1 Acids and Bases Chapter 17: 5, 6, 7, 9, 11, 13, 17, 18, 43, 67a-d, 71 Chapter 18: 5-9, 26, 27a-e, 32 Arrhenius Theory of Acids an acid-base reaction involves the reaction of hydrogen ions and hydroxide ions to form water. All bases contain OH -. All acids contain H + : H + (aq) + OH - (aq) H 2 O (l) problem with this theory is that it fails to address the role of the solvent, i.e. H 2 O. Obvious example is NH 3, which does not contain OH -, but is nonetheless a base.
2 Bronsted-Lowry Theory an acid is a proton donor; a base is a proton acceptor. problem with the Arrhenius theory is now taken care of. We now recognize that NH 3 acts as a base because of it s role as a hydrogen atom acceptor in the reaction: K Base Ionization Constant for the previous reaction, we can write the following equilibrium expression, called the base ionization constant. Note that water does not explicitly appear in the equilibrium expression because the reaction is taking place in water + - [NH4 ] [OH ] 5 b [NH3] In the reaction, NH 3 acts as a base (proton acceptor) and H 2 O acts as an acid (proton donator). The conjugate acid of NH 3 is NH 4+. The conjugate base of water in the reaction is OH -. We would refer to NH 4+ /NH 3 as a conjugate acid/base pair.
3 A Weak Acid water can also act as base (proton acceptor), when it reacts with an acid such as acetic acid, to form the hydronium ion, H 3 O + Acetic acid gives up a proton to form the acetate ion. CH 3 CO 2 H/CH 3 CO 2- form a conjugate acid base pair. K [CH CO Acid Ionization Constants for the acid reaction we can write the acid ionization equilibrium constant, K a ][H O a [CH3CO2H] ] the size of K a is a measure of the strength of the acid in water. The larger the value of K a, the further the equilibrium lies to the right, the stronger the acid. For a base, the larger the value of K b, the stronger the base. Regardless of the value of K a or K b, if the equilibrium for the acid or base reactions do not lie completely to the right, we call them weak acids and weak bases
4 A Strong Acid Hydrochloric acid will react in the following way in water: HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) We write a single arrow for the reaction since the reaction is complete. The K a for the reaction is about The large value of K a implies that the equilibrium lies completely to the right. Because the acid dissociates completely, we call this a strong acid. Comparison of Extent of Ionization of Acetic Acid and HCl What does complete dissociation mean? To what extent does HCl dissociate in water, and how does this compare to a weak acid such as acetic acid. Let s calculate the fraction of the acid in the A - form after the reaction: HA (aq) + H 2 O (l) W H 3 O + (aq) + A - (aq) I nitial (mol/l) C hange (mol/l) E quilibrium (mol/l) HA C HA -x C HA -x W H 3 O + 0 +x +x A - 0 +x +x C HA is the initial formal concentration of the acid
5 K - + [A ][H3O ] [HA] a x [C HA Cont d 2 - x] x 2 + K x K C a a HA b ± b 4ac K a ± K a + 4K ac x 2a 2 - [A ] x x fraction of ionized acid - [HA] + [A ] (C - x) + x C CH 3 CO 2 H C HA 0.01M HCl C HA 0.01M K a 1.8 x 10-5 ~ 1 x 10 6 x 4.15 x 10-4 M HA x 10-3 M 4.15% HA HA X/C HA *100% % Relative Strength of Acids and Bases the stronger an acid, the weaker it s conjugate base e.g. HCl dissociates completely in water. It s conjugate base, Cl -, has virtually no tendency to take a proton from H 3 O + (or H 2 O).
