Presentation on theme: "Acids and Bases Chapter 16 Johannes N. Bronsted Thomas M. Lowry"— Presentation transcript:
1 Acids and Bases Chapter 16 Johannes N. Bronsted Thomas M. Lowry Both independently developed Bronsted-Lowry theory of acids and bases.1111
2 Acids and Bases: A Brief Review Classical Acids:Taste sourDonate H+ (called “H-plus” or “proton”)Turn litmus redGenerally formed from H-Z, where Z = nonmetalClassical Bases:Taste bitter and feel soapy.Donate OH- (called “O-H-minus” or “hydroxide”)Turn litmus blueGenerally formed from MOH, where M = metalNeutralization:Acid + Base Salt + waterH-Z + MOH MZ + HOHH+ in water is actuallyin the form of H3O+,“hydronium”
3 Brønsted-Lowry Acids and Bases Proton Transfer ReactionsBrønsted-Lowry acid/base definition:acid donates H+base accepts H+.Brønsted-Lowry base does not need to contain OH-.Consider HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq):HCl donates a proton to H2O. Therefore, HCl is an acid.H2O accepts a proton from HCl. Therefore, H2O is a base.Water can behave as either an acid or a base.Amphoteric substances can behave as acids and bases.
4 Brønsted-Lowry Acids and Bases Conjugate Acid-Base PairsWhatever is left of the acid after the proton is donated is called its conjugate base.Similarly, whatever remains of the base after it accepts a proton is called a conjugate acid.ConsiderAfter H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs.After HCl (acid) loses its proton it is converted into Cl- (base). Therefore HCl and Cl- are conjugate acid-base pairs.Conjugate acid-base pairs differ by only one proton.
5 Brønsted-Lowry Acids and Bases Conjugate Acid-Base PairsIn each of following Bronsted-Lowry Acid/Base reactions,Which is acid? base? conjugate acid? conjugate base?HCl + H2O H3O Cl-acidbaseconj. acidconj. baseNH3 + H2O NH OH-baseacidconj. acidconj. baseCH3NH2 + H2SO4 CH3NH HSO4-conj. basebaseacidconj. acid
6 Brønsted-Lowry Acids and Bases Relative Strengths of Acids and BasesThe stronger the acid, the weaker the conjugate base.The stronger the base, the weaker the conjugate acid.H+ is the strongest acid that can exist in equilibrium in aqueous solution.OH- is the strongest base that can exist in equilibrium in aqueous solution.
7 Strong and Weak Acids and Bases Strong acids: completely ionized in water:HCl, HNO3, H2SO4 (also HBr, HI)Strong bases: completely ionized in water:MOH, where M = alkaliM(OH)2, where M = alkaline earthWeak acids: incompletely ionized in water:any acid that is not strong - acetic acid, etc.Ka is finite.Weak bases: incompletely ionized in water:any base that is not strong – NH3, etc.Kb is finite.
8 The Autoionization of Water The Ion Product of WaterIn pure water the following equilibrium is establishedbut [H2O]2 = constant(at 25oC)This is called the autoionization of water
9 The Autoionization of Water In pure water at 25oC, [H3O+][OH-] = 1 x 10-14and also, [H3O+] = [OH-] = 1 x 10-7(From now on, for simplification, let’s use the abbreviation:[H+] = [H3O+]which means [H]+ = [OH-] = 1 x 10-7We define pH = -log [H+]and pOH = -log [OH-]In pure water at 25oC, pH = pOH = 7.00pH + pOH = 14pKw = 14Acidic solutions have pH < 7.00Basic solutions have pH > 7.00
11 The pH Scale For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-] pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10(b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+]pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12(c) For pH= 8.9, calculate, pOH, [H+], [OH-]pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6(d) For pOH= 3.2, calculate pH, [H+], [OH-]pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4The pH meter is the most accurate way to measure pH values of solutions.
12 Use this “decision tree” to calculate pH values of solutions of specific solutions.Is it pure water? If yes, pH = 7.00.Is it a strong acid? If yes, pH = -log[HZ]Is it a strong base? If yes, pOH = -log[MOH]or pOH = -log (2 x [M(OH)2])Is it a weak acid? If yes, use the relationshipKa = x2/(HZ – x), where x = [H+]Is it a weak base? If yes, use the relationshipKb = x2/(base – x), where x = [OH-]Is it a salt (MZ)? If yes, then decide if it is neutral, acid,or base; calculate its K value by the relationshipKaKb = Kw, where Ka and Kb are for a conjugate system;then treat it as a weak acid or base.Is it a mixture of a weak acid and its weak conjugate base?It is a buffer; see next chapter.
13 2. Strong Acids 3. Strong Bases Calculate the pH of 0.2 M HCl . pH = -log[H+] = -log[0.2] = 0.703. Strong BasesCalculate the pH of 0.2 M NaOH .pOH = -log[OH-] = -log[0.2] = 0.70pH = 14 – 0.70 = 13.30Calculate the pH of 0.2 M Ba(OH)2.pOH = -log[OH-] = -log[0.4] = 0.40pH = 13.60
14 4. Weak AcidsWeak acids are only partially ionized in solution, and are in equilibrium:(shorthand):OR:(shorthand)Ka is the acid dissociation constant.
16 4. Weak Acids Calculate pH of 0.2 M solution of acetic acid, HC2H3O2 From previous slide, Ka= 1.8 x 10-5From now on, we’ll use the shorthand notation for HC2H3O2 = HAcDenote the acetate ion, C2H3O2- as “Ac-”Equilibrium is then: HAc H+ + Ac-Initial conc (M):-x x xchange:at equilibrium:0.2-x x xKa=But, this is a quadratic equation!!We can assume x is small if Ka < 10-4.Then, x2/0.2= 1.8 x 10-5x = 1.90 x 10-3 = [H+] = [Ac-]pH = -log (1.90 x 10-3) = 2.72
17 4. Weak Acids Polyprotic Acids Polyprotic acids have more than one ionizable proton.The protons are removed in steps not all at once:It is always easier to remove the first proton in a polyprotic acid than the second.Therefore, Ka1 > Ka2 > Ka3 etc.Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).
19 Carbonic acid, H2CO3, Ka1=4.3 x 10-7 H2CO3 H+ + HCO Ka1 = 4.3 x 10-7HCO3- H CO Ka2 = 5.6 x 10-11Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium.Therefore, the 1st equilibrium determines the pH.Calculate pH of 0.4 M H2CO3 solution.Ka = x2/HZ-x4.3 x 10-7 = x2/0.4x = [H+] = 4.15 x 10-4pH = -log (4.1 x 10-4) = 3.38
20 5. Weak Bases Weak bases remove protons from substances. There is an equilibrium between the base and the resulting ions:Example:The base dissociation constant, Kb is defined as:
21 5. Weak BasesThe larger Kb the stronger the base.
22 5. Weak Bases x = 8.12 x 10 = [OH ] For convenience, denote it as “B” What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base?For convenience, denote it as “B”Then: B H2O BH OH Kb =4.4 x 10-4Initially:change: x x xAt equil: x x xThen:x=8.12 x 10-3=[OH-]
23 Relationship Between Ka and Kb For a conjugate acid-base pairKa Kb = Kw (constant)Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base.Taking negative logarithms:-log Ka- log Kb= -log KwpKa + pKb = pKwFor HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74.Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26
24 6. Salts Salts may be acidic, basic or neutral. Salts made by a strong acid and a strong base are neutral,e.g., NaCl, KNO3.Salts made by a weak acid and a strong base are weakly basic,e.g., sodium acetate, NaAc, NaHCO3.Salts made by a strong acid and a weak base are weakly acidic,e.g., NH4Cl.Calculate the pH of a 0.35 M solution of sodium acetate.Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid,Then Kb for NaAc (sodium acetate) isKb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 xNow treat this as a weak base problem,Kb = x2/base = 5.56 x = x2/0.35x = [OH-] = 1.39 x 10-5pOH = 4.85 and pH = 9.14