CHAPTER 16 Acids and Bases Acids and Bases

CHAPTER 16 Acids and Bases

Acids and Bases (Arrhenius) There are several definitions we can use for acids and bases: Arrhenius definition acid - A substance which, when added to water, forms H+ ion base - A substance which, when added to water, forms OH- ion Examples: HCl(aq) H+(aq) + Cl-(aq) HF(aq) H+(aq) + F-(aq) Na. OH(s) Na+(aq) + OH-(aq) NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) The advantages of the Arrhenius definition are that it is simple and easy to implement. The disadvantages are that it is tied in to a particular solvent (water) and is not a general definition.

Acids and Bases (Bronsted-Lowry) Bronsted-Lowry definition acid - a proton (H+) donor; forms a conjugate base - a proton (H+) acceptor; forms a conjugate acid H+ HF(aq) + H 2 O( ) H 3 O+(aq) + F-(aq) acid base conjugate acid conjugate base In the Bronsted theory, in an acid-base reaction an acid donates a proton to form a conjugate base, while a base accepts a proton to form a conjugate acid. In addition, in Bronsted theory acids form hydronium ion (H 3 O+ ion) instead of hydrogen ion (H+ ion) when added to water.

Examples of Acids HCl(aq) + H 2 O( ) H 3 O+(aq) + Cl-(aq) acid: HCl base: H 2 O conjugate base: Clconjugate acid: H 3 O+ CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) acid: CH 3 COOH base: H 2 O conjugate base: CH 3 COOconjugate acid: H 3 O+ The second reaction goes in both directions, so has an equilibrium constant, the acid ionization constant KC = Ka = [H 3 O+] [CH 3 COO-] [CH 3 COOH]

Examples of Bases Na. OH(s) Na+(aq) + OH-(aq) Considered an ionization reaction in Bronsted theory, not an acidbase reaction. NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) base: NH 3 conjugate acid: NH 4+ acid: conjugate base: OH- H 2 O The reaction goes in both directions, so has an equilibrium constant, the base ionization constant KC = Kb = [NH 4+] [OH-] [NH 3]

The reaction of a Bronsted acid with a Bronsted base always involves the transfer of a proton (H+) from the donor (Bronsted acid) to the acceptor (Bronsted base). The substance that forms when the Bronsted acid donates a proton is the conjugate base of that acid, and the substance that forms when the Bronsted base accepts a proton is the conjugate acid of that base. So HF is a Bronsted acid; F- is the conjugate base of HF H 2 O is a Bronsted base; H 3 O+ is the conjugate acid of H 2 O

Amphoteric Substances Some substances can act as either Bronsted acids or Bronsted bases. Substances that can act as either acids or bases depending on what they are reacting with are called amphoteric. For example: H 2 O (water) (acid) NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) (base) CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) HCO 3 - (hydrogen carbonate ion) (acid) NH 3(aq) + HCO 3 -(aq) NH 4+(aq) + CO 32 -(aq) (base) CH 3 COOH(aq) + HCO 3 -(aq) H 2 CO 3(aq) + CH 3 COO-(aq)

Given any substance, the conjugate base for the substance is formed by removing a proton (H+), and the conjugate acid is formed by adding a proton (H+). Example: H 2 O The conjugate base of H 2 O is OHThe conjugate acid of H 2 O is H 3 O+ Example: HSO 4 The conjugate base of HSO 4 - is SO 42 The conjugate acid of HSO 4 - is H 2 SO 4

Example Chlorous acid (HCl. O 2) is a weak acid. Pyridine (C 5 H 5 N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory.