6 Nonaqueous Solvents Note that HClO 4, HI, HBr, HCl, H 2 SO 4 and HNO 3 are all strong acids. We cannot tell which acid is stronger because they all dissociate completely in water to yield H 3 O + water is said to have a leveling effect on the acids the strong acids all appear to have the same strength since H 3 O + is the strongest acid that can exist in H 2 O. to differentiate them, we need use a solvent that is a weaker base then water (acetic acid or diethyl ether) HClO 4 dissociates completely in diethylether. It is a stronger acid than HCl. Autoprotolysis of Water Electrical conductivity measurements indicate that even the purest water has a finite electrical conductivity. Electrical conduction in water requires the presence of ions. Finite conductivity remains due to the self ionization of water (autoprotolysis) which stems from it s amphiprotic nature (ability to act both as an acid and a base): H 2 O (l) + H 2 O (l) WH 3 O + (aq) + OH - (aq) K w [H 3 O + ][OH - ] ~ 1.0 x (25 o C) K w is referred to as the ion product of water
7 Pure water What is the [H 3 O + ] in pure water at 25 o C(0 o C,50 o C,100 o C)? 2 H 2 O (l) W H 3 O + (aq) + OH - (aq) 55M -X x x [H 3 O + ] [OH - ] x K w [H 3 O + ] [OH - ] x 2 T ( o C) K w x [H O + 3 ] K w [H 3 O + ] Hydronium ion in solution In solution, the hydronium ion is likely highly solvated. This figure shows a H 3 O + surrounded by 4 water molecules, H 11 O 5+. Other hydrated species, such as H 9 O 4+ are also postulated to exist.
8 ph Soren Sorenson defined ph back in The potential of hydrogen ion, or ph of a solution is defined to be: ph -log [H 3 O + ] (Strictly ph -log a H3O+ ) poh -log [OH - ] (Strictly ph -log a H3O+ ) How are ph and poh related? K w [H 3 O + ][OH - ] 1.0 x log K w -log [H 3 O + ][OH - ] -log 1.0 x pk w -log [H 3 O + ] log [OH - ] 14 ph + poh 14 ph Scale
9 Strong Acids and Bases Strong acids and bases completely dissociate in water. HNO 3(aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) Mg(OH) 2 Mg +2 (aq) + 2 OH - (aq) Calculating the ph, poh, or other quantities is usually trivial in such cases. There are exceptions in dilute solution. Calculations for Strong Acids & Bases What is the ph, poh, [H 3 O + ], [OH - ] and [NO 3- ] for a 0.01 M solution of HNO 3? HNO 3(aq) W H 3 O + NO 3 - (aq) I (mol/l) C (mol/l) E (mol/l) [H 3 O + ] 0.01M; [NO 3- ] 0.01, ph -log [OH - ] K w /[H 3 O + ] 1 x / x M poh 14 ph
10 Dilute Solutions Q: What is the ph of a M solution of HNO 3? A: ph -log [0.001] 3.00 Q: What is the ph of a 1 x 10-5 M solution of HNO 3? A: ph -log [1 x 10-5 ] 5.0 Q: What is the ph of a 1 x 10-7 M solution of HNO 3? A: ph -log [1 x 10-7 ] 7.0?? Can this be correct? Q: What is the ph of a 1 x 10-9 M solution of HNO 3? A: ph -log [1 x 10-9 ] 9.0?? Definitely incorrect!! A dilute solution of an acid cannot be basic. What s the problem? Solution to dilute acid problem Problem: we ignored the autoprotolysis of H 2 O!! HNO 3(aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) 2 H 2 O W H 3 O + (aq) + OH - (aq) There are two source of H 3 O +. Information we can use to solve the problems is as follows: C HNO3 [NO 3- ] 1.0 x 10-9 M K w [H 3 O + ][OH - ] 1.0 x (1) Charge Balance Equation: for a neutral solution the number of moles of positive charge must be equal to the number of moles of negative charge. In the current case: [H 3 O + ] [NO 3- ] + [OH - ] (2)
