CHAPTER 16 (MOORE) ACIDS, BASES AND ACID-BASE EQUILIBRIA This chapter deals with acids and bases and their equilibria. We will treat acids as proton (H + ) donors in solution and bases as proton acceptors. According to this approach, acid-base reactions are merely proton transfer reactions from A (an acid) to B (a base). We will examine how the structure of an acid influences its strength or ability to give up its proton in aqueous solution. Recall that, in Chapter 5, acids and bases were classified as strong or weak, based upon their degree of ionization. See Moore, pp. 171-179 to review what you should already know about acids and bases as you begin this chapter. Key Concepts: electrolytes strong and weak strong acids and weak acids acid-base reactions Brønsted Lowry Theory Arrhenius theory: an acid forms H + in water; a base forms OH 1 in water. NaOH (+ H 2 O) Na + (aq) + OH 1- (aq) base hydroxide ion HCl (+ H 2 O) H + (aq) + Cl 1- (aq) acid proton But not all acid base reactions involve water, and many bases (NH3, CO32-) don t contain any OH1, so Brønsted Lowry theory defines acids and bases in terms of a proton (H+) donor acceptor model. A Brønsted Lowry acid is a proton donor: HCl(aq) + H 2 O H 3 O + (aq) + Cl 1- (aq) donor acceptor A Brønsted Lowry base is a proton acceptor. NH 3 (aq) + H 2 O NH 4 1+ + OH 1- (aq) acceptor donor Ionization of Acetic Acid in Water HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 1- (aq) Ionization of Ammonia in Water NH 3 (aq) + H 2 O(l) NH 4 1+ (aq) + OH 1- (aq) Ammonia is a weak base in water.
Strength of Conjugate Acid Base Pairs Stronger acids donate H + more readily than a weaker acid. The stronger an acid, the weaker is its conjugate base. The stronger a base, the weaker is its conjugate acid. Conjugate Acids and Conjugate Bases conjugate acid: a base plus an attached H + conjugate base: an acid minus an attached H + HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl 1- (aq) conjugate acid conjugate base What is the conjugate base of HCl? We say that Cl1- is the conjugate base of HCl. HCl and Cl 1- are an example of a conjugate acid-conjugate base pair. Example 16.1. Identify the Brønsted Lowry acids and bases and their conjugates in: (a) H 2 S + NH 3 NH + 4 + HS 1 (b) OH 1 1 2 + H 2 PO 4 H 2 O + HPO 4 Acid Strengths. See Table 15.1 (Hill, p. 620) K a and K b The equilibrium constant for a Brønsted acid (HA) is represented by K a, and that for a Brønsted base (B) is represented by K b. Weak Acid HA(aq) + H 2 O(l) H 3 O + (aq) + A 1 (aq) Weak Base B(aq) + H 2 O(l) BH + (aq) + OH 1 (aq) K a = ([H 3 O + ][A 1- ]) / [HA] K b = ([BH + ][OH 1- ]) / [B] Acid/Base Strength and Direction of Equilibrium Consider the following equilibrium. Since HBr is a stronger acid than CH 3 COOH, the equilibrium for the reaction lies to the left. weaker base stronger base CH 3 COOH + Br 1- HBr + CH 3 COO 1- weaker acid stronger acid equilibrium lies in this direction Strong Acids. Strong acids ionize completely in water. The strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4 and HClO 4. Note that the strong acids all appear above (i.e. are stronger acids than) H 3 O + in the table. Strong acids are all leveled to the same strength that of H 3 O + when they are in water. They are all essentially 100% dissociated (ionized) in water. (HI> HBr> HCl > H 2 SO 4 > HNO 3 > H 3 O + )
Periodic Trends in Acid Strength The greater the tendency for HX (general acid) to transfer a proton to H 2 O, the more the forward reaction is favored and the stronger the acid. Any factor that makes it easier for the H + to leave will increase the strength of the acid. Acid strength is inversely proportional to H X bonddissociation energy. Weaker H X bond means a stronger acid. Periodic Trends in Acid Strength (see pp.790-794, Moore) Strength of Oxoacids Acid strength increases with the electronegativity of the central atom and with the number of terminal oxygen atoms. Strength of Carboxylic Acids Carboxylic acids all have the COOH group in common. Differences in acid strength come from differences in the R group attached to the carboxyl group. In general, the more electronegative atoms there are in the R group, the stronger the acid will