Example Chlorous acid (HCl. O 2) is a weak acid. Pyridine (C 5 H 5 N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory. HCl. O 2(aq) + H 2 O( ) H 3 O+(aq) + Cl. O 2 -(aq) C 5 H 5 N(aq) + H 2 O( ) C 5 H 5 NH+(aq) + OH-(aq)

Strong and Weak Acids There are seven common strong acids: Binary strong acids Ternary strong acids (oxyacids) HCl. O 3 HNO 3 HBr HCl. O 4 H 2 SO 4 (1 st proton) HI Sulfuric acid is special in that it is a strong acid with respect to the first proton and a weak acid with respect to the second proton. H 2 SO 4(aq) + H 2 O( ) HSO 4 -(aq) + H 3 O+(aq) (1 st proton) HSO 4 -(aq) + H 2 O( ) SO 42 -(aq) + H 3 O+(aq) (2 nd proton) Ka 1 = [HSO 4 -] [H 3 O+] = “large” [H 2 SO 4] Ka 2 = [SO 42 -] [H 3 O+] = 1. 2 x 10 -2 [HSO 4 -]

All other acids are weak acids. HNO 2(aq) + H 2 O( ) H 3 O+(aq) + NO 2 -(aq) Ka = [H 3 O+] [NO 2 -] = 4. 5 x 10 -4 [HNO 2] CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) Ka = [H 3 O+] [CH 3 COO-] = 1. 8 x 10 -5 [CH 3 COOH] The larger the value of Ka, the stronger the acid. Therefore, HNO 2 is a stronger acid than CH 3 COOH.

Strong and Weak Bases There are eight common strong soluble bases: Group 1 A strong bases Group 2 A strong bases Li. OH Ca(OH)2 Na. OH Sr(OH)2 KOH Ba(OH)2 Rb. OH Cs. OH All other metal hydroxides are insoluble bases. They do not dissolve in water to an appreciable extent, but react as bases in acid-base reactions. Examples: Cu(OH)2, Al(OH)3, Pb(OH)2. Cu(OH)2(s) + 2 HCl(aq) Cu 2+(aq) + 2 Cl-(aq) + 2 H 2 O( )

Weak bases establish an equilibrium in water. NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) Kb = [NH 4+] [OH-] = 1. 8 x 10 -5 [NH 3] N 2 H 4(aq) + H 2 O( ) N 2 H 5+(aq) + OH-(aq) Kb = [N 2 H 5+] [OH-] = 8. 9 x 10 -7 [N 2 H 4] Since Kb is larger for NH 3 than for N 2 H 4, NH 3 is a stronger base than N 2 H 4.

Autoionization of Water In pure water there will be a small number of H 3 O+ and OH- ions. This is due to the autoionization reaction: H 2 O( ) + H 2 O( ) H 3 O+(aq) + OH-(aq) Kw = [H 3 O+] [OH-] = 1. 0 x 10 -14 (at T = 25 C) Because the above reaction is endothermic ( Hrxn = + 55. 8 k. J/mol) the value for Kw increases as T increases. For example, at physiological temperatures (T = 37 C) Kw = 2. 4 x 10 -14.

Equilibrium in Pure Water Since H 2 O( ) + H 2 O( ) H 3 O+(aq) + OH-(aq) Kw = [H 3 O+] [OH-] = 1. 0 x 10 -14 (at T = 25 C) Substance Initial Change Equilibrium H 3 O+ 0. 0 x x OH- 0. 0 x x So (x) = x 2 = 1. 0 x 10 -14 x = 1. 0 x 10 -7 So in pure water at equilibrium at T = 25 C, [H 3 O+] = [OH-] = 1. 0 x 10 -7 M.

Acidic, Basic, and Neutral Solutions We can use the above results to define what we mean by an acidic, basic, and neutral solution, using [H 3 O+] [OH-] = 1. 0 x 10 -14. acidic solution [H 3 O+] > [OH-] [H 3 O+] > 1. 0 x 10 -7 M [OH-] < 1. 0 x 10 -7 M neutral solution [H 3 O+] = [OH-] [H 3 O+] = 1. 0 x 10 -7 M [OH-] = 1. 0 x 10 -7 M basic solution [H 3 O+] < [OH-] [H 3 O+] < 1. 0 x 10 -7 M [OH-] > 1. 0 x 10 -7 M

p. H represents a convenient way of representing the concentration of hydronium ion in an aqueous solution. p. H is defined as follows: p. H = - log 10[H+] = - log 10[H 3 O+] For a neutral solution at T = 25 C p. H = - log 10[H 3 O+] = - log 10(1. 0 x 10 -7) = 7. 00 Note that the number of digits to the right of the decimal point is equal to the number of significant figures in the H 3 O+ concentration. For acidic and basic solutions acidic solution [H 3 O+] > 1. 0 x 10 -7 M means p. H < 7. 00 basic solution [H 3 O+] < 1. 0 x 10 -7 M means p. H > 7. 00 The further away the p. H is from 7. 00 the more acidic (if less than 7. 00) or basic (if greater than 7. 00) the solution.

p. OH and p. K We can use the “p” notation as a general symbol to indicate that we are taking - log 10 of something. In particular p. H = - log 10[H 3 O+] p. OH = - log 10[OH-] p. K = - log 10 K By reversing the above definitions we get the following relationships [H 3 O+] = 10 -p. H [OH-] = 10 -p. OH K = 10 -p. K

Relationship Between p. H and p. OH There is a simple relationship between p. H and p. OH. [H 3 O+] [OH-] = 1. 0 x 10 -14 -log 10[H 3 O+] + ( - log 10[OH-]) = - log 10(1. 0 x 10 -14) p. H + p. OH = 14. 00 (at 25 C) The more general relationship, true at all temperatures, is p. H + p. OH = p. Kw

Relationship Between p. H and Concentration If we know the concentration of hydronium ions in solution we can find the p. H of the solution (and vice versa). We can also find p. OH and the concentration of hydroxide ions in solution. Example: A particular solution has p. H = 4. 62 at T = 25. C. What are the concentrations of H 3 O+ and OH- in the solution?

Relationship Between p. H and Concentration Example: A particular solution has p. H = 4. 62 at T = 25. C. What are the concentrations of H 3 O+ and OH- in the solution? [H 3 O+] = 10 -p. H = 10 -4. 62 = 2. 4 x 10 -5 M [H 3 O+] [OH-] = 1. 0 x 10 -14, so [OH-] = 1. 0 x 10 -14 = 4. 2 x 10 -10 M [H 3 O+] 2. 4 x 10 -5 We could also find the OH- concentration as follows: p. H + p. OH = 14. 00 - p. H = 14. 00 - 4. 62 = 9. 38 [OH-] = 10 -p. OH = 10 -9. 38 = 4. 2 x 10 -10 M

p. H for Solutions of Strong Acids or Strong Bases Because strong acids and strong soluble bases are strong electrolytes, and so completely dissociate in solution, finding the value for p. H is relatively simple. We may use the following procedure: 1) Use the information in the problem to find the concentration of H 3 O+ (strong acid) or OH- (strong base). Assuming the concentration is larger than ~ 10 -6 mol/L, then: 2) Find the p. H a) For strong acids, find the p. H directly b) For strong bases, find the p. OH, then use (at T = 25 C) p. H + p. OH = 14. 00 ; p. H = 14. 00 - p. OH to find the p. H

Example: Find the p. H for the following solutions, at T = 25 C a) A 3. 5 x 10 -3 M solution of HBr, a strong acid b) A solution formed by dissolving 0. 200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500. 0 m. L.

a) A 3. 5 x 10 -3 M solution of HBr, a strong acid Reaction is HBr(aq) + H 2 O( ) H 3 O+(aq) + Br-(aq) [H 3 O+] = 3. 5 x 10 -3 mol HBr L soln p. H = - log 10(3. 5 x 10 -3) = 2. 46 1 mol H 3 O+ = 3. 5 x 10 -3 M 1 mol HBr

b) A solution formed by dissolving 0. 200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500. 0 m. L. Reaction is Sr(OH)2(s) Sr 2+(aq) + 2 OH-(aq) [OH-] = 0. 200 mol Sr(OH)2 2 mol OH- = 0. 800 M 0. 5000 L soln 1 mol Sr(OH)2 p. OH = - log 10(0. 800) = 0. 097 p. H = 14. 00 - p. OH = 14. 00 - 0. 097 = 13. 90

Weak Acids or Weak Bases For problems involving solutions containing a single weak acid or weak base we proceed as we do other equilibrium problems. 1) Start with the following information Balanced chemical reaction Initial conditions Value for Ka (weak acid) or Kb (weak base) 2) Set up the problem using the “ICE” method 3) Find [H 3 O+] (weak acid) or [OH-] (weak base) 4) Find p. H (for weak base, first find p. OH, then use p. H + p. OH = 14. 00 to find p. H)

Example: Find the p. H of a 0. 100 M solution of HNO 2, a weak acid (Ka = 4. 5 x 10 -4), at T = 25 C

Find the p. H of a 0. 100 M solution of HNO 2, a weak acid (Ka = 4. 5 x 10 -4), at T = 25 C HNO 2(aq) + H 2 O( ) H 3 O+(aq) + NO 2 -(aq) Ka = [H 3 O+] [NO 2 -] = 4. 5 x 10 -4 [HNO 2] Substance Initial Change Equilibrium H 3 O+ 0. 00 x x NO 2 - 0. 00 x x HNO 2 0. 100 -x 0. 100 - x (x) (0. 100 - x) = 4. 5 x 10 -4 There are two ways to proceed. . .

1) Assume x is small (in this case, x << 0. 100) (x) (0. 100 - x) x 2 = 4. 5 x 10 -4 0. 100 so x 2 = (0. 100)(4. 5 x 10 -4) = 4. 5 x 10 -5 x = (4. 5 x 10 -5)1/2 = 6. 7 x 10 -3 Is 6. 7 x 10 -3 << 0. 100? Yes (at least 10 times smaller). [H 3 O+] = 6. 7 x 10 -3 M ; p. H = - log 10(6. 7 x 10 -3) = 2. 17 2) Solve the quadratic x 2 = (0. 100 - x)(4. 5 x 10 -4) = (4. 5 x 10 -5) - (4. 5 x 10 -4) x x 2 + (4. 5 x 10 -4) x - (4. 5 x 10 -5) = 0 x = 6. 5 x 10 -3 ; - 6. 9 x 10 -3 [H 3 O+] = 6. 5 x 10 -3 M ; p. H = - log 10(6. 5 x 10 -3) = 2. 19

Percent Dissociation The percent dissociation of a weak acid is defined as % dissociation = amount dissociated x 100% initial amount For strong acids the percent dissociation is ~ 100%. For weak acids we can use the equilibrium concentrations to find the percent dissociation. For the above example, initial HNO 2 = 0. 100 M amount dissociated = 6. 7 x 10 -3 M % dissociation = 6. 7 x 10 -3 x 100% = 6. 7 % 0. 100

p. H Calculations involving Weak Bases Just as we can find concentrations and p. H values for solutions of weak acids, we can do the same for solutions of weak bases. Example: Find the p. H of a 0. 100 M solution of NH 3, a weak base (Kb = 1. 8 x 10 -5), at T = 25 C

Find the p. H of a 0. 100 M solution of NH 3, a weak base (Kb = 1. 8 x 10 -5), at T = 25 C NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) Kb = [NH 4+] [OH-] = 1. 8 x 10 -5 [NH 3] Substance Initial Change Equilibrium NH 4+ 0. 00 x x OH- 0. 00 x x NH 3 0. 100 -x 0. 100 - x (x) (0. 100 - x) = 1. 8 x 10 -5

Assume x is small (in this case, x << 0. 100) (x) x 2 = 1. 8 x 10 -5 (0. 100 - x) 0. 100 so x 2 = (0. 100)(1. 8 x 10 -5) = 1. 8 x 10 -6 x = (1. 8 x 10 -6)1/2 = 1. 3 x 10 -3 Is 1. 3 x 10 -3 << 0. 100? Yes (at least ten times smaller) [OH-] = 1. 3 x 10 -3 M ; p. OH = - log 10(1. 3 x 10 -3) = 2. 87 p. H = 14. 00 - 2. 87 = 11. 13 If we solve using the quadratic formula, we get p. H = 11. 12

Polyprotic Acid A polyprotic acid has two or more ionizable protons that can be donated in an acid-base reaction. Monoprotic (one ionizable proton) HCl, HNO 2, CH 3 COOH Diprotic (two ionizable protons) H 2 CO 3, H 2 SO 4 Triprotic (three ionizable protons) H 3 PO 3 For polyprotic acids one can talk about the acid dissociation constant for each proton.

Example: H 2 CO 3 1 st proton H 2 CO 3(aq) + H 2 O( ) H 3 O+(aq) + HCO 3 -(aq) Ka 1 = [H 3 O+][HCO 3 -] = 4. 3 x 10 -7 [H 2 CO 3] 2 nd proton HCO 3 -(aq) + H 2 O( ) H 3 O+(aq) + CO 32 -(aq) Ka 2 = [H 3 O+][CO 32 -] = 5. 6 x 10 -11 [HCO 3 -] For a polyprotic acid Ka 1 > Ka 2 > Ka 3. . .

Calculations Involving Polyprotic Acids It would seem like calculations with polyprotic acids should be complicated, since there are several sources of H 3 O+ ions. However, there is usually a big enough difference in the values of the Kas (acid dissociation constants) that only the first dissociation needs to be considered for polyprotic acid solutions. Procedure (diprotic acid): 1) Calculate equilibrium concentrations using the first ionization constant. 2) Calculate equilibrium concentrations using the second ionization constant, using the results from the first calculation for the initial conditions. 3) If there any significant changes in concentrations involving the first ionization, go back and recalculate concentrations using the previous results for the initial conditions.

Example: 0. 100 M H 2 CO 3 solution 1 st proton H 2 CO 3(aq) + H 2 O( ) H 3 O+(aq) + HCO 3 -(aq) Ka 1 = [H 3 O+][HCO 3 -] = 4. 3 x 10 -7 [H 2 CO 3] Initial Change H 3 O+ 0 x HCO 30 x H 2 CO 3 0. 100 -x Solving using this ICE table gives x = 2. 1 x 10 -4. Equilibrium x x 0. 100 - x

Example: 0. 100 M H 2 CO 3 solution 2 nd proton HCO 3 -(aq) + H 2 O( ) H 3 O+(aq) + CO 32 -(aq) Ka 2 = [H 3 O+][CO 32 -] = 5. 6 x 10 -11 [HCO 3 -] Initial Change H 3 O+ 2. 1 x 10 -4 y CO 320 y HCO 32. 1 x 10 -4 -y Equilibrium 2. 1 x 10 -4 + y y 2. 1 x 10 -4 - y Solving using this ICE table gives y = 5. 6 x 10 -11. Since y is so small compared to 2. 1 x 10 -4, the second equilibrium had little effect on the first equilibrium, particularly on the H 3 O+ concentration.

Relationship Between Ka and Kb We may find a general relationship between the values for Ka and Kb for a Bronsted acid/conjugate base pair of substances. We proceed as follows: Let HA be a weak monoprotic acid. A- is the conjugate base. HA(aq) + H 2 O( ) H 3 O+(aq) + A-(aq) Ka = [H 3 O+] [A-] [HA] A-(aq) + H 2 O( ) HA(aq) + OH-(aq) Kb = [HA] [OH-] [A-] Ka Kb = [H 3 O+] [A-] [HA] [OH-] = [H 3 O+] [OH-] = Kw [HA] [A-] Ka Kb = K w p. Ka + p. Kb = p. Kw Ka Kb = 1. 0 x 10 -14 p. Ka + p. Kb = 14. 00, at T = 25. C

Example: The value for the acid ionization constant for acetic acid (CH 3 COOH) is Ka = 1. 8 x 10 -5 at T = 25. C. What is the value for Kb for the acetate ion (CH 3 COO-)?

Example: The value for the acid ionization constant for acetic acid (CH 3 COOH) is Ka = 1. 8 x 10 -5 at T = 25. C. What is the value for Kb for the acetate ion (CH 3 COO-)? Ka Kb = Kw = 1. 0 x 10 -14 for an acid/conjugate base pair Kb = Kw = (1. 0 x 10 -14) = 5. 6 x 10 -10 Ka (1. 8 x 10 -5) Note that we could prepare a solution that initially only contained the conjugate base if we added a salt such as Na. CH 3 COO, KCH 3 COO, etc. Na. CH 3 COO(aq) Na+(aq) + CH 3 COO-(aq) Calculations would be done in the same way as any other weak acid or weak base solution.

General Statements About Acid and Base Strength We may use the above results to make the following general statement concerning the relative strengths of acids and their conjugate bases. This is based on the relationship Ka Kb = Kw = 1. 0 x 10 -14. The stronger the acid the weaker the conjugate base. Example: Which is a stronger base, F- or CN-? Ka(HF) = 3. 5 x 10 -4 Ka(HCN) = 4. 9 x 10 -10 Since HF is a stronger acid than HCN, F- is a weaker base than CN-. Kb(F-) = (1. 0 x 10 -14)/(3. 5 x 10 -4) = 2. 9 x 10 -11 Kb(CN-) = (1. 0 x 10 -14)/(4. 9 x 10 -10) = 2. 0 x 10 -5

Factors Affecting Acid Strength We can often predict the relative strengths of weak acids by focussing on the factors that control acid strength. Binary acids 1) In the same column (group). Acid strength increases from top to bottom. Reason: The H - X bond strength decreases from top to bottom, so it is easier to break the bond and form H+ (or H 3 O+). Example: Group 7 binary acids. HF 567 k. J/mol Ka = 3. 5 x 10 -4 HCl 431 k. J/mol “strong acid” (Ka ~ 107) HBr 366 k. J/mol “strong acid” (Ka ~ 109) HI 299 k. J/mol “strong acid” (Ka ~ 1011)

2) In the same row. Acid strength increases from left to right. Reason: The electronegativity of the nonmetal increases from left to right, making the conjugate base more stable. Example: Second row. acid conj. base nonmetal EN CH 4 CH 3 - EN(C) = 2. 5 NH 3 NH 2 - EN(N) = 3. 0 “weak base” (Ka ~ 10 -33) H 2 O OH- EN(O) = 3. 5 “amphoteric” (Ka = 1. 0 x 10 -14) HF F- EN(F) = 4. 0 insoluble in water (Ka ~ 10 -49) “weak acid” (Ka = 3. 5 x 10 -4)

Ternary acids (oxyacids) - A ternary acid (oxyacid) contains oxygen, hydrogen, and a central nonmetal atom. Many oxyacids are polyprotic.

Ternary acids (oxyacids) 1) Same nonmetal, different number of oxygens. The more oxygens, the stronger the acid. Reason: The more oxygens in the oxyacid the more stable the conjugate base, and so the easier it will form. acid conj. base HCl. O- Ka = 3. 5 x 10 -8 HCl. O 2 - Ka = 1. 1 x 10 -2 HCl. O 3 - “strong acid” (Ka ~ 5 x 102) HCl. O 4 - “strong acid” (Ka = 1 x 103)

2) Same number of oxygens, different nonmetal in the same column (group). Acid strength increases from bottom to top. Reason: The electronegativity of the nonmetal increases from bottom to top, making the conjugate base more stable, and so easier to form. acid conj. base nonmetal EN HCl. O- EN(Cl) = 3. 0 Ka = 3. 5 x 10 -8 HBr. O- EN(Br) = 2. 8 Ka = 2. 0 x 10 -9 HIO IO- EN(I) = 2. 5 Ka = 2. 3 x 10 -11

Acid-Base Properties of Salts Recall that a salt is an ionic compound formed from the reaction of an acid with a base. We have the following four possibilities: 1) Salt of a strong acid and a strong base. Example: HCl + Na. OH Na. Cl + H 2 O (Na+ and Cl-) No acid-base properties for the salt. Solutions will be neutral. 2) Salt of a strong acid and a weak base. Example: HCl + NH 3 NH 4 Cl (NH 4+ and Cl-) NH 4+ will act as a weak acid. Solutions will be acidic. 3) Salt of a weak acid and a strong base. Example: HF + KOH KF + H 2 O (K+ and F-) F- will act as a weak base. Solutions will be basic.

4) Salt of a weak acid and a weak base. Example: HF + NH 3 NH 4 F (NH 4+ and F-) NH 4+ will act as a weak acid. F- will act as a weak base. Solutions will be approximately neutral. Ka(HF) = 3. 5 x 10 -4 so Kb(F-) = 2. 9 x 10 -11 Kb(NH 3) = 1. 8 x 10 -5 so Ka(NH 4+) = 5. 6 x 10 -10 Since NH 4+ is a stronger acid than F- is a base, the solution will be slightly acidic.

Acid-Base Calculations for Salts We do acid-base problems for salts the same way as we do other weak acid or weak base problems. Example: What is the p. H of a 0. 100 M solution of ammonium chloride (NH 4 Cl), the salt of a strong acid and a weak base. Kb(NH 3) = 1. 8 x 10 -5. Assume T = 25 C

Example: What is the p. H of a 0. 100 M solution of ammonium chloride (NH 4 Cl), the salt of a strong acid and a weak base. Kb(NH 3) = 1. 8 x 10 -5. Assume T = 25 C NH 4 Cl(s) NH 4+(aq) + Cl-(aq) NH 4+(aq) + H 2 O( ) H 3 O+(aq) + NH 3(aq) Ka Kb = 1. 0 x 10 -14, so Ka = 1. 0 x 10 -14 = (1. 0 x 10 -14) = 5. 6 x 10 -10 Kb (1. 8 x 10 -5) Ka = [H 3 O+] [NH 3] = 5. 6 x 10 -10 [NH 4+] Substance Initial Change Equilibrium H 3 O+ 0. 00 x x NH 3 0. 00 x x NH 4+ 0. 100 -x 0. 100 - x

Ka = [H 3 O+] [NH 3] = (x) = 5. 6 x 10 -10 [NH 4+] (0. 100 - x) Assume x << 0. 100. Then x 2 = 5. 6 x 10 -10 (0. 100) x 2 = (0. 100) (5. 6 x 10 -10) = 5. 6 x 10 -11 x = (5. 6 x 10 -11)1/2 = 7. 5 x 10 -6 p. H = - log 10(7. 5 x 10 -6) = 5. 13 Note that 7. 5 x 10 -6 << 0. 100, so x is in fact small.

Acid-Base Properties of Cations Metal cations can act as weak acids in solution. There are several ways in which these reactions can be written Example: Al 3+ ion Arrhenius Al 3+(aq) + H 2 O( ) H+(aq) + Al(OH)2+(aq) Bronsted Al(H 2 O)63+(aq) + H 2 O( ) H 3 O+(aq) + Al(H 2 O)5(OH)2+

Ka 1 = [Al(OH)(H 2 O)52+][H 3 O+] = 1. 3 x 10 -5 [Al(H 2 O)63+] Generally speaking, metal cations act as weak acids when they are small and have multiple positive charges. Common examples of metal cations acting as weak acids are Co 2+, Ni 2+, Zn 2+, Fe 3+, and Al 3+ ions.

Oxides (Main Group Elements) There are two general types of oxides: Metal oxides (K 2 O, Mg. O) - Usually basic. K 2 O(s) + H 2 O(l) 2 KOH(aq) Mg. O(s) + H 2 O(l) Mg(OH)2(s) Nonmetal oxides (CO 2, N 2 O 5) - Usually acidic. CO 2(g) + H 2 O(l) H 2 CO 3(aq) N 2 O 5(g) + H 2 O(l) 2 HNO 3(aq)

Lewis Acids and Bases There is a third way in which acids and bases are defined, developed by G. N. Lewis definition acid - an electron pair acceptor base - an electron pair donor Example: BF 3(g) + NH 3(g) BF 3 NH 3(s)

In the above reaction BF 3 is an electron pair acceptor (Lewis acid) and NH 3 is an electron pair donor (Lewis base). Note that the above reaction would not be an acid-base reaction under the Bronsted definition. Also note that in the above reaction a coordinate covalent bond is formed (a bond where both electrons in the covalent bond come from the same atom).

Identifying Lewis Acids and Bases To identify the Lewis acid and Lewis base in a reaction one needs only to identify the species that is accepting an electron pair (Lewis acid) and donating an electron pair (Lewis base). Example: CO 2 + OH- HCO 3 - CO 2 accepts an electron pair and so is a Lewis acid; OH- donates an electron pair and so is a Lewis base.

End of Chapter 16 “Things should be made as simple as possible, but not any simpler. ” – Albert Einstein “Never express yourself more clearly than you think. ” – Neils Bohr