Acids and Bases
Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors
Acid Dissociation HA H + + A - Acid Proton Conjugate base [ H + ][ A ] K a = [ HA] Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion)
Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large K a or small K a? Reactant favored or product favored?
Dissociation Constants: Strong Acids Acid Formula Conjugate Base K a Perchloric HClO 4 ClO 4 - Very large Hydriodic HI I - Very large Hydrobromic HBr Br - Very large Hydrochloric HCl Cl - Very large Nitric HNO 3 NO 3 - Very large Sulfuric H 2 SO 4 HSO 4 - Very large Hydronium ion H 3 O + H 2 O 1.0
Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large K a or small K a? Reactant favored or product favored?
Dissociation Constants: Weak Acids Iodic Oxalic Sulfurous Acid Phosphoric Citric Nitrous Hydrofluoric Formic Benzoic Acetic Carbonic Hypochlorous Hydrocyanic Formula HIO 3 H 2 C 2 O 4 H 2 SO 3 H 3 PO 4 H 3 C 6 H 5 O 7 HNO 2 HF HCOOH C 6 H 5 COOH CH 3 COOH H 2 CO 3 HClO HCN Conjugate Base IO 3 - HC 2 O 4 - HSO 3 - H 2 PO 4 - H 2 C 6 H 5 O 7 - NO 2 - F - HCOO - C 6 H 5 COO - CH 3 COO - HCO 3 - ClO - CN - K a 1.7 x 10-1 5.9 x 10-2 1.5 x 10-2 7.5 x 10-3 7.1 x 10-4 4.6 x 10-4 3.5 x 10-4 1.8 x 10-4 6.5 x 10-5 1.8 x 10-5 4.3 x 10-7 3.0 x 10-8 4.9 x 10-10
Self-Ionization of Water H 2 O + H 2 O H 3 O + + OH - At 25, [H 3 O + ] = [OH - ] = 1 x 10-7 K w is a constant at 25 C: K w = [H 3 O + ][OH - ] K w = (1 x 10-7 )(1 x 10-7 ) = 1 x 10-14
Calculating ph, poh ph = -log 10 (H 3 O + ) poh = -log 10 (OH - ) Relationship between ph and poh ph + poh = 14 Finding [H 3 O + ], [OH - ] from ph, poh [H 3 O + ] = 10 -ph [OH - ] = 10 -poh
ph and poh Calculations [OH - ] = 1 x 10-14 [H + ] H + OH - [H + ] = 1 x 10-14 [OH - ] [H + ] = 10 -ph ph = -log[h + ] ph + poh = 14 Kw = [H + ] [OH - ] = 1 x 10-14 [OH - ] = 10 -poh poh = -log[oh - ] ph poh = 14 - ph ph = 14 - poh poh
ph Scale
A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10-5? Step #1: Write the dissociation equation HC 2 H 3 O 2 H + + C2H3O2-1
A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10-5? Step #2: ICE it! HC 2 H 3 O 2 H + + C 2 H 3 O 2 - I 0.50 0 0 C -x +x +x E 0.50 - x x x
A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10-5? Step #3: Set up the law of mass action E HC 2 H 3 O 2 H + + C 2 H 3 O 2-0.50 - x x x 1.8 x10 ( x)( x) x 5 = (0.50 x) (0.50) 2
A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10-5? Step #4: Solve for x, which is also [H + ] E HC 2 H 3 O 2 C 2 H 3 O 2- + H + 0.50 - x x x 1.8 x 10 = x 2 5 (0.50) [H + ] = 3.0 x 10-3 M ph = log(3.0 x10 3 ) = 2.52
pk a problem The ph of 0.015 M HNO 2 (nitrous acid) aqueous solution was measured to be 2.63. What is the value of Ka and pka of nitrous acid? I C E 10-2.63 = 0.00234423 mol/l =x
Dissociation of Strong Bases MOH(s) M + (aq) + OH - (aq) Strong bases are metallic hydroxides Group I hydroxides (NaOH, KOH) are very soluble Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble ph of strong bases is calculated directly from the concentration of the base in solution
Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O CH 3 NH 3+ + OH - K b = 4.38 x 10-4 K b CH NH + OH 4 3 3 [ ][ ] = 4.38 x10 = [ CH NH ] 3 2
K b for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids. What patterns do you see that may help you? Ammonia Base Methylamine Ethylamine Diethylamine Triethylamine Hydroxylamine Hydrazine Aniline Pyridine Formula NH 3 CH 3 NH 2 C 2 H 5 NH 2 (C 2 H 5 ) 2 NH (C 2 H 5 ) 3 N HONH 2 H 2 NNH 2 C 6 H 5 NH 2 C 5 H 5 N Conjugate Acid NH 4 + CH 3 NH 3 + C 2 H 5 NH 3 + (C 2 H 5 ) 2 NH 2 + (C 2 H 5 ) 3 NH + HONH 3 + H 2 NNH 3 + C 6 H 5 NH 3 + C 5 H 5 NH + K b 1.8 x 10-5 4.38 x 10-4 5.6 x 10-4 1.3 x 10-3 4.0 x 10-4 1.1 x 10-8 3.0 x 10-6 3.8 x 10-10 1.7 x 10-9