be. Example 16.2 Select the stronger acid in each pair: (a) nitrous acid, HNO 2, and nitric acid, HNO 3 [HNO 3 is the stronger acid (more O atoms on N)] (b) Cl 3 CCOOH and BrCH 2 COOH [Cl 3 CCOOH = stronger acid - Cl more electronegative than Br)] Self-Ionization of Water ( autoprotolysis ) Fact: Pure water conducts some electricity, since water self-ionizes: H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH 1- (aq) The equilibrium constant for this process is called the ion product of water (K w ). At 25 C, K w = 1.0 x 10 14 = [H 3 O + ][OH 1 ] This equilibrium applies to all aqueous solutions acids, bases, salts, and nonelectrolytes and not just to pure water. The ph Scale Concentration of H 3 O + can vary over a wide range in aqueous solution, from about 10 M to about 1 x 10 14 M. A more convenient expression for H 3 O + is ph. ph = log [H 3 O + ] and [H 3 O + ] = 10 ph p-functions: p( ) means take the log( ) poh = log [OH 1 ] and [OH 1 ] = 10 poh pk w = log K w = log (1.0 x 10-14 ) = 14.00 Recall that K w = 1.0 x 10 14 = [H 3 O + ][OH 1 ] at 25 C [H 3 O + ][OH 1 ] = 1.0 x 10 14
Taking log of each side: log { [H 3 O + ][OH 1 ] } = log (1.0 x 10 14 ) log [H 3 O + ] log [OH 1 ] = 14.00 = pk w ph = log [H 3 O + ] poh = log [OH 1 ] pk w = ph + poh = 14.00 The ph Scale (see Figure 16.2, Moore, p. 782) Example 16.3 A Conceptual Example Is the solution 1.0 x 10 8 M HCl acidic, basic, or neutral? ph = - log(1.0 x 10-8 ) = 8.00 Basic! (but only slightly basic) Equilibria in Solutions of Weak Acids and Weak Bases 1. Write the equation for the reversible reaction. 2. Set up the ICE table. 3. Find the changes that occur at equilibrium 4. Use simplifying assumptions ( 5% rule ). 5. Calculate equilibrium concentrations / constant. Weak Acid and Weak Base Ionization Constants. See Table 16.2 (Moore, p.788) Example 16.4 Ordinary vinegar is approximately 1 M CH 3 COOH and as shown in Figure 16.2, it has a ph of about 2.4. Calculate the expected ph of 1.00 M CH 3 COOH(aq), and show that the calculated and measured ph values are in good agreement. CH 3 COOH(aq) + H 2 O H 3 O + (aq) + CH 3 COO 1- (aq) K a = 1.8 x 10-5 = [H 3 O + ][CH 3 COO 1- ] / [CH 3 COOH] Calculate the expected ph of 1.00 M CH 3 COOH(aq): [CH 3 COOH] [H 3 O + ] [CH 3 COO 1- ] I 1.00 0 0 C x + x + x E 1.00 x x x Substituting equilibrium values into the K a expression, K a = 1.8 x 10-5 = [H 3 O + ][CH 3 COO 1- ] / [CH 3 COOH] K a = 1.8 x 10-5 = [(x)(x) / (1.00 x), so, x 2 / (1.00 x) = 1.8 x 10-5 Assume that x << 1.00 1.00 x 1.00, so, (x 2 / 1.00) 1.8 x 10-5 x 2 1.8 x 10-5 and [H 3 O + ] = x (1.8 x 10-5 ) 1/2 = 4.2 x 10-3 M
Check assumption: (1.00-4.2 x 10-3 = 0.9958 = 1.00 (so assumption is valid) ph = log (4.2 x 10-3 ) = 2.38 (very close to 2.4) Example 16.5 What is the ph of 0.00200 M ClCH 2 COOH(aq)? K a = 1.4 x 10-3 K a = 1.4 x 10-3 = [H3O + ][ClCH 2 COO 1- ] / [ClCH 2 COOH] Now set up the ICE table: [ClCH 2 COOH] [H 3 O + ] [ClCH 2 COO 1- ] I 0.00200 0 0 C x + x + x E 0.00200 x x x Substituting equilibrium values into the K a expression, K a = 1.4 x 10-3 = [H 3 O + ][ClCH 2 COO 1- ] / [ClCH 2 COOH] = (x)(x) / (0.0200 x) so, x 2 / (0.0200 x) = 1.4 x 10-3 or, x 2 = (0.0200 x)(1.4 x 10-3 ) = 2.8 x 10-6 1.4 x 10-3 x Collecting terms on the left, x 2 + 1.4 x 10-3 x 2.8 x 10-6 = 0 ax 2 + bx + c = 0 (quadratic expression) a = 1; b = 1.4 x 10-3 ; c = 2.8 x 10-6 Quadratic Solution: x = [( b ± SQRT(b 2 4ac)] /2a Roots: x(1) = 1.1 x 10-3 and x(2) = 0.0025 (negative concentration not possible) [H 3 O + ] = x = 1.1 x 10-3 M ph = log(1.1 x 10-3 ) = 2.96 Example 16.6 What is the ph of 0.500 M NH 3 (aq)? Example 16.7 The ph of a 0.164 M aqueous solution of dimethylamine is 11.98. What are the values of K b and pk b? The ionization equation is (CH 3 ) 2 NH + H 2 O (CH 3 )2NH 2 + + OH 1 K b =? dimethylamine dimethylammonium ion Polyprotic Acids monoprotic acid: one ionizable H + per molecule polyprotic acid: more than one ionizable H + per molecule Examples: sulfuric acid, H 2 SO 4 carbonic acid, H 2 CO 3 phosphoric acid, H 3 PO 4 diprotic acid diprotic acid triprotic acid The protons of a polyprotic acid dissociate in steps. Each step has its own K a. K a1 > K a2 > K a3, etc. for steps (1), (2) and (3) and so on.
