 x 10-7 M.In a neutral solution, [H3O+ ] = [OH-] = 1.0 x 10-7 M.In a basic solution, [H3O+ ] < 1.0 x 10-7 M.2
24The pH scale pH = −log10[H3O+] Can describe acidity in terms of [H3O+], but a logarithmic scale is more convenient:pH = −log10[H3O+]At 25°C a neutral aqueous solution has:pH = −log10[1.0 x 10-7] = −(−7.00) = 7.00acidic solutions: [H3O+]>1.0 x 10-7, pH < 7.00basic solutions: [H3O+]<1.0 x 10-7, pH > 7.00
25The pH of a SolutionFor a solution in which the hydronium-ion concentration is 1.0 x 10-3, the pH is:Note that the number of decimal places in the pH equals the number of significant figures in the hydronium-ion concentration.2
26Example – pH given [H3O+] ? A sample of orange juice has a hydronium-ion concentration of 2.9 x 10-4 M. What is the pH?2
27Example - [H3O+] given pH? The pH of human arterial blood is 7.40 (acidic or basic?). What is the hydronium-ion concentration? (Guesss….> or < 1.0x10-7M?)10logx = x2
29we first calculate the pOH: ExampleAn ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution?we first calculate the pOH:Then the pH is:2
30pH Calculations Given two aqueous solutions (25°C). Solution A: [OH-] = 4.3 x 10-4 M,Solution B: [H3O+] = 7.5 x 10-9 M.Which has the higher pH? Which is more acidic?Solution A: pOH = −log[ OH-] = 3.37 pH + pOH = pKw = pH = Solution B: pH = −log[ H3O+] = 8.12 A has higher pH, B is more acidic
32pH of Aqueous Solutions Stomach FluidsLemon JuiceVinegarWineTomatoesBlack CoffeeMilkPure waterBlood, seawaterSodium bicarbonateBorax solutionMilk of MagnesiaDetergentsAqueous ammonia1 M NaOH[H3O+] [OH-]ExampleBattery AcidpH1234567891011121314BleachpH of Aqueous Solutions
33Measuring pH H3O+ concentrations can be measured with an: Electronic pH meter:fast and accuratepreferred method.Acid-base Indicator:substance that changes color within a narrow pH rangemay have multiple color change (e.g. bromthymol blue)one “color” may be colorless (e.g. phenolphthalein)cheap and convenient.
34Ionization Constants of Acids and Bases When an acid ionizes in water:HA(aq) + H2O(l) H3O+(aq) + A-(aq)The acid ionization constant is used to report the degree of ionization:[A-][H3O+][HA]Ka =(Water omitted, as usual)Strong acids have large Ka valuesWeak acid have small Ka values
35Base Ionization Constants For a base in water:B(aq) + H2O(l) BH+(aq) + OH-(aq)The base ionization constant, Kb, is:Kb =[BH+][OH-][B]If the base is an anion:A-(aq) + H2O(l) HA(aq) + OH-(aq)Kb =[HA][OH-][A-]
36Ionization Constants Larger Ka = stronger acid Larger Kb = stronger base
37Ionization Constants Larger Ka = stronger acid Larger Kb = stronger base
38Acid and Base StrengthSmall value of Ka indicates that the numerator is smaller thanthe denominator. Equilibrium favors reactants or only a small amount dissociates – a weak acid.Small value of Kb indicates that the numerator is smaller thanthe denominator. Equilibrium favors reactants or only a small amount dissociates – a weak base.
39ExampleWrite the ionization equation and ionization constant expression for the following acids and bases: a.) HF b.) HBrO c.)NH3 d.) CH3NH2
40Ka Values for Polyprotic Acids Some acids can donate more than one H+Formula Name Acidic H’sH2S Hydrosulfuric Acid 2H3PO4 Phosphoric Acid 3H2CO3 Carbonic Acid 2HOOC-COOH Oxalic acid 2C3H5(COOH)3 Citric acid 3Each H+ ionization has a different Ka.The 1st proton is easiest to removeThe 2nd is harder, etc.
41Ka Values for Polyprotic Acids Phosphoric acid (H3PO4) is weak. It has three acidic protons:H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4- (aq)Ka = 7.5 x 10-3H+ is harder to removeKa decreasingH2PO4-(aq) + H2O(l) H3O+(aq) + HPO42- (aq)Ka = 6.2 x 10-8HPO42-(aq) + H2O(l) H3O+(aq) + PO43- (aq)Ka = 3.6 x 10-13
42Practice – Relative Strength (Ka and Kb) The following reactions all have Keq>1.NO HF HNO2 + F-CH3COO- + HF CH3COOH + F-HNO2 + CH3COO HNO CH3COOHArrange the substances based on their relative base strength.F-CH3COOHHFHNO2-CH3COO-HNO234211 – strongest base2 – intermediate base3 – weakest base4 – not a B-L base
43Molecular Structure and Acid Strength What makes a strong acid?A weak H–A bond, so H+ can be easily removed!Consider the binary acids HF, HCl, HBr, HI.HX Bond Energy (kJ) KaHF weak acid x 10-4HCl x 107HBr x 108HI x 1010strong acidssmaller bond energylarger acid strengthH-A, bond energy decreases down a group.Acid strength increases.
