Chapter 16 - Acids and Bases 16.1 Acids and Bases: The Brønsted Lowry Model 16.2 ph and the Autoionization of Water 16.3 Calculations Involving ph, K a and K b 16.4 Polyprotic Acids
16.1 Acids and Bases: Basic Definitions
Properties of Acids Sour Taste React with active metals (Al, Zn, Fe) to yield H 2 gas: Corrosive React with carbonates to produce CO 2 : Change color of vegetable dyes Turn blue litmus red React with bases to form salts
Properties of Bases Bitter Taste Also Known as Alkalies Solutions feel slippery Change color of vegetable dyes Turn red litmus blue React with acids to form salts
Acids and Bases in Solution Acids ionize in water to form H + ions. (More precisely, the H+ from the acid molecule is donated to a water molecule to form hydronium ion, H 3 O + ) Bases dissociate in water to form OH- ions. (Bases, such as NH 3, that do not contain OH- ions, produce OHby pulling H + off water molecules.)
Acids Acids are molecular compounds that ionize when they dissolve in water. The molecules are pulled apart by water. The percentage of molecules that ionize varies. Acids that ionize virtually 100% are called strong acids. HCl(aq) H + (aq) + Cl (aq) Acids that only ionize a small percentage are called weak acids. HF(aq) H + (aq) + F (aq)
Strong Acid Weak Acid
Molecular Models of Selected Acids Binary Acid Oxy" Acids
Common Acids Name Formula Uses Strength Perchloric HClO4 explosives, catalysts Strong Nitric HNO3 explosives, fertilizers, dyes, glues Strong Sulfuric H2SO4 explosives, fertilizers, dyes, glue, batteries Strong Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid Strong Phosphoric H3PO4 fertilizers, plastics, food preservation Moderate Chloric HClO3 explosives Moderate Acetic HC2H3O2 plastics, food preservation, vinegar Weak Hydrofluoric HF metal cleaning, glass etching Weak Carbonic H2CO3 soda water, blood buffer Weak Hypochlorous HClO sanitizer Weak Boric H3BO3 eye wash Weak
Strong Acids and Their Ionization in Water
Common Weak Acids and Their Ionization in Water 4.74 3.75 3.17 7.54 3.15 Why do we not define K a for a strong acid?? pk a = -log K a
Structure of Bases Most contain OH- ions Some contain CO 3 2- ions Molecular bases contain structures which react with H+
Common Inorganic Bases Name Formula Common Name Uses Strength Sodium Hydroxide NaOH Lye, Caustic Soda soap, plastic production, petroleum refining Strong Potassium Hydroxide KOH Caustic Potash soap, cotton processing, electroplating Strong Calcium Hydroxide Ca(OH)2 Slaked Lime cement Strong Sodium Bicarbonate NaHCO3 Baking Soda food preparation, antacids Weak Magnesium Hydroxide Mg(OH)2 Milk of Magnesia antacids Weak Ammonium Hydroxide NH4OH Ammonia Water fertilizers, detergents, explosives Weak
Strong Bases and Their Ionization in Water
Weak Bases and Their Ionization in Water pk b = -log K b
Arrhenius Theory Bases dissociate in water to produce OH- and cations: Acids ionize in water to produce H+ and anions: acetic acid acetate anion
Arrhenius Theory An Arrhenius Acid-Base Reaction
Problems with Arrhenius Theory Does not explain why some molecular substance dissolve in water to give basic solutions, but do not contain OH-. (NH 3 ) Does not explain why some ionic substances dissolve in water to give basic solutions, but do not contain OH-. (NaCO 3 ) Does not explain why some molecular substance dissolve in water to give acidic solutions, but do not contain H+. (CO 2 ) Only applicable to aqueous solutions.
Brønsted-Lowry Theory Acid are proton donors Bases are proton acceptors Acid-Base Reactions involve proton transfer Base structure includes a pair of unshared electrons
Brønsted-Lowry Theory
Conjugate Pairs In a Brønsted-Lowry reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process. Each reactant and product are a conjugate pair. The original base becomes a conjugate acid, and the original acid becomes a conjugate base.
Conjugate Pairs H-A + :B :A - + H-B + acid base conjugate conjugate base acid HCHO2 + H2O CHO2 - + H3O + acid base conjugate conjugate base acid H2O + NH3: HO - + NH4 + acid base conjugate conjugate base acid
Conjugate Pairs H2O and OH- are an acid/base conjugate pair. NH3 and NH4+ are an base/acid conjugate pair.
Lewis Acid-Base Theory A Lewis base is an electron donor or nucleophile. A Lewis acid is an electron acceptor or electrophile. When a nucleophile donates a pair of electrons to an electrophile, a covalent bond forms.
16.2 ph and the Autoionization of Water
Autoionization of Water Water is an extremely weak electrolyte. One out of every 10 million water molecules forms ions. All aqueous solutions, therefore, contain some H+ and OH-.
Autoionization of Water The Ion Product of Water is Defined:
Autoionization of Water at 25ºC in the absence of any other acids or bases. A change in [H3O+] causes and inverse change in [OH-].