11 Cont d Mass Balance Equation: there is usually a mass balance that can be written in equilibrium problems. In the current case, we could write: total H 3 O + H 3 O + from nitric acid + moles of H 3 O + from dissociation of water. But since water dissociates 1:1 (1 H + for each OH - ), the amount of H 3 O + from water dissociation must be equal to [OH - ] [H 3 O + ] C HNO3 + [OH - ] (3) OR [H 3 O + ] [NO 3- ] + [OH - ] Note that in this problem, the charge balance (2) and mass balance equations (3) are identical. Solution to problem Substitute (1) into (3): + K w [H3 O ] CHNO3 + [H + 3 O + ] 2 -C HNO3 [H 3 O + ] K w 0 [H3O ] 2 + CHNO3 + CHNO3 + 4K w [H3O ] 2 If C HNO3 1 x 10-9 M, we get [H 3 O + ] x 10-7 M ph -log [1.005 x 10-7 M] ph Exact solutions as a function of C HNO3 are shown on the following graph. ph Dilute Nitric Acid 1E-09 1E-08 1E-07 1E-06 1E-05 1E-04 1E-03 C_HNO3 [M] Approximation Exact
12 Formulas of Weak Acids Example: Chloroacetic acid Empirical Formula: C 2 H 3 O 2 Cl Molecular Formula: C 2 H 3 O 2 Cl Acid Formula: HC 2 H 2 O 2 Cl Condensed Formula: CH 2 ClCOOH OR CH 2 ClCO 2 H Lewis Structure: H O Cl C C H O H K a and K b The following are general formulas for the K a s and K b s of monofunctional weak acids and weak bases + [BH ][OH B + H 2 O BH + + OH - K b [B] ] HA + H 2 O H 3 O + + A - [A ][H O [HA] 3 K a + ] pk a -log K a pk b -log K b The stronger the weak acid, the larger it s K a, and thus the smaller it s pk a. Similar argument for strength of bases
13 K a s of Weak Acids Weak Acid/Weak Base Problems 1. calculate K b or K a for a solution of known concentration if ph is measured 2. calculate ph of a solution when amount of dissociation is minimal 3. calculate ph of a solution when amount of dissociation is significant
14 Finding K a Hypochlorous acid, HOCl, is used as a disinfectant in pools and water treatment. A 0.150M solution of HOCl has a ph Find K a Create an ICE table HOCl + H 2 O H 3 O + + OCl - Initial Change -x + x + x Equilib 0.15-x x x K a [H 3 O + ][OCl - ] [HOCl] But, we should know [H 3 O + ] Cont d ph -log [H 3 O + ] -ph log [H 3 O + ] [H 3 O + ] 10 -ph [H 3 O + ] x 10-5 M x K a [H 3 O + ][OCl - ] x 2 (6.61 x 10-5 M) 2 [HOCl] 0.15-x ( 0.15M x 10-5 M) K a 2.9 x 10-8!
15 ph of Weak Acid Solution Boric acid, B(OH) 3, is used as a mild antiseptic. What is the ph of a M aqueous solution of boric acid? What is the degree of ionization of boric acid in this solution? Although it s molecular formula does not imply this to be an acid, the hydrogen ion arises principally from the reaction B(OH) 3 (aq) + 2H 2 O W B(OH) 4- + H 3 O + K a 5.9 x To solve, create ICE table again. Use HBo as the symbol for boric acid and Bo - as the symbol for B(OH) 4-. Continued HBo H + + Bo - Initial Change -x +x +x Equilibrium x x x [H + ][Bo - ] [HBo] K a [x] x ( x) APPROXIMATION Solve the equation for x, assuming that x is much smaller than 0.025, so that ( x) is approximately
16 Boric acid example - continued x 2 (5.9 x )(0.025) x x 3.84 x 10-6 M [H + ] CHECK ASSUMPTION! ( x) Error ( )/0.025*100% % When this error is less than 5%, assumption is valid. ph -log[h + ] -log (3.84 x 10-6 ) degree of ionization (3.84 x 10-6 )/ % Generalization Go back to previous problem. We had: x 2 [H + ] 2 (5.9 x )(0.025) (K a C HA ) where: [H C HA formal concentration of weak acid, HA (i.e. initial concentration before dissociation) K A acid dissociation constant This approximation only holds if dissociation is insignificant. When would it not be true? + 3O ] K acha
17 failure of approximation 1. when acid is relatively strong; (i.e. large K a ) HA + H 2 O H 3 O + + A - K a 2. when acid is very very dilute ph of weak acid, significant dissociation From a weak acid dissociation ICE table, we get: K + 2 [H3O ] C [H O a + HA 3 If dissociation is not insignificant, we CANNOT assume that [H + ] << C HA. Must solve a quadratic! The above expression becomes: ] [H + ] 2 + K a [H + ] K a C HA 0 where a 1, b K a, c K a C HA K a + K a + 4K acha [H3O ] 2
18 Problem - weak acid, significant dissociation Find the ph of, and find the concentration of all species in, a 0.015M solution of fluoroacetic acid. FCH 2 COOH. K a 2.6 x 10-3 Try quick solution. If [H + ] << C HA, then + 3 [ HO ] C K ( )( 2. 6x10 ) M 3 HA a [H + ]/C HA M /0.015M (i.e. 42%) Approximation is not valid. Use quadratic [H O ] M -3 ( ) ( )(0.015) Problem - continued [H 3 O + ] M ph -log [H + ] 2.29 Other species? [OH - ] K w / [H 3 O + ] 1x10-14 / M 1.97x10-12 M [CH 2 FCOOH] C HA [H 3 O + ] M [CH 2 FCOO - ] [H 3 O + ] M
19 Dissociation of Weak Bases By analogy, create an ICE table for dissociation of a weak base, B. B + H 2 O BH + + OH - K b [BH + ][OH - ] [B] Initial C B 0 0 Change -x +x + x Equilib C B -x x x If [BH + ] << [B] [OH ] K b C B If [BH + ] [B] [OH - K ] b K b 2 4K b C B Polyprotic Acids (and Bases) Acids (or bases) that generate more than one H + (OH - ) upon dissolving in aqueous solution typically undergo a series of acid/base equilibria. Consider H 3 PO 4 : H 3 PO 4 H + + H 2 PO - 4 K a1 7.5 x 10-3 H 2 PO 4 - H + + HPO 4 2- K a2 6.2 x 10-8 HPO 4 2- H + + PO 4 3- K a3 4.8 x Intermediate species can act as both an acid and a base. We say they are amphiprotic. PO H 2 O HPO OH - K b1 2.1 x 10-2 HPO H 2 O H 2 PO 4- + OH - K b2 1.6 x 10-7 H 2 PO 4- + H 2 O H 3 PO 4 + OH - K b3 1.3 x 10-12
20 Other Polyprotic Acids Relationship between an acid and it s conjugate base CH 3 CO 2 H (aq) + H 2 O (l) W H 3 O + (aq) + CH 3 CO 2 - (aq) acetic acid acetate + [CH3CO2 ][H3O ] 5 K a [CH3CO2H] But let us also consider the hydrolysis reaction of acetate, where acetate acts as a base: CH 3 CO 2 - (aq) + H 2 O (l) W OH - (aq) + CH 3 CO 2 H (aq) acetate acetic acid [CH3CO2H][OH ] 10 K b [CH3CO2 ]
21 Acid/Base Relationships-cont d + [CH3CO2 ][H3O ] [CH3CO2H][OH ] K ak b [CH3CO2H] [CH3CO2 ] K ak b [H3O + ] [OH ] K ak b K w pk + pk pk OR pk + pk 14 a b w a b This is a general result, the K a of an acid and the K b of it s conjugate base are related. From this we can write three equivalent statements The higher the K a of an acid, the lower the K b of the base. The lower the pk a of an acid, the higher the pk b of the base. The stronger an acid is, the weaker is it s conjugate base! Salts Solutions of salts are very common in chemistry, biological systems, environmental matrices, etc. We can now predict in a qualitative sense (and in some cases quantitatively) the ph of solutions of acids, bases and salts. Salts of strong acids/strong bases -do not hydrolyze (react with water), ph 7 Example solution of MgBr 2, salt of strong acid + strong base 2HBr (aq) + Mg(OH) 2 (aq) 2 H 2 O (l) + MgBr 2(aq) MgBr 2 (aq) Mg 2+ (aq) + 2 Br - (aq) Mg +2 (aq) + H 2 O?? Br - (aq) + H 2 O?? formation dissolution reaction? reaction?
22 Salt of Strong Acid/Weak Base Salts of strong acids/weak bases -conjugate acid of the weak base will hydrolyze, ph <7 Example solution of NH 4 NO 3, salt of strong acid + weak base HNO 3(aq) + NH 3(aq) NH 4 NO 3(aq) formation NH 4 NO 3(aq) NH + 4 (aq) + NO - 3 (aq) dissolution NO - 3 (aq) + H 2 O?? reaction? NH + 4 (aq) + H 2 O H 3 O + (aq) + NH 3 (aq) reaction! Salt of Weak Acid/Strong Base Salts of weak acids/strong bases -conjugate base of the weak acid will hydrolyze, ph > 7 Example solution of NaF, salt of weak acid + strong base NaOH (aq) + HF (aq) H 2 O (l) +NaF (aq) formation NaF Na + (aq) + F - (aq) dissolution Na + (aq) + H 2 O?? reaction? F - (aq) + H 2 O HF (aq) + OH - (aq) reaction!
23 Salt of Weak Acid/Weak Base Salts of weak acids/weak bases -conjugate base of the weak acid will hydrolyze, as will the conjugate acid of the weak base. One must look at the pk a and pk b to predict the ph of solution. Example solution of C 2 H 5 NH 3 C 7 H 5 O 2, (ethylammonium benzoate), salt of weak acid + weak base C 7 H 5 O 2 H (aq) + C 2 H 5 NH 2(aq) C 2 H 5 NH 3 C 7 H 5 O 2 (aq) form C 2 H 5 NH 3 C 7 H 5 O 2 (aq) C 2 H 5 NH 3 + (aq) +C 7 H 5 O 2 - (aq) dissolution C 2 H 5 NH 3 + (aq) + H 2 O H 3 O + (aq) + C 2 H 5 NH 2 (aq) reaction! C 7 H 5 O 2 - (aq) + H 2 O C 7 H 5 O 2 H (aq) + OH - (aq) reaction! How do we predict which wins out in this competition? Conjugate Acid/Base competition pk a of C 7 H 5 O 2 H 4.20; pk b of C 7 H 5 O pk b of C 2 H 5 NH ; pk a of C 2 H 5 NH Since the pk b of the basic part of the salt (i.e. benzoate) is lower than the pk a of the acidic part of the salt (ethylammonium), benzoate s base strength is stronger than ethylammonium s acid strength. The base should win out and we predict the final solution will be basic, ph>7. (Another way to look at this is to compare the pk a and pk b of the acid and base that form the salt. Since ethylamine is a stronger base than benzoic acid is an acid, then in the formation reaction, the base wins out and the resultant solution of the salt must be basic.)
24 Acid-Base Properties of Salts Summary Cation Anion Acidic or Basic Example neutral neutral neutral NaCl neutral conj base of basic NaF weak acid conj acid of neutral acidic NH 4 Cl weak base conj acid of weak base conj base of weak acid Al 2 (SO 4 ) 3 depends on K a & K b values Structure, Acidity & Basicity If E has low electronegativity, E-O bond breaks to release OH -. It behaves as base! Example: KOH, NaOH Oxides can be considered to be acid or base anhydrides. If E has high electronegativity, strong covalent E-O bond, but O-H bond is weakened. It behaves as acid! Example: O n ClOH, O n SOH
25 Structure related acidity K a 2.9x10-8 Hypochlorous acid K a 1.1x10-2 Chlorous acid K a strong? Chloric acid K a strong! Perchloric acid Very strong oxidizing Agent, explosive! Acetic acid ethanol K a 1.8 x 10-5 K a 1.3 x Acidity is not only related to structure of acid, but also to the stability of anion product. Resonance in acetate ion stabilizes this structure weak acid. For ethoxide ion (other alcohols are similar), the charge is localized on O -, it behaves as relatively strong base. Ethanol is therefore a very weak acid. Structure Related Basicity of Amines Hydrocarbon attachments have little electron withdrawing ability. Alkyl amines are stronger base than NH 3. The effect is illustrated when comparing base strength of cyclohexylamine (nonaromatic) and aniline (aromatic). Lone pair electrons on N are responsible for base behaviour of amines, by binding to H +. Electronegative groups attached to N lower electron density, reducing strength of base. Aromatic amines have additional electron withdrawing effects due to delocalization on aromatic ring. An additional illustration of the effect of an electron withdrawing group on base strength of an aromatic amine.
26 Lewis Acids & Bases Lewis Acid: electron pair acceptor Lewis Base: electron pair donor To identify Lewis acids, look for species (ions, atoms, etc ) that have valence shells that can accept electron pairs. To identify Lewis bases, look for species that have lone pair electrons that can be donated to form a covalent bond. H + Lewis acid.. :NH 3 :OH.. - Lewis bases Al O Lewis Acid Formation of complex ions is an important application of Lewis acid-base theory. Lewis acid Lewis base H H Al O H H 6 3+ Hydrated metal ion in solution