Chapter 16 AcidBase Equilibria 16 1 Acids and

Chapter 16 Acid-Base Equilibria

16. 1 Acids and Bases: A Brief Review

Properties of Acids & Bases Acids: Sour taste p. H less than 7 Litmus paper - red Bases Bitter taste Slippery p. H greater than 7 Litmus paper - blue

Strong Acids & Bases - dissociate completely HCl HBr HI HNO 3 H 2 SO 4 HCl. O 3 HCl. O 4 Li. OH Na. OH KOH Rb. OH Cs. OH Ca(OH)2 Sr(OH)2 Ba(OH)2

Arrhenius Definition of Acid & Base Svante Arrhenius An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. p. 668 GIST: What two ions are central to the Arrhenius definitions of acids and bases?

16. 2 Brønsted-Lowry Acids and Bases

Bronsted-Lowry Definition of Acid & Base Johannes Brønsted & Thomas Lowry An acid is a proton donor. A base is a proton acceptor. p. 670 GIST: In the forward reaction, which substance acts as the Bronsted-Lowry base: HSO 4 -(aq) + NH 3(aq) ↔ SO 42 -(aq) + NH 4+(aq)?

A Brønsted-Lowry acid… …must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

If it can be either… …it is amphiprotic. HCO 3 HSO 4 H 2 O

What Happens When an Acid Dissolves in Water? Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

Conjugate Acids and Bases The term conjugate comes from the Latin word “conjugare, ” meaning “to join together. ” Reactions between acids and bases always yield their conjugate bases and acids.

Example Determine the conjugate acid-base pairs in each reaction: NH 4+(aq) + CN-(aq) ⇄ HCN(aq) + NH 3(aq) A B CB CA (CH 3)3 N(aq) + H 2 O(l) ⇄ (CH 3)3 NH+(aq) + OH-(aq) B A CA CB HCHO 2(aq) + PO 43 -(aq) ⇄ CHO 2 -(aq) + HPO 42 -(aq) A B CB CA

Sample Exercise 16. 1 Identifying Conjugate Acids and Bases (a)What is the conjugate base of each of the following acids: HCl. O 4, H 2 S, PH 4+, HCO 3–? (b) What is the conjugate acid of each of the following bases: CN–, SO 42–, H 2 O, HCO 3– ? Practice Exercise Write the formula for the conjugate acid of each of the following: HSO 3–, F– , PO 43–, CO.

Sample Exercise 16. 2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO 3–) is amphiprotic. (a) Write an equation for the reaction of HSO 3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO 3– with water, in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. Practice Exercise When lithium oxide (Li 2 O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O 2–) with water. Write the reaction that occurs, and identify the conjugate acid–base pairs.

Acid and Base Strength The more easily an acid gives up a proton, the less easily a conjugate base accepts a proton Strong acids are completely dissociated in water. n Their conjugate bases are quite weak. Weak acids only dissociate partially in water. n Their conjugate bases are weak, but not as weak as the conjugate base of a strong acid.

Acid and Base Strength Substances with negligible acidity do not dissociate in water. n n For example, CH 4 contains hydrogen, but does not show any acidic behaviors. Their conjugate bases are exceedingly strong. p. 672 GIST: Specify the strength of HNO 3 and the strength of its conjugate base, NO 3 -.

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l) H 3 O+ (aq) + Cl- (aq) • H 2 O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K is not measured (K>>1).

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. CH 3 CO 2 H (aq) + H 2 O (l) H 3 O+ (aq) + CH 3 CO 2 - (aq) • Acetate ion is a stronger base than H 2 O, so the equilibrium favors the left side (K<1).

Sample Exercise 16. 3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction, use Figure 16. 4 to predict whether the equilibrium lies predominantly to the left (that is, Kc < 1 ) or to the right (Kc > 1): Practice Exercise For each of the following reactions, use Figure 16. 4 to predict whether the equilibrium lies predominantly to the left or to the right:

16. 3 The Autoionization of Water

Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. H 2 O (l) + H 2 O (l) H 3 O+ (aq) + OH- (aq) This is referred to as autoionization.