44Molecular Structure and Acid Strength What makes a strong acid?A weak H–A bond, so H+ can be easily removed!Consider the binary acids SiH4, PH3, H2S, HCl.The higher electronegativity, across a period, makes the bond more polar and hence more ionic – a stronger acid.
45Molecular Structure and Acid Strength Oxoacidshydrogen, oxygen and one other elementH O-Ehigher the electronegativity on E, stronger the acid as this weakens the bond between the O and HHOCl HOBr HOIKa: × × × 10-11Electronegativity: Cl = 3.0, Br = 2.8, I = 2.5
46Molecular Structure and Acid Strength The acid strength increases as the number of oxygen atoms bonded to E increases. Strong oxoacid has at least two oxygen atoms per acidic H atoms.HClO HClO HClO *HClO4Ka: 3.5 × × ≈ ≈ 108
47Problem Solving Using Ka and Kb Example: Ka from pH Lactic acid is monoprotic. The pH of a M solution was 2.43 at 25°C. Determine Ka for this acid.Using the pH information:- log [H3O+] = 2.43 [H3O+] = = MThese hydronium ions are produced by:HA(aq) + H2O(l) H3O+(aq) + A-(aq)and from the autoionization of water:H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
48Problem Solving Using Ka and Kb HA(aq) + H2O(l) H3O+(aq) + A-(aq)[ ]initial x(water autoionization)change[ ]equil –Ka =[H3O+][A-][HA]Ka = = 1.4 x 10-4(0.0037)(0.0037)(0.100 – )
49Problem Solving Using Ka and Kb Example: pH from Ka Determine the pH of a M propanoic acid solution at 25°C. Ka = 1.4 x What % of acid is ionized?C2H5COOH + H2O(l) H3O+(aq) + C2H5COO-(aq)Ka = = 1.4 x 10-5[H3O+][C2H5COO-][C2H5COOH]
50Problem Solving Using Ka and Kb Determine the pH of M propanoic acid soln. at 25°C. Ka = 1.4 x What % of acid is ionized?HA(aq) + H2O(l) H3O+(aq) A-(aq)[ ]initial * 0change -x x x[ ]equil – x x* x*Ignore the water contribution[H3O+][A-][HA]Ka = 1.4 x 10-5 =1.4 x 10-5=x2(0.100 – x)
51Problem Solving Using Ka and Kb Because Ka is small, assume x < [HA]Ka = 1.4 x = ≈x2(0.100 – x)x20.100x = (1.4 x 10-5 )(0.100) = MpH = -log( ) = 2.93( << M; the approximation is good)%-ionized = x100 = 1.18 %0.100<5% rulecan assume[HA]eq~[HA]0
52Example 3 – calculating pH from known Kb What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.Initial0.20Change-x+xEquilibrium0.20-xx2
53Example 3 – calculating pH from known Kb What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.2
54Example 3 – calculating pH from known Kb What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.Solving for x we get2
55Example 3 – calculating pH from known Kb What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.Solving for pOH2
56PracticeKa from pHPara-aminobenzoic acid (PABA), HC7H6NO2, is used in some sunscreen agents. A solution is made by dissolving mol of PABA in enough water to make mL of solution. The solution’s pH = What is the Ka for PABA? (1.9x10-5)pH from KaBarbituric acid (Ka=1.1x10-4) is used in the manufacture of some sedatives. What is the pH for a M solution of barbituric acid? (2.07)
57Relationship between Ka and Kb values For an acid-base conjugate pair: HA and A-[HA][OH-][A-]Ka x Kb =[H3O+][A-][HA]=[H3O+][OH-]= KwKaKbWhen two reactions are added, their equilibrium constantsare multiplied (chapter 14).
58Relationship between Ka and Kb values Phenol, C6H5OH, is a weak acid, Ka = 1.3 x at 25°C. Calculate Kb for the phenolate ion C6H5O-Ka x Kb = 1.0 x 10-14Kb = = 7.7 x 10-51.0 x 10-141.3 x 10-10
59Acid Base Neutralization Reactions Salt = ionic compound formed in acid + base reactionHX(aq) + MOH(aq) MX(aq) + H2O(l)acid base saltStrong acid + strong base form neutral salts.HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)Na+(aq) + Cl-(aq) + H2O(l)Net ionic equation: H+(aq) + OH-(aq) H2O(l)H3O+(aq) + OH-(aq) H2O(l)
60Salts of Strong Bases and Strong Acids H3O+(aq) + OH-(aq) H2O(l) + H2O(l)Keq = 1/Kw = 1x1014strongacidstrongbaseweakbaseweakacid100%H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)Na+(aq) + Cl-(aq) + 2 H2O(l)Na+ and Cl- are spectator ions.Na+ is the v. weak conjugate acid of a v. strong base.Cl- is the v. weak conjugate base of a v. strong acid.Final solution has pH = 7.