Relationship between [H 3 O+] and [OH-] [H 3 O+] > [OH-] [H 3 O+] = [OH-] [H 3 O+] < [OH-] Acidic Solution Neutral Solution Basic Solution
Relationship between [H+] & [OH-] [H+] 10 0 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 H+ H+ H+ H+ H+ OH- OH- OH- OH- OH- [OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 10 0 Kw =[H3O+][OH-]=1.0 x 10-14
ph as a Measure of Acidity/Basicity ph = -log [H 3 O + ] ph < 7, acidic ph > 7, basic ph = 7, neutral
ph of Common Substances
ph vs poh ph 0 1 3 5 7 9 11 13 14 [H+] 10 0 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 H+ H+ H+ H+ H+ OH- OH- OH- OH- OH- [OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 10 0 poh 14 13 11 9 7 5 3 1 0 ph + poh =14
1) What are the [OH ] and ph of household ammonia, an aqueous solution that has an [H3O + ] of 1.99 x 10-12 M? Kw =[H3O+][OH-]=1.0 x 10-14 (1.99 x10-12 )[OH-]=1.0 x 10-14 [OH-]=(1.0 x 10-14 )/(1.99 x10-12 ) [OH-]=5.02 x 10-3 ph = -log[h3o+] =11.70 poh = -log[oh-] = 2.30
pka = -logka
16.3 Calculations Involving ph, K a and K b
2) Calculate and compare [H + ] and ph for a 0.100 M solution of HClO4 and a 0.100 M solution of HClO (Ka = 2.9 10 8 ). HClO4 ClO - + H + For a strong acid, [H+] = 0.100 M, ph = -log(0.10) = 1 HClO ClO - + H + For the weak acid, a little more work is needed. x 2 0.100-x = 2.9 x 10-8 [H+] = x = 5.4 x 10-5 ph = 4.3
3) The ph of a 0.050 M solution of a weak organic acid is 2.23. Calculate [H+], Ka and % ionization for the acid. [H+] = 10 -ph = 10-2.23 = 5.89 x 10-3 [A-][H+] Ka = = [5.89 x 10-3 ] 2 [HA] [0.044] = 7.9 x 10-4 % ionization = (5.9 x 10-3 /0.050) x 100 = 12%
4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 10 4 ). [A-][H+] Ka = = [x][x] [HA] [1.28-x] = 4.0 10 4 [x] 2 [1.28] = 4.0 10 4 x = 2.3 10 2 % ionization = (2.3 x 10-2 /1.28) x 100 = 1.8%
4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 10 4 ). [A-][H+] Ka = = [x][x] [HA] [0.0150-x] = 4.0 10 4 [x] 2 [0.0150] = 4.0 10 4 x = 2.3 10 3 % ionization = (2.4 x 10-3 /0.0150) x 100 = 16% (2.6 x 10 3 ) (17%)
5) Calculate and compare [OH-] and ph for 0.200 M LiOH and a 0.200 M solution of methylamine, CH3-NH2 (Kb = 4.4 10 4 ). LiOH Li + + OH - For a strong base, [OH-] = 0.200 M, poh = -log(0.20) = 0.7, ph=13.3 CH3-NH2 + H2O CH3-[NH3] + + OH - For the weak base, once again, a little more work is needed. [BH+][OH-] Kb = = [B] [x][x] [0.200-x] = 4.4 10 4 [x][x] [0.200] = 4.4 10 4 x = 9.4 10 3 = [OH-] poh = -log(9.4 10 3 ) = 2.03, ph = 11.97
6) Calculate [H+] and ph for 1.0 10 8 M HCl. HCl H + + Cl - For a strong acid, [H+] = 1.0 10 8 M, ph = 8.0, BUT THIS DOES NOT MAKE SENSE!!!
THERE IS ANOTHER EQUILIBRIUM TO CONSIDER H2O + H2O H3O + + OH - KW = [H3O+][OH-] = 1.0 x 10-14 IF x REPRESENTS [H+] AND [OH-] FROM AUTOIONIZATION (x + 1.00 x 10-8 ) (x) = 1.0 x 10-14 solve quadratic x 2 + (1.00 x 10-8 )(x)-1.0 x 10-14 = 0 x = 9.5 x 10-8 = [OH-] [H+] = 9.5 x 10-8 + 1.0 x 10-8 = 10.5 x 10-8 = 1.05 x 10-7 ph = 6.98
16.4 Polyprotic Acids
Polyprotic Acids
7) Calculate the ph of 0.034 M carbonic acid solution (Ka1 = 4.3 10 7 and Ka2 = 4.7 10 11 ). Ka1 = [HCO3 - ][H+] [HA] 4.3 x 10-7 = [x][x] [0.034] 1.2 x 10-4 = x = [H+] 1.2 x 10-4 = x = [HCO3 - ] 0.034 = [H2CO3]
7) Calculate the ph of 0.034 M carbonic acid solution (Ka1 = 4.3 10 7 and Ka2 = 4.7 10 11 ). Ka2 = [CO3 2- ][H+] [HCO3 - ] 4.7 x 10-11 = [x][x +1.2 x 10-4 ] [1.2 x 10-4 ] 4.7 x 10-11 = x = [CO3 2- ] 1.2 x 10-4 = [HCO3 - ] ph = -log [H+] = -log (1.2 x 10-4 + 4.7 x 10-11 ) = 3.92
Polyprotic Acids
Polyprotic Acids