Ion-Product Constant The equilibrium expression for this process is Kc = [H 3 O+] [OH-] This special equilibrium constant is referred to as the ion-product constant for water, Kw. At 25 C, Kw = 1. 0 10 -14

Sample Exercise 16. 4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH-] in a neutral solution at 25 ºC. Practice Exercise Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] = 4 × 10– 9 M ; (b) [H+] = 4 × 10– 9 M ; (c) [OH–] = 7 × 10– 13 M.

Sample Exercise 16. 5 Calculating [H+] from [OH-] Calculate the concentration of H+(aq) in (a) a solution in which [OH–] is 0. 010 M, (b) a solution in which [OH–] is 1. 8 × 10– 9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 ºC. Practice Exercise Calculate the concentration of OH–(aq) in a solution in which (a) [H+] = 2 × 10– 6 M; (b) [H+] = [OH–]; (c) [H+] = 100× [OH–].

16. 4 The p. H Scale

p. H is defined as the negative base-10 logarithm of the concentration of hydronium ion. p. H = -log [H 3 O+]

p. H In pure water, Kw = [H 3 O+] [OH-] = 1. 0 10 -14 Since in pure water [H 3 O+] = [OH-], [H 3 O+] = 1. 0 10 -14 = 1. 0 10 -7

p. H Therefore, in pure water, p. H = -log (1. 0 10 -7) = 7. 00 An acid has a higher [H 3 O+] than pure water, so its p. H is <7. A base has a lower [H 3 O+] than pure water, so its p. H is >7. p. 676 GIST: What is the significance of p. H = 7? How does the p. H change as OH- is added to the solution?

p. H These are the p. H values for several common substances.

Other “p” Scales The “p” in p. H tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are n n p. OH: -log [OH-] p. Kw: -log Kw

Watch This! Because [H 3 O+] [OH-] = Kw = 1. 0 10 -14, we know that -log [H 3 O+] + -log [OH-] = -log Kw = 14. 00 or, in other words, p. H + p. OH = p. Kw = 14. 00

Sample Exercise 16. 6 Calculating p. H from [H+] Calculate the p. H values for the two solutions described in Sample Exercise 16. 5. Practice Exercise (a) In a sample of lemon juice [H+] is 3. 8 × 10– 4 M. What is the p. H? (b) A commonly available window-cleaning solution has [OH–] = 1. 9 × 10 – 6 M. What is the p. H?

Sample Exercise 16. 7 Calculating [H+] from p. H A sample of freshly pressed apple juice has a p. H of 3. 76. Calculate [H+]. Practice Exercise A solution formed by dissolving an antacid tablet has a p. H of 9. 18. Calculate [H+].

How Do We Measure p. H? For less accurate measurements, one can use n n Litmus paper “Red” paper turns blue above ~p. H = 8 “Blue” paper turns red below ~p. H = 5 Or an indicator.

How Do We Measure p. H? For more accurate measurements, one uses a p. H meter, which measures the voltage in the solution.

16. 4 GIST If the p. OH for a solution is 3. 00, what is the p. H of the solution? Is the solution acidic or basic? If phenolphthalein turns pink when added to a solution, what can we conclude about the p. H of the solution?

16. 5 Strong Acids and Bases

Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HCl. O 3, and HCl. O 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O+] = [acid].

Sample Exercise 16. 8 Calculating the p. H of a Strong Acid What is the p. H of a 0. 040 M solution of HCl. O 4? Practice Exercise An aqueous solution of HNO 3 has a p. H of 2. 34. What is the concentration of the acid?

Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+). Again, these substances dissociate completely in aqueous solution.

