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Chapter 14 Aqueous Equilibria: Acids and Bases
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Polyprotic Acids Acids that contains more than one dissociable proton Dissociate in a stepwise manner ◦ Each dissociation step has its own Ka Stepwise dissociation constants decreases in the order Ka 1 > Ka 2 > Ka 3 ◦ More difficult to remove a positively charge proton from negative ion
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Polyprotic Acids Diprotic acid solutions contain a mixture of acids: H 2 A, HA, H 2 O ◦ Strongest acid – HA Principle reaction – dissociation of H 2 A All of H 3 O + come from the first ionization H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) + H 2 O(l) H 3 O + (aq) + SO 4 2-
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Polyprotic acids H 2 CO 3 (aq) + H 2 O(l) H 3 O + ( aq) + HCO 3 - (aq) K a1 = 4.3 x 10 -7 HCO 3 - (aq) + H 2 O(l) H 3 O + ( aq) + CO 3 2- (aq) K a2 = 4.8 x 10 -11
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Polyprotic Acids
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Equilibria in Solutions of Weak Bases NH 4 1+ (aq) + OH 1- (aq)NH 3 (aq) + H 2 O(l) Base Acid Base [NH 4 1+ ][OH 1- ] [NH 3 ] K b = BH 1+ (aq) + OH 1- (aq)B(aq) + H 2 O(l) [BH 1+ ][OH 1- ] [B] K b = Base-Dissociation Constant:
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Equilibria in Solutions of Weak Bases
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Calculate the [ - OH] and pH of a 0.40 M NH 3 solution. At 25 °C, K b = 1.8 x 10 -5.
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Example Morphine (C 17 H 19 NO 3 ), a narcotic used in painkillers, is a weak organic base. If the pH of a 7.0 x 10 -4 M solution of morphine is 9.5, what is the value of K b ?
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Relation Between K a and K b H 3 O 1+ (aq) + OH 1- (aq)2H 2 O(l) H 3 O 1+ (aq) + NH 3 (aq)NH 4 1+ (aq) + H 2 O(l) NH 4 1+ (aq) + OH 1- (aq)NH 3 (aq) + H 2 O(l) [NH 4 1+ ][OH 1- ] [NH 3 ] [H 3 O 1+ ][NH 3 ] [NH 4 1+ ] x KwKw KaKa KbKb = [H 3 O 1+ ][OH 1- ] = K w = (5.6 x 10 -10 )(1.8 x 10 -5 ) = 1.0 x 10 -14 K a x K b =
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Relation Between K a and K b pK a + pK b = pK w = 14.00 K a x K b = K w Kb =Kb = KaKa KwKw K a = KbKb KwKw conjugate acid-base pair
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Example Find the pH of a 0.100 M NaCHO 2 solution. The salt completely dissociate into Na + (aq) and CHO 2 - (aq) and Na + ion has no acid or base properties. Ka (HCHO 2 )= 1.8 x 10 -4
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Example What is the pH of 0.10M sodium nicotinate at 25 o C? The K a for nicotinic acid is 1.4 x 10 -5.
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Acid-Base Properties of Salts pH of a salt solution is determined by the acid-base properties of the consistuent cations and anions ◦ In an acid-base reaction, the influence of the stronger partner is predominant ◦ Strong acid + Strong Base Neutral solution ◦ Strong acid + Weak Base Basis solution ◦ Weak acid + Strong Base Acidic solution
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Acid-Base Properties of Salts Acidic cation + neutral anion Acidic salt NH 4 + + Cl - NH 4 Cl Neutral cation + neutral anion neutral salt Na + Cl - NaCl Neutral cation + basic anion basic salt Na + CN - NaCN
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Acid-Base Properties of Salts Acidic cation + basic anion (50 :50 mixture) must compare K a and K b K a > K b : The solution will contain an excess of H 3 O 1+ ions (pH < 7). K a 7). K a = K b : The solution will contain approximately equal concentrations of H 3 O 1+ and OH 1- ions (pH ≈ 7).
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Acid-Base Properties of Salts Salts That Yield Acidic Solutions Hydrated cations of small, highly charged metal ions, such as Al 3+.
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Acid-Base Properties of Salts
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Examples Classify each of the following salt solution as acidic basic or neutral. Write a hydrolysis equation for each ion. ◦ KBr ◦ NaNO 2 ◦ NH 4 Br
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Example Calculate the pH of a 0.10M solution of sodium fluoride (NaF) at 25 o C. K a = 7.1 x 10 -4
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Examples Calculate K a for the cation, and Kb for the anion in an aqueous NH 4 CN solution. Is the solution acidic, basic or neutral? Write the hydrolysis reaction of the salt (K b for NH 3 = 1.8 x 10 -5, K a for HCN = 4.9 x 10 -10 )
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Example Predict whether 0.35M NH 4 Br solution is acidic, basic or neutral. Calculate its pH. K b = 1.8 x10 -5
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Lewis Acids and Bases Lewis Base: An electron-pair donor. All Lewis bases are Bronsted-Lowry bases Lewis Acid: An electron-pair acceptor. Include cations and neutral molecule having vacant valence orbitals that can accept a share in a pair of electrons from a Lewis Base
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Lewis Acids and Bases
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Examples For each of the following reactions, identify the Lewis acid and the Lewis base ◦ CO 2 (g) + - OH(aq) HCO 3 - (aq) ◦ AlCl 3 (aq) + Cl - (aq) AlCl 4 - (aq)
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