61Salts of Strong Bases and Weak Acids Weak acid + strong base form basic salts.CH3COOH(aq) + NaOH(aq)CH3COONa(aq) + H2O(l)CH3COOH(aq) + Na+(aq) + OH-(aq)Na+(aq) + CH3COO-(aq) + H2O100%Net ionic: CH3COOH(aq) + OH-(aq)CH3COO- + H2O(l)final solution
62Salts of Strong Bases and Weak Acids 100%CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l)baseacidacetate is a weak base (much stronger than water)solution is basic (pH > 7.0).CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)Calculate pH of solution from Kb of CH3COO-.A hydrolysis reaction – water is broken apart.
63pH of a Salt SolutionFor a weak acid + strong base, pH depends on Kb Larger Kb = stronger baseExampleWhat is the pH of a 1.50 M aqueous solution of Na2CO3? Kb (CO32-) = 2.1 x 10-4.(or equal volumes of 1.5M H2CO3 and 3.0 M NaOH are mixed, what is the pH of the mixture?)
64pH of a Salt Solution Na+ is the conjugate acid of NaOH (strong base). What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4Na+ is the conjugate acid of NaOH (strong base).It remains 100% ionizedNa+ has no effect on pH.CO32- is the conjugate base of HCO3- (weak acid)Most CO32- ions undergo hydrolysisGenerates HCO3- and OH-Changes the pHCO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq)
65pH of a Salt Solution Kb = 2.1 x 10-4 = = ≈ [HCO3-][OH-] [CO32-] x2 What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4CO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq)[ ]initialchange x x x[ ]equil x x xKb = 2.1 x 10-4 = = ≈[HCO3-][OH-][CO32-]x2(1.50 – x)1.50x = 1.77 x 10-2 pOH = −log(1.77 x 10-2) = 1.75pH = = 12.25
66Salts of Weak Bases and Strong Acids NH3(aq) + HCl(aq) NH4+(aq) + Cl-(aq)NH3(aq) + H3O+(aq) + Cl-(aq)NH4+(aq) + Cl-(aq) + H2O(l)Net ionic: NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l)The products of the neutralization reaction form an acidic solution:NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)The final pH depends on the Ka , larger Ka = more acidic
67pH of a Salt SolutionFind the pH of a M NH4Br. Is this solution acidic, basic or neutral? Ka(NH4+) = 5.6 xThe solution contains NH4+ ions and Br- ions.NH4+ is the conjugate acid of NH3 (a weak base).It will partially ionize in solution giving NH3 and H3O+Forms an acidic solution.Br- is the conjugate base of HBr (a strong acid)Br- ions stay fully ionizedDoes not change the pH.
68pH of a Salt Solution Ka = 5.6 x 10-10 = = ≈ [NH3][H3O+] [NH4+] x2 pH of a M aq. solution of NH4Br? Is this acidic, basic or neutral? Ka(NH4+) = 5.6 xNH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)[ ]initialchange x x x[ ]equil x x xKa = 5.6 x = = ≈[NH3][H3O+][NH4+]x2(0.132 – x)0.132x = 8.57 x pH = −log(8.57 x 10-6) = 5.07Acidic!
69Salts of Weak Bases and Weak Acids The most difficult. Consider qualitative results. e.g.NH3 (aq) + HF(aq) NH4+ (aq) + F- (aq)NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)F-(aq) + H2O(l) HF(aq) OH-(aq)Ka(NH4+) = 5.6 x ; Kb(F-) = 1.4 x 10-11Ka (weak acid) > Kb (weak base)The ammonium reaction is more favorable.The acid wins! The solution will be acidic!
70Acids, Bases and SaltsStrong acid + strong base → salt solution, pH = 7 Strong acid + weak base → salt solution, pH < 7 Weak acid + strong base → salt solution, pH > 7 Weak acid + weak base → salt solution, pH = ? Need K’s “strongest wins”
71Neutral Ions in Water (Table 16.5) Anions Cl- NO3-Br- ClO4-I-Cations Li+ Ca+2Na+ Sr+2K+ Ba+2
72ExampleWhat is the pH of a 0.10 M NaCN solution at 25 oC? The Kb for CN- is 2.5 x 10-5.Sodium cyanide gives Na+ ions and CN- ions in solution.Only the CN- ion hydrolyzes.pH =11.22
73ExampleWhat is the pH of a 0.15 M NH4NO3 solution at 25 oC? The Ka for NH4+ is 5.6 xNH4NO3 gives NH4+ ions and NO3- ions in solution.Only the NH4+ ion hydrolyzes.2
74The Common Ion EffectThe common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.Consider a solution of acetic acid (HC2H3O2),in which you have the following equilibrium.2
75The Common Ion EffectIf we were to add NaC2H3O2 to this solution, it would provide C2H3O2- ions which are present on the right side of the equilibrium.The equilibrium composition would shift to the left(LeChatelier Principle) and the degree of ionization of the acetic acid would decrease.This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.2
76ExampleWhat is the pH of a solution that is 0.025M in formic acid? The Ka for formic acid is 1.7 x What is the pH of the same solution after M in sodium formate, NaCH2O is added?InitialChangeEquilibrium0.018-x x x0.025-x x xFrom NaCH2O2