Sample Exercise 16. 9 Calculating the p. H of a Strong Base What is the p. H of (a) a 0. 028 M solution of Na. OH, (b) a 0. 0011 M solution of Ca(OH)2? Practice Exercise What is the concentration of a solution of (a) KOH for which the p. H is 11. 89; (b) Ca(OH)2 for which the p. H is 11. 68? p. 681 GIST: The CH 3 - ion is the conjugate base of CH 4, and CH 4 shows no evidence of being an acid in water. What happens when CH 3 - is added to water?

Calculate p. H involving strong acids & bases 0. 425 g of HCl. O 4 in 2. 00 L of solution 10. 0 m. L of 0. 0105 M Ca(OH)2 diluted to 500. 0 m. L A mixture formed by adding 50. 0 m. L of 0. 020 M HCl to 150 m. L of 0. 010 M HI A solution formed by mixing 10. 0 m. L of 0. 015 M Ba(OH)2 with 40. 0 m. L of 7. 5 x 10 -3 M Na. OH

16. 6 Weak Acids

Weak Acids & Dissociation Constants For a generalized acid dissociation, HA (aq) + H 2 O (l) A- (aq) + H 3 O+ (aq) the equilibrium expression would be [H 3 O+] [A-] Kc = [HA] This equilibrium constant is called the acid-dissociation constant, Ka.

Dissociation Constants The greater the value of Ka, the stronger is the acid.

Calculating Ka from the p. H The p. H of a 0. 10 M solution of formic acid, HCOOH, at 25 C is 2. 38. Calculate Ka formic acid at this temperature. We know that [H 3 O+] [HCOO-] Ka = [HCOOH]

Sample Exercise 16. 10 Calculating Ka from the p. H The p. H of a 0. 10 M solution of formic acid, HCOOH, at 25 C is 2. 38. Calculate Ka formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H 3 O+], which is the same as [HCOO-], from the p. H.

Calculating Ka from the p. H = -log [H 3 O+] 2. 38 = -log [H 3 O+] -2. 38 = log [H 3 O+] 10 -2. 38 = 10 log [H 3 O+] = [H 3 O+] 4. 2 10 -3 = [H 3 O+] = [HCOO-]

Calculating Ka from p. H Now we can set up a table… [HCOOH], M [H 3 O+], M [HCOO-], M Initially 0. 10 0 0 Change - 4. 2 10 -3 + 4. 2 10 -3 0. 10 - 4. 2 10 -3 = 0. 0958 = 0. 10 4. 2 10 -3 At Equilibrium

Calculating Ka from p. H = 1. 8 10 -4

Practice Exercise – p. 683 Niacin, one of the B vitamins has the molecular structure shown here. A 0. 020 M solution of niacin has a p. H of 3. 26. n A) What is the acid dissociation constant, Ka, for niacin?

Calculating Percent Ionization = [H 3 O+]eq 100 [HA]initial In this example [H 3 O+]eq = 4. 2 10 -3 M [HCOOH]initial = 0. 10 M 4. 2 10 -3 Percent Ionization = 100 0. 10 = 4. 2%

Sample Exercise 16. 11 A 0. 10 M solution of formic acid (HCOOH) contains 4. 2 x 10 -3 M H+(aq). Calculate the percentage of the acid that is ionized. Practice: A 0. 020 M solution of niacin has a p. H of 3. 26. Calculate the percent ionization of niacin.

Sample Exercise 16. 12 Calculate the p. H of a 0. 20 M solution of HCN (refer to Table 16. 2 or Appendix D for the value of Ka).

Practice Exercise The Ka for niacin (Practice Exercise 16. 10) is 1. 5 × 10 -5. What is the p. H of a 0. 010 M solution of niacin?

Calculating p. H from Ka Calculate the p. H of a 0. 30 M solution of acetic acid, HC 2 H 3 O 2, at 25 C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O+ (aq) + C 2 H 3 O 2 - (aq) Ka for acetic acid at 25 C is 1. 8 10 -5.