Polyprotic Acids (see Appendix p. A31) Example of stepwise ionization for a diprotic acid (1) H 2 CO 3 + H 2 O H 3 O + 1- + HCO 3 K a1 = 4.4 x 10-7 1- (2) HCO 3 + H 2 O H 3 O + 2- + CO 3 K a2 = 4.7 x 10-11 H 2 CO 3 + 2 H 2 O 2 H 3 O + 2- + CO 3 (2-step, overall ionization) K overall = K a1 x K a2 = (4.4 x 10-7 )(4.7 x 10-11 ) = 2.1 x 10-17 Example 16.8 What is the approximate ph of 0.71 M H 2 SO 4? Ions as Acids and Bases HCl is a strong acid: therefore Cl is so weakly basic in water that a solution of chloride ions (e.g. in NaCl (aq)) is virtually neutral. Acetic acid, HC 2 H 3 O 2, is a weak acid, so C 2 H 3 O 2 1-, is significantly basic in water. A solution of sodium acetate (NaC 2 H 3 O 2 ) dissociates completely (into sodium and acetate ions in water) is therefore slightly basic: C 2 H 3 O 2 1-, + H 2 O HC 2 H 3 O 2 + OH 1 1. Salts of strong acids and strong bases form neutral solutions: NaCl, KNO 3 2. Salts of weak acids and strong bases form basic solutions: KNO 2, NaClO 3. Salts of strong acids and weak bases form acidic solutions: NH 4 NO 3 4. Salts of weak acids and weak bases form solutions that may be acidic, neutral, or basic; it depends on the relative strengths of the cations and the anions: NH 4 NO 2, CH 3 COONH 4. Example 16.9 A Conceptual Example: Is NH 4 I(aq) acidic, basic, or neutral? HI(aq) + NH 3 (aq) NH 4 I(aq) strong weak acid base Recall: Salts of strong acids and weak bases form acidic solutions, e.g. NH 4 NO 3 Thus, we predict that this solution should be ACIDIC!! In order to make quantitative predictions of ph for a salt solution, we need an equilibrium constant for the hydrolysis reaction. The relationship between K a and K b of a conjugate acid base pair is: K a x K b = K w or, K b = K w / K a If, instead, we have values of pk a or pk b : pk a + pk b = pkw = 14.00 Example 16.10 Calculate the ph of a solution that is 0.25 M NaC 2 H 3 O 2 aq). Example 16.11 What is the molarity of an NH 4 NO 3 (aq) solution that has a ph = 4.80? The Common Ion Effect Consider a solution of a weak monoprotic acid, HA, like acetic acid below: CH 3 COOH (aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO 1- (aq)
If we add a sodium salt containing the conjugate base (A 1- ) ion as a second solute (i.e., NaA), the ph of the solution increases. Let s see why: When a salt supplies A 1-, the equilibrium shifts to the left. Then apply LeChâtelier s principle: what happens to [H 3 O + ] when the equilibrium shifts to the left? (it decreases!) common ion effect: the inhibition of the ionization of a weak acid or a weak base by the presence of a common ion from a strong electrolyte The equilibrium is shifted back to the left this is just another case of LeChatelier s principle! Buffer Solutions Many biochemical reactions especially enzyme-catalyzed reactions are sensitive to ph. For working with such reactions we often need a solution that maintains a nearly-constant ph, i.e. a ph stat. We use a buffer solution! a solution that changes ph only slightly when small amounts of a strong acid or a strong base are added. Buffers resist changes in ph when we add small amounts of strong acid or strong base. A buffer contains significant concentrations (like 0.1 or even 1.0 M) of both a weak acid and its conjugate base (an weak acid buffer), or of both a weak base and its conjugate acid (a weak base buffer). Weak acid + conjugate base: HA and A 1- or, Weak base + conjugate acid: B and BH + The acid component of the buffer neutralizes small added amounts of OH 