A Weak Base Equilibrium Problem What is the ph of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10-5? Step #1: Write the equation for the reaction NH 3 + H 2 O NH 4+ + OH -
A Weak Base Equilibrium Problem C What is the ph of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10-5? Step #2: ICE it! NH 3 + H 2 O NH 4+ + OH - I 0.50 0 0 -x +x +x E 0.50 - x x x
A Weak Base Equilibrium Problem What is the ph of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10-5? Step #3: Set up the law of mass action E NH 3 + H 2 O NH 4+ + OH - 0.50 - x x x 1.8 x10 ( x)( x) x 5 = (0.50 x) (0.50) 2
A Weak Base Equilibrium Problem What is the ph of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10-5? Step #4: Solve for x, which is also [OH - ] E 1.8 NH 3 + H 2 O NH 4+ + OH - 0.50 - x x x x 10 = x 2 5 (0.50) [OH - ] = 3.0 x 10-3 M
A Weak Base Equilibrium Problem What is the ph of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10-5? Step #5: Convert [OH - ] to ph E NH 3 + H 2 O NH 4+ + OH - 0.50 - x x x poh = log(3.0 x10 3 = ) 2.52 ph = 14.00 poh = 11.48
I C E
Assume x is small compared to 0.0044
% protonation is the % of H + or OH - that forms when the acid or base dissociates in water.
Acid-Base Properties of Salts Type of Salt Examples Comment ph of solution Cation is from a strong base, anion from a strong acid KCl, KNO 3 NaCl NaNO 3 Both ions are neutral Neutral These salts simply dissociate in water: KCl(s) K + (aq) + Cl - (aq)
Acid-Base Properties of Salts Type of Salt Examples Comment ph of solution Cation is from a strong base, anion from a weak acid NaC 2 H 3 O 2 KCN, NaF Cation is neutral, Anion is basic Basic The basic anion can accept a proton from water: C 2 H 3 O - 2 + H 2 O HC 2 H 3 O 2 + OHbase acid acid base
Acid-Base Properties of Salts Type of Salt Examples Comment ph of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH 4 Cl, NH 4 NO 3 Cation is acidic, Anion is neutral Acidic The acidic cation can act as a proton donor: NH 4+ (aq) NH 3 (aq) + H + (aq) Acid Conjugate Proton base
Acid-Base Properties of Salts Type of Salt Examples Comment ph of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH 4 C 2 H 3 O 2 NH 4 CN Cation is acidic, Anion is basic See below IF K a for the acidic ion is greater than K b for the basic ion, the solution is acidic IF K b for the basic ion is greater than K a for the acidic ion, the solution is basic IF K b for the basic ion is equal to K a for the acidic ion, the solution is neutral
Acid-Base Properties of Salts Type of Salt Examples Comment ph of solution Cation is a highly charged metal ion; Anion is from strong acid Al(NO 3 ) 2 FeCl 3 Hydrated cation acts as an acid; Anion is neutral Acidic Step #1: AlCl 3 (s) + 6H 2 O Al(H 2 O) 6 3+ (aq) + Cl - (aq) Salt water Complex ion anion Step #2: Al(H 2 O) 6 3+ (aq) Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) Acid Conjugate base Proton
Nonmetallic oxides become acidic in water. SO 2 + H 2 O H 2 SO 3 Metallic oxides become basic in water. Na 2 O + H 2 O Amphoteric Substances 2 NaOH An amphoteric oxide is an oxide that is amphoteric, that is, it can act either as an acid or a base. In a strongly acidic environment, these oxides will act as bases; whereas in a strongly basic environment, these oxides will act as acids. Examples: Aluminum oxide» in acid: Al 2 O 3 + 6HCl 2AlCl 3 + 3H 2 O» in base: Al 2 O 3 + 2NaOH + 3H 2 O 2NaAl(OH) 4
Oxide Hydrated acid Cl 2 O 7 HClO 4 Cl 2 O 5 HClO 3 Cl 2 O 3 HClO 2 SO 3 H 2 SO 4 Oxide Hydrated base MgO Mg(OH) 2 ZnO Zn(OH) 2 Fe 2 O 3 Fe(OH) 3
Effect of Structure on Oxy-Acids The more O s, the stronger the acid and the lower the ph