Calculating p. H from Ka The equilibrium constant expression is [H 3 O+] [C 2 H 3 O 2 -] Ka = [HC 2 H 3 O 2]

Calculating p. H from Ka We next set up a table… [C 2 H 3 O 2], M [H 3 O+], M [C 2 H 3 O 2 -], M Initially 0. 30 0 0 Change -x +x +x 0. 30 - x 0. 30 x x At Equilibrium We are assuming that x will be very small compared to 0. 30 and can, therefore, be ignored.

Calculating p. H from Ka Now, 2 (x) 1. 8 10 -5 = (0. 30) (1. 8 10 -5) (0. 30) = x 2 5. 4 10 -6 = x 2 2. 3 10 -3 = x �

Calculating p. H from Ka p. H = -log [H 3 O+] p. H = -log (2. 3 10 -3) p. H = 2. 64

The active ingredient in aspirin is acetylsalicylic acid (HC 9 H 7 O 4), a monoprotic acid with a Ka of 3. 3 x 10 -4 at 25 o. C. What is the p. H of a solution obtained by dissolving two extra strength tablets, containing 500 mg of acetylsalicylic acid each, in 250 m. L of water?

Sample Exercise 16. 13 Using Ka to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0. 10 M HF solution, (b) a 0. 010 M HF solution.

Practice Exercise In Practice Exercise 16. 11, we found that the percent ionization of niacin (Ka = 1. 5 × 10 -5) in a 0. 020 M solution is 2. 7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0. 010 M, (b) 1. 0 × 10 -3 M.

Polyprotic Acids… …have more than one acidic proton If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the p. H generally depends only on the first dissociation. p. 688 GIST: What is meant by the symbol Ka 3 for H 3 PO 4?

Sample Exercise 16. 14 Calculating the p. H of a Polyprotic Acid Solution The solubility of CO 2 in pure water at 25 ºC and 0. 1 atm pressure is 0. 0037 M. The common practice is to assume that all of the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3), which is produced by reaction between the CO 2 and H 2 O: What is the p. H of a 0. 0037 M solution of H 2 CO 3?

Practice Exercise (a) Calculate the p. H of a 0. 020 M solution of oxalic acid (H 2 C 2 O 4). (See Table 16. 3 for Ka 1 and Ka 2. ) (b) Calculate the concentration of oxalate ion [C 2 O 4 2–] , in this solution.

16. 7 Weak Bases

Weak Bases react with water to produce hydroxide ion.

Weak Bases The equilibrium constant expression for this reaction is [HB] [OH-] Kb = [B-] where Kb is the base-dissociation constant.

Weak Bases Kb can be used to find [OH-] and, through it, p. H.

Sample Exercise 16. 15 p. H of Basic Solutions What is the p. H of a 0. 15 M solution of NH 3? NH 3 (aq) + H 2 O (l) NH 4+ (aq) + OH- (aq) [NH 4+] [OH-] Kb = = 1. 8 10 -5 [NH 3]

p. H of Basic Solutions Tabulate the data. Initially At Equilibrium [NH 3], M [NH 4+], M [OH-], M 0. 15 - x 0. 15 0 x

p. H of Basic Solutions 2 (x) 1. 8 10 -5 = (0. 15) (1. 8 10 -5) (0. 15) = x 2 2. 7 10 -6 = x 2 1. 6 10 -3 = x

p. H of Basic Solutions Therefore, [OH-] = 1. 6 10 -3 M p. OH = -log (1. 6 10 -3) p. OH = 2. 80 p. H = 14. 00 - 2. 80 p. H = 11. 20

Practice Exercise 16. 15 Which of the following compounds should produce the highest p. H as a 0. 05 M solution: pyradine, methylamine, or nitrous acid?

Sample Exercise 16. 16 Using p. H to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (Na. Cl. O) to enough water to make 2. 00 L of solution has a p. H of 10. 50. Using the information in Equation 16. 37, calculate the number of moles of Na. Cl. O that were added to the water.

Practice Exercise A solution of NH 3 in water has a p. H of 11. 17. What is the molarity of the solution?

16. 8 Relationship Between Ka and Kb

Ka and Kb are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other.

Sample Exercise 16. 17 Calculating Ka or Kb for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, Kb, for the fluoride ion (F–); (b) the acid dissociation constant, Ka, for the ammonium ion (NH 4+).

Practice Exercise (a) Which of the following anions has the largest basedissociation constant: NO 2–, PO 43–, or N 3–? (b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a p. Ka of 4. 90. What is the base dissociation constant for quinoline?

16. 9 Acid-Base Properties of Salt Solutions

Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH- and the conjugate acid: X- (aq) + H 2 O (l) HX (aq) + OH- (aq)

Reactions of Cations with Water Cations with acidic protons (like NH 4+) will lower the p. H of a solution. Most metal cations that are hydrated in solution also lower the p. H of the solution.

Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic.

Effect of Cations and Anions 1. An anion that is the conjugate base of a strong acid will not affect the p. H. 2. An anion that is the conjugate base of a weak acid will increase the p. H. 3. A cation that is the conjugate acid of a weak base will decrease the p. H.

Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the p. H. 5. Other metal ions will cause a decrease in p. H. 6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on p. H depends on the Ka and Kb values.

16. 9 GIST What effect will each of the following ions have on the p. H of a solution: NO 3 - and CO 32 -? Which of the following cations has no effect on the p. H of a solution: K+, Fe 2+, or Al 3+?

Sample Exercise 16. 18 Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: (a) Ba(CH 3 COO)2, (b) NH 4 Cl, (c) CH 3 NH 3 Br, (d) KNO 3, (e) Al(Cl. O 4)3.

Practice Exercise In each of the following, indicate which salt will form the more acidic (or less basic) 0. 010 M solution: n A) Na. NO 3, Fe(NO 3)3 n B) KBr, KBr. O n C) CH 3 NH 3 Cl, Ba. Cl 2 n D) NH 4 NO 2, NH 4 NO 3

Sample Exercise 16. 19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na 2 HPO 4 will form an acidic solution or a basic solution on dissolving in water.

Practice Exercise Predict whether the dipotassium salt of citric acid (K 2 HC 6 H 5 O 7) will form an acidic or basic solution in water (see Table 16. 3 for data).

16. 10 Acid-Base Behavior and Chemical Structure

Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group.

Factors Affecting Acid Strength In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

16. 10 GIST What is the major factor determining the increase in acidity of binary acids going down a column of the periodic table? What is the major factor going across a period? What group of atoms is present in all carboxylic acids?

Sample Exercise 16. 20 Predicting Relative Acidities from Composition and Structure Arrange the compounds in each of the following series in order of increasing acid strength: (a) As. H 3, HI, Na. H, H 2 O; (b) H 2 SO 4, H 2 Se. O 3, H 2 Se. O 4.

Practice Exercise In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution: (a) HBr, HF; (b) PH 3, H 2 S; (c) HNO 2, HNO 3; (d) H 2 SO 3, H 2 Se. O 3.

16. 11 Lewis Acids and Bases

Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

Lewis Bases Lewis bases are defined as electron-pair donors. Anything that could be a Brønsted-Lowry base is a Lewis bases can interact with things other than protons, however.

16. 11 GIST What feature must any molecule or ion have to act as a Lewis base? Which of the following cations will be most acidic and why: Ca 2+, Fe 3+?

Sample Integrated Exercise Putting Concepts Together Phosphorous acid (H 3 PO 3) has the following Lewis structure (a) Explain why H 3 PO 3 is diprotic and not triprotic. (b) A 25. 0 -m. L sample of a solution of H 3 PO 3 is titrated with 0. 102 M Na. OH. It requires 23. 3 m. L of Na. OH to neutralize both acidic protons. What is the molarity of the H 3 PO 3 solution? (c) The original solution from part (b) has a p. H of 1. 59. Calculate the percent ionization and Ka 1 for H 3 PO 3, assuming that Ka 1 >> Ka 2. (d) How does the osmotic pressure of a 0. 050 M solution of HCl compare qualitatively with that of a 0. 050 M solution of H 3 PO 3? Explain.