1, forming the weaker conjugate base, which does not affect ph much: HA + OH 1 H 2 O + A 1 Pure water does not buffer at all More Buffer Solutions HA(aq) + H 2 O H 3 O + (aq) + A 1 (aq) K a = [H 3 O + ] x {[A 1-] / [HA]} pka = ph log {[A 1- ] / [HA]} Example. Consider 1000 A 1- and 1000 HA in a solution: Initial: {[A 1- ] / [HA]} = 1000/1000 = 1.000 Final: {[A 1- ] / [HA]} = 1010/990 = 1.020 Only a 2.0% change in the ratio Comparison with a Buffer and an Unbuffered Solution. Pure water increases in ph by about 5 ph units when the OH is added, and decreases by about 5 ph units when the H 3 O + is added. In contrast, the same amounts of OH and H 3 O+ added to a buffer solution barely change the ph. An Equation for Buffer Solutions. The Henderson Hasselbalch equation ph = pka + log {[A 1-] / [HA] } In order to use this equation, the following must be true: (1) 0.10 < [A 1- ]/[HA] < 10 and (2) [conjugate base] / Ka >100 and [weak acid] / Ka > 100
Example 16.12 Calculate the ph of an aqueous solution that is both 1.00 M HC 2 H 3 O 2 and 1.00 M NaC 2 H 3 O 2. This is an acetic acid-acetate buffer, so the equilibrium we need is HC 2 H 3 O 2 (aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 1- (aq), where K a = {[H 3 O + ][C 2 H 3 O 2 1- ]}/[HC 2 H 3 O 2 ] = 1.8 x 10-5 For acid and conjugate base concentrations > 0.1 M, we can neglect x, so let x = [H 3 O + ]: K a = {(x)(1.00)} / (1.00) = 1.8 x 10-5 x = 1.8 x 10-5 M = [H 3 O + ] ph = - log(1.8 x 10-5 ) = 4.74 Alternate solution using Henderson-Hasselbalch: ph = pk a + log{[a 1- ]/[HA]} ph = - log(1.8 x 10-5 ) + log{(1.00)/(1.00)} = 4.74 + 0 = 4.74 Example 16.13 What concentration of acetate ion in 0.500 M HC 2 H 3 O 2 (aq) produces a buffer solution with ph = 5.00? (ph 5.00 [H 3 O + ] = 1.00 x 10-5 M HC 2 H 3 O 2 (aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 1- (aq) 1.00 x 10-5 M x 0.500 M K a = {[H 3 O + ][C 2 H 3 O 2 1- ]}/[HC 2 H 3 O 2 ] = 1.8 x 10-5 x = [(1.8 x 10-5 )(0.500)]/(1.00 x 10-5 ) = 0.900 M Buffer Capacity and Buffer Range Buffer - most effective if the [A 1- ] and [HA] are equal or nearly equal. This is when ph pk a of HA in the buffer. buffer capacity (β): the number of mol of strong acid or strong base that must be added to change the ph of 1000 mol of a buffer 1.00 ph unit Acid Base Indicators acid base indicator: a weak organic acid or base the acid form (HA) of the indicator has one color, the conjugate base (A ) form has a different color. Note: one of the forms may be colorless. In acid solutions, [H 3 O + ] is large. Because H 3 O + is a common ion, it suppresses the ionization of the indicator acid, shifting the equilibrium to the left, and we see the color of HA. HA + H 2 O H 3 O + + A Different indicators have different values of K a and different color transition ranges (different values of ph across which they change) Titration Curve, Strong Acid with Strong Base A plot of ph vs. ml titrant added, e.g. from a buret. When the titrant is NaOH, the ph increases since we re adding a base
Neutralization Reactions: terminology titration curve: a graph of ph versus volume of titrant a neutralization titration equivalence point (in an acid base titration): precise stoichiometric proportions of acid and base endpoint: volume of added titrant at the point at which the indicator changes color Ideally, the indicator is selected so that V(endpoint) = V(equivalence point) Strong acid titrated with strong base: ph = 7.00 at equivalence point HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq)