The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and +

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1 CHAPTER FOURTEEN ACIDS AND BASES Questions 16. The Arrhenius definitions are: acids produce H in water and bases produce OH in water. The difference between strong and weak acids and bases is the amount of H and OH produced. a. A strong acid is 100% dissociated in water. b. A strong base is 100% dissociated in water. c. A weak acid is much less than 100% dissociated in water (typically 110% dissociated). d. A weak base is one that has only a small percentage of molecules react with water to form OH. 17. a. Arrhenius acid: produce H in water b. BrønstedLowry acid: proton (H ) donor c. Lewis acid: electron pair acceptor The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and BrønstedLowry acids; H has an empty 1s orbital and forms bonds to all bases by accepting a pair of electrons from the base. In addition, the Lewis definition incorporates other reactions not typically considered acidbase reactions, e.g., BF 3(g) NH 3(g) FBSNH 3 3(s). NH 3is something we usually consider a base and it is a base in this reaction using the Lewis definition; NH 3 donates a pair of electrons to form the NSB bond. 18. When a strong acid (HX) is added to water, the reaction HX H2O H3O X basically goes to completion. All strong acids in water are completely converted into H3O and X. Thus, no acid stronger than H3O will remain undissociated in water. Similarly, when a strong base (B) is added to water, the reaction B H2O BH OH basically goes to completion. All bases stronger than OH are completely converted into OH and BH. Even though there are acids and bases stronger than H3O and OH, in water these acids and bases are completely converted into H O and OH H2O H OH K w = [H ] [OH ] = Neutral solution: [H ] = [OH ]; [H ] = M; ph = log ( ) = (4 S.F.); 6.78 (3 S.F.); 0.78 (2 S.F.); A ph value is a logarithm. The numbers to the left of the decimal place identy the power of ten to which [H ] is expressed in scientific notation, e.g., , 10, 10. The number of decimal places in a ph value identifies the number of significant figures in [H ]. In all three ph values, the [H ] should be expressed only to two significant figures since these ph values have only two decimal places. 21. Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak 345

2 346 CHAPTER 14 ACIDS AND BASES bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons is used to form a bond to H. 22. a. The weaker the X S H bond, the stronger the acid. b. As the electronegativity of neighboring atoms increases, the strength of the acid increases. c. As the number of oxygen atoms increases, the strength of the acid increases. 23. In general, the weaker the acid, the stronger the conjugate base and vice versa. a. Since acid strength increases as the X S H bond strength decreases, conjugate base strength will increase as the strength of the X S H bond increases. b. Since acid strength increases as the electronegativity of neighboring atoms increases, conjugate base strength will decrease as the electronegativity of neighboring atoms increases. c. Since acid strength increases as the number of oxygen atoms increases, conjugate base strength decreases as the number of oxygen atoms increase. 24. A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have an unshared pair of electrons. 25. a. H2O and CH3CO 2 b. An acidbase reaction can be thought of as a competition between two opposing bases. Since this equilibrium lies far to the left (K < 1), CH CO is a stronger base than H O. a c. The acetate ion is a better base than water and produces basic solutions in water. When we put acetate ion into solution as the only major basic species, the reaction is: CH3CO 2 H2O CH3CO2H OH Now the competition is between CH3CO 2 and OH for the proton. Hydroxide ion is the strongest base possible in water. The equilibrium above lies far to the left, resulting in a K bvalue less than 14 one. Those species we specifically call weak bases (10 < K b < 1) lie between H2O and OH in base strength. Weak bases are stronger bases than water but are weaker bases than OH. 26. The NH 4 ion is a weak acid because it lies between H2O and H3O (H ) in terms of acid strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids because they are not as strong as H3O (H ). Weak acids only partially dissociate in water and have 14 K values between 10 and 1. a 27. When an acid dissociates, ions are produced. The conductivity of the solution is a measure of the number of ions. In addition, the colligative properties, which are discussed in Chapter 11, depend on the number of particles present. So measurements of osmotic pressure, vapor pressure lowering, freezing point depression or boiling point elevation will also allow us to determine the extent of ionization. 28. K a K b = K w, log (K a K b) = log Kw

3 CHAPTER 14 ACIDS AND BASES 347 log K a log K b = log K w, pk a pk b = pk w = (at 25 C) Exercises Nature of Acids and Bases 29. a. HClO 4(aq) H2O(l) H3O (aq) ClO 4(aq). Only the forward reaction is indicated since HClO4 is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reaction is commonly written without water as a reactant. The common abbreviation for this reaction is: HClO 4(aq) H (aq) ClO 4(aq). This reaction is also called the K a reaction as the equilibrium constant for this reaction is called K. a b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation reaction is: CH3CH2CO2H(aq) H2O(l) H3O (aq) CH3CH2CO 2(aq) or CH CH CO H(aq) H (aq) CH CH CO (aq) c. NH 4 is a weak acid. Similar to propanoic acid, the dissociation reaction is: NH 4 (aq) H2O(l) H3O (aq) NH 3(aq) or NH 4 (aq) H (aq) NH 3(aq) 30. The dissociation reaction (the K a reaction) of an acid in water commonly omits water as a reactant. We will follow this practice. All dissociation reactions produce H and the conjugate base of the acid that is dissociated. a. HC2H3O 2(aq) H (aq) C2H3O 2(aq) K a = 3 2 b. Co(H2O) 6 (aq) H (aq) Co(H2O) 5(OH) (aq) K a = c. CH3NH 3 (aq) H (aq) CH3NH 2(aq) K a = 31. An acid is a proton (H ) donor and a base is a proton acceptor. A conjugate acidbase pair differs by only a proton (H ). Conjugate Conjugate Acid Base Base of Acid Acid of Base a. HF H O F H O b. H SO H O HSO H O c. HSO H O SO H O Conjugate Conjugate Acid Base Base of Acid Acid of Base

4 348 CHAPTER 14 ACIDS AND BASES a. Al(H O) H O Al(H O) (OH) H O b. HONH H O HONH H O c. HOCl C H NH OCl C H NH Strong acids have a K a >> 1 and weak acids have K a < 1. Table 14.2 in the text lists some K a values for weak acids. K a values for strong acids are hard to determine so they are not listed in the text. However, there are only a few common strong acids so if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO, HClO and H SO a. HClO 4 is a strong acid. 8 b. HOCl is a weak acid (K a = ). c. H2SO 4 is a strong acid. d. H SO is a weak diprotic acid since the K and K values are less than one. 2 3 a1 a2 34. The beaker on the left represents a strong acid in solution; the acid, HA, is 100% dissociated into the H and A ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid, HB, dissociates into ions, so the acid exists mostly as undissociated HB molecules in water. a. HNO 2: weak acid beaker b. HNO 3: strong acid beaker c. HCl: strong acid beaker d. HF: weak acid beaker e. HC H O : weak acid beaker The K a value is directly related to acid strength. As K aincreases, acid strength increases. For water, use K when comparing the acid strength of water to other species. The K values are: w a HNO 3: strong acid (K a >> 1); HOCl: K a = NH 4 : K a= ; H2O: K a = K w = From the K a values, the ordering is: HNO 3> HOCl > NH 4 > H2O. 36. Except for water, these are the conjugate bases of the acids in the previous exercise. In general, the weaker the acid, the stronger the conjugate base. NO 3 is the conjugate base of a strong acid; it is a terrible base (worse than water). The ordering is: NH > OCl > H O > NO a. 14 HCl is a strong acid and water is a very weak acid with K a = K w = HCl is a much stronger acid than H O. 2 b. H O, K = K = ; HNO, K = ; HNO is a stronger acid than H O since K for HNO > K for H O a w 2 a 2 2 a 2 a c. HOC6H 5, K a = ; HCN, K a= ; HCN is a stronger acid than HOC6H 5 since K for HCN > K for HOC H. a a a. H2O; The conjugate bases of strong acids are terrible bases (K b < 10 ).

5 CHAPTER 14 ACIDS AND BASES b. NO 2; The conjugate bases of weak acids are weak bases (10 < K b < 1). c. OC6H 5; For a conjugate acidbase pair, K a K b = K w. From this relationship, the stronger the acid the weaker the conjugate base (K b decreases as K aincreases). Since HCN is a stronger acid than HOC H (K for HCN > K for HOC H ), OC H will be a stronger base than CN. 6 5 a a Autoionization of Water and the ph Scale At 25 C, the relationship: [H ] [OH ] = K w = always holds for aqueous solutions. When 7 7 [H ] is greater than M, the solution is acidic; when [H ] is less than M, the 7 solution is basic; when [H ] = M, the solution is neutral. In terms of [OH ], an acidic 7 7 solution has [OH ] < M, a basic solution has [OH ] > M, and a neutral solution 7 has [OH ] = M. 7 a. [OH ] = = M; The solution is neutral. 11 b. [OH ] = = M; The solution is acidic. 4 c. [OH ] = = M; The solution is basic. 15 d. [OH ] = = M; The solution is acidic a. [H ] = = M; basic 6 b. [H ] = = M; acidic 12 c. [H ] = = M; basic 7 d. [H ] = = M; neutral 41. a. Since the value of the equilibrium constant increases as the temperature increases, the reaction is endothermic. In endothermic reactions, heat is a reactant so an increase in temperature (heat) shifts the reaction to produce more products and increases K in the process. 14 b. H2O(l) H (aq) OH (aq) K w= = [H ][OH ] at 50 C In pure water [H ] = [OH ], so = [H ], [H ] = M = [OH ] a. H2O(l) H (aq) OH (aq) K w = = [H ] [OH ]

6 350 CHAPTER 14 ACIDS AND BASES In pure water [H ] = [OH ], so = [H ], [H ] = M = [OH ] 7 b. ph = log [H ] = log ( ) = c. [H ] = K w/[oh ] = /0.10 = M; ph = log ( ) = ph = log [H ]; poh = log [OH ]; At 25 C, ph poh = 14.00; For Exercise 13:39: 7 a. ph = log [H ] = log ( ) = 7.00; poh = ph = = b. ph = log ( ) = 3.17; poh = = c. ph = log ( ) = 10.72; poh = = 3.28 d. ph = log (2.3) = 0.36; poh = (0.36) = Note that ph is less than zero when [H ] is greater than 1.0 M (an extremely acidic solution). For Exercise 13.40: a. poh = log [OH ] = log (3.6) = 0.56; ph = poh = (0.56) = b. poh = log ( ) = 8.01; ph = = c. poh = log ( ) = 2.66; ph = = d. poh = log ( ) = 7.00; ph = = 7.00 Note that ph is greater than when [OH ] is greater than 1.0 M (an extremely basic solution). ph a. [H ] = 10, [H ] = 10 = M poh poh = ph = = 6.60; [OH ] = 10 = 10 = M 7 or [OH ] = = M; This solution is basic since ph > (1.3) b. [H ] = 10 = 5 10 M; poh = = 1.3; [OH ] = 10 = 20 M; basic (1.0) c. [H ] = 10 = 10 M; poh = 14.0 (1.0) = 15.0; [OH ] = 10 = 1 10 M; acidic d. [H ] = 10 = M; poh = = 10.80; [OH ] = 10 = M; acidic e. [OH ] = 10 = 1 10 M; ph = 14.0 poh = = 9.0; [H ] = 10 = 1 10 M; basic f. [OH ] = 10 = M; ph = = 4.40; [H ] = 10 = M; acidic ph poh = ph = = 7.23; [H ] = 10 = 10 = M

7 CHAPTER 14 ACIDS AND BASES poh [OH ] = = M or [OH ] = 10 = 10 = M The sample of milk is slightly acidic since the ph is less than 7.00 at 25 C. ph ph = poh = = 8.26; [H ] = 10 = 10 = M 6 poh [OH ] = = M or [OH ] = 10 = 10 = M The solution of baking soda is basic since the ph is greater than 7.00 at 25 C. Solutions of Acids 47. All the acids in this problem are strong acids that are always assumed to completely dissociate in water. The general dissociation reaction for a strong acid is: HA(aq) H (aq) A (aq) where A is the conjugate base of the strong acid HA. For M solutions of these strong acids, M H and M A are present when the acids completely dissociate. The amount of H donated from water will be insignificant in this problem since H O is a very weak acid. a. Major species present after dissociation = H, ClO 4 and H2O; ph = log [H ] = log (0.250) = b. Major species = H, NO 3 and H2O; ph = Strong acids are assumed to completely dissociate in water: HCl(aq) H (aq) Cl (aq) a. A 0.10 M HCl solution gives 0.10 M H and 0.10 M Cl since HCl completely dissociates. The amount of H from H O will be insignificant. ph = log [H ] = log (0.10) = b. 5.0 M H is produced when 5.0 M HCl completely dissociates. The amount of H from H2O will be insignificant. ph = log (5.0) = 0.70 (Negative ph values just indicate very concentrated acid solutions) c M H is produced when M HCl completely dissociates. If you take the 11 negative log of this gives ph = This is impossible! We dissolved an acid in water and got a basic ph. What we must consider in this problem is that water by itself donates M H. We can normally ignore the small amount of H from H2O except when we have a very dilute solution of an acid (as is the case here). Therefore, the ph is that of neutral water (ph = 7.00) since the amount of HCl present is insignificant. 49. Both are strong acids L mol/l = mol HCl = mol H mol Cl L 0.10 mol/l = mol HNO 3 = mol H mol NO3 13 [H ] = = M; [OH ] = = M

8 352 CHAPTER 14 ACIDS AND BASES [Cl ] = = M; [NO 3 ] = = M L = mol H from HCl L = mol H from HNO 3 [H ] = = M; ph = log (0.690) = poh = = ; [OH ] = 10 = M ph [H ] = 10 = 10 = M. Since HCl is a strong acid, a M HCl solution will 3 produce M H giving a ph = ph [H ] = 10 = 10 = M. Since HNO 3 is a strong acid, a M HNO 3 solution is necessary to produce a ph = 5.10 solution. 53. a HNO 2 (K a = ) and H2O (K a = K w = ) are the major species. HNO 2 is a much stronger acid than H2O so it is the major source of H. However, HNO 2 is a weak acid (K a < 1) so it only partially dissociates in water. We must solve an equilibrium problem to determine [H ]. In the Solutions Guide, we will summarize the initial, change and equilibrium concentrations into one table called the ICE table. Solving the weak acid problem: HNO 2 H NO2 Initial M ~0 0 x mol/l HNO2 dissociates to reach equilibrium Change x x x Equil x x x 4 K a = = = ; If we assume x << 0.250, then: , x = = M We must check the assumption: 7 All the assumptions are good. The H contribution from water (10 M) is negligible, and x is small compared to (percent error = 4.0%). If the percent error is less than 5% for an assumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem: x = M = [H ]; ph = log(0.010) = b. CH3CO2H (K a = ) and H2O (K a = K w = ) are the major species. CH3CO2H is the major source of H. Solving the weak acid problem:

9 CHAPTER 14 ACIDS AND BASES 353 CH3CO2H H CH3CO 2 Initial M ~0 0 x mol/l CH3CO2H dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = = (assuming x << 0.250) 3 x = M; Checking assumption: 100= 0.84%. Assumptions good. 3 3 [H ] = x = M; ph = log ( ) = a HOC6H 5 (K a = ) and H2O (K a = K w = ) are the major species. The major equilibrium is the dissociation of HOC H. Solving the weak acid problem: 6 5 HOC6H 5 H OC6H5 Initial M ~0 0 x mol/l HOC6H 5 dissociates to reach equilibrium Change x x x Equil x x x 10 K a = = (assuming x << 0.250) 6 3 x = [H ] = M; Checking assumption: x is % of 0.250, so assumption is valid by the 5% rule. 6 ph = log( ) = b. HCN (K a = ) and H2O are the major species. HCN is the major source of H. HCN H CN Initial M ~0 0 x mol/l HCN dissociates to reach equilibrium Change x x x Equil x x x 10 K a = = (assuming x << 0.250)

10 354 CHAPTER 14 ACIDS AND BASES 5 3 x = [H ] = M; Checking assumption: x is % of Assumptions good. ph = log ( ) = ml glacial acetic acid = mol HC2H3O2 Initial concentration of HC2H3O 2 = = 1.40 M 5 HC2H3O 2 H C2H3O 2 K a = Initial 1.40 M ~0 0 x mol/l HC2H3O 2 dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = 3 x = [H ] = M; ph = 2.30 Assumptions good (x is 0.36% of 1.40) HC3H5O 2 (K a = ) and H2O (K a = K w= ) are the major species present. HC3H5O2 will be the dominant producer of H since HC3H5O 2 is a stronger acid than H2O. Solving the weak acid problem: HCHO H CHO Initial M ~0 0 x mol/l HC3H5O 2 dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = 3 3 x = [H ] = M; ph = log ( ) = 2.96 Assumption follows the 5% rule (x is 1.1% of 0.100) [H ] = [C3H5O 2] = M; [OH ] = K w/[h ] = M 3 [HC3H5O 2] = = M Percent dissociation = 100 = 1.1%

11 CHAPTER 14 ACIDS AND BASES This is a weak acid in water. Solving the weak acid problem: 4 HF H F K a = Initial M ~0 0 x mol/l HF dissociates to reach equilibrium Change x x x Equil x x x 4 K a = = (assuming x << 0.020) 3 x = [H ] = M; Check assumptions: The assumption x << is not good (x is more than 5% of 0.020). We must solve 2 4 x /(0.020 x) = exactly by using either the quadratic formula or by the method of successive approximations (see Appendix 1.4 of text). Using successive approximations, we let M be a new approximation for [HF]. That is, in the denominator, try x = (the value of x we calculated making the normal assumption), so = 0.016, then solve for a new value of x in the numerator. 4 3 = , x = We use this new value of x to further refine our estimate of [HF], i.e., x = = (carry extra significant figure). 4 3 = , x = We repeat until we get an answer that repeats itself. This would be the same answer we would get solving exactly using the quadratic equation. In this case it is: x = So: [H ] = [F ] = x = M; [OH ] = K w/[h ] = M [HF] = x = = M; ph = 2.46 Note: When the 5% assumption fails, use whichever method you are most comfortable with to solve exactly. The method of successive approximations is probably fastest when the percent error is less than ~25% (unless you have a calculator that can solve quadratic equations). 58. Major species: HIO 3, H2O; Major source of H : HIO 3 (a weak acid, K a = 0.17) HIO 3 H IO3 Initial 0.20 M ~0 0 x mol/l HIO 3 dissociates to reach equilibrium Change x x x Equil x x x

12 356 CHAPTER 14 ACIDS AND BASES K a = 0.17 =, x = 0.18; Check assumption. Assumption is horrible (x is 90% of 0.20). When the assumption is this poor, it is generally quickest to solve exactly using the quadratic formula (see Appendix 1.4 in text). The method of successive approximations will require many trials to finally converge on the answer. For this problem, 5 trials were required. Using the quadratic formula and carrying extra significant figures: =, x = 0.17(0.20 x), x 0.17 x = 0 x =, x = 0.12 or 0.29 Only x = 0.12 makes sense. x = 0.12 M = [H ]; ph = log (0.12) = Major species: HC2H2ClO 2 (K a = ) and H2O; Major source of H : HC2H2ClO2 HCHClO H CHClO Initial 0.10 M ~0 0 x mol/l HC2H2ClO 2 dissociates to reach equilibrium Change x x x Equil x x x K = = a, x = M 3 2 Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve = x /(0.10 x) exactly using either the method of successive approximations or the 2 quadratic equation. Using either method gives x = [H ] = M. ph = log [H ] = log ( ) = HClO 2 H ClO 2 K a = Initial 0.22 M ~0 0 x mol/l HClO 2 dissociates to reach equilibrium Change x x x Equil x x x 2 2 K a = =, x = The assumption that x is small is not good (x is 23% of 0.22). Using the method of successive approximations and carrying extra significant figures:

13 CHAPTER 14 ACIDS AND BASES = , x = = , x = ; x = repeats 2 [H ] = [ClO 2 ] = x = M; % dissociation = 100 = 21% 61. a. HCl is a strong acid. It will produce 0.10 M H. HOCl is a weak acid. Let's consider the equilibrium: 8 HOCl H OCl K a = Initial 0.10 M 0.10 M 0 x mol/l HOCl dissociates to reach equilibrium Change x x x Equil x 0.10 x x K = = a, x = M Assumptions are great (x is % of 0.10). We are really assuming that HCl is the only important source of H, which it is. The [H ] contribution from HOCl, x, is negligible. Therefore, [H ] = 0.10 M; ph = 1.00 b. HNO 3 is a strong acid, giving an initial concentration of H equal to M. Consider the equilibrium: 5 HC2H3O 2 H C2H3O 2 K a = Initial 0.50 M M 0 x mol/l HC2H3O 2 dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = 4 x = ; Assumptions are good (well within the 5% rule). [H ] = x = M and ph = HF and HOC6H 5are both weak acids with K avalues of and , respectively. Since the K avalue for HF is much greater than the K avalue for HOC6H 5, HF will be the dominant producer of H (we can ignore the amount of H produced from HOC H since it will be insignificant). 6 5 HF H F Initial 1.0 M ~0 0 x mol/l HF dissociates to reach equilibrium

14 358 CHAPTER 14 ACIDS AND BASES Change x x x Equil. 1.0 x x x 4 K a = = 2 2 x = [H ] = M; ph = log ( ) = 1.57 Assumptions good. Solving for [OC6H 5] using HOC6H 5 H OC6H 5 equilibrium: 10 9 K a = =, [OC6H 5] = M 9 Note that this answer indicates that only M HOC6H 5 dissociates, which indicates that HF is truly the only significant producer of H in this solution. 63. In all parts of this problem, acetic acid (HC2H3O 2) is the best weak acid present. We must solve a weak acid problem. a. HCHO H CHO Initial 0.50 M ~0 0 x mol/l HC2H3O 2 dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = x = [H ] = [C H O ] = M Assumptions good. Percent dissociation = 100 = 100 = 0.60% b. The setups for solutions b and c are similar to solution a except the final equation is slightly different, reflecting the new concentration of HC H O. 5 K a = = x = [H ] = [C2H3O 2] = M Assumptions good. % dissociation = 100 = 1.9% 5 c. K a = = 4 x = [H ] = [C2H3O 2] = M; Check assumptions. Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the

15 CHAPTER 14 ACIDS AND BASES 359 method of successive approximations (see Appendix 1.4 of text): =, x = Next trial also gives x = % dissociation = 100 = 5.8% d. As we dilute a solution, all concentrations decrease. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water, then: Q = Q < K a, so the equilibrium shifts to the right or towards a greater percent dissociation. e. [H ] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions ac the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus, [H ] decreases. 64. a. HNO 3 is a strong acid; it is assumed 100% dissociated in solution. 4 b. HNO 2 H NO2 K a = Initial 0.20 M ~0 0 x mol/l HNO 2 dissociates to reach equilibrium Change x x x Equil x x x 4 K a = = 3 x = [H ] = [NO 2 ] = M; Assumptions good. % dissociation = 100 = 4.5% 10 c. HOC6H 5 H OC6H5 K a =

16 360 CHAPTER 14 ACIDS AND BASES Initial 0.20 M ~0 0 x mol/l HOC6H 5 dissociates to reach equilibrium Change x x x Equil x x x 10 K a = = x = [H ] = [OC H ] = M; Assumptions good. % dissociation = 100 = % d. For the same initial concentration, the percent dissociation increases as the strength of the acid increases (as K increases). a 65. Let HX symbolize the weak acid. Setup the problem like a typical weak acid equilibrium problem. HX H X Initial 0.15 M ~0 0 x mol/l HX dissociates to reach equilibrium Change x x x Equil x x x If the acid is 3.0% dissociated, then x = [H ] is 3.0% of 0.15: x = (0.15 M) = M. Now that we know the value of x, we can solve for K. K a = = HF H F Initial M ~0 0 x mol/l HF dissociates to reach equilibrium Change x x x Equil x x x a K a = 3 ; x = [H ] = [F ] = (0.100 M) = M 3 4 [HF] = = M; K a = = Setup the problem using the K a equilibrium reaction for HOCN. HOCN H OCN Initial M ~0 0 x mol/l HOBr dissociates to reach equilibrium

17 CHAPTER 14 ACIDS AND BASES 361 Change x x x Equil x x x ph K a = ; Since ph = 2.77, then: x = [H ] = 10 = 10 = M K a = = HClO 4 is a strong acid with [H ] = M. This equals the [H ] in the trichloroacetic acid solution. Now setup the problem using the K equilibrium reaction for CCl CO H. a 3 2 CCl3CO2H H CCl3CO2 Initial M ~0 0 Equil x x x K a = 2 ; x = [H ] = M K a = = Major species: HCOOH and H2O; Major source of H : HCOOH HCOOH H HCOO Initial C ~0 0 where C = [HCOOH] o x mol/l HCOOH dissociates to reach equilibrium Change x x x Equil. C x x x 4 K a = = where x = [H ] = ; Since ph = 2.70, then: [H ] = 10 = M =, C ( ) =, C = M A M formic acid solution will have ph = [HA] = = 0.50 mol/l; Solve using the K equilibrium reaction. o a HA H A Initial 0.50 M ~0 0 Equil x x x

18 362 CHAPTER 14 ACIDS AND BASES K a = ; In this problem, [HA] = 0.45 M so: [HA] = 0.45 M = 0.50 M x, x = 0.05 M; K a = = Solutions of Bases 71. a. NH 3(aq) H2O(l) NH 4 (aq) OH (aq) K b = b. C5H5N(aq) H2O(l) C5H5NH (aq) OH (aq) K b = 72. a. C6H5NH 2(aq) H2O(l) C6H5NH 3 (aq) OH (aq) K b = b. (CH 3) 2NH(aq) H2O(l) (CH 3) 2NH 2(aq) OH (aq) K b= 73. NO 3: K b << K w since HNO 3 is a strong acid. All conjugate bases of strong acids have no base strength. H O: K = K = ; NH : K = ; C H N: K = b w 3 b 5 5 b NH 3 > C5H5N > H2O > NO 3 (As K b increases, base strength increases.) 74. Excluding water, these are the conjugate acids of the bases in the previous exercise. In general, the stronger the base, the weaker the conjugate acid. Note: Even though NH 4 and C5H5NH are conjugate acids of weak bases, they are still weak acids with K avalues between K w and 1. Prove this to yourself by calculating the K values for NH and C H NH (K = K /K ). a a w b HNO > C H NH > NH > H O 75. a. NH b. NH c. OH d. CH NH The base with the largest K b value is the strongest base. OH is the strongest base possible in water. 76. a. HNO3 b. NH4 c. NH4 The acid with the largest K a value is the strongest acid. To calculate K a values for NH 4 and CH NH, use K = K /K where K refers to the bases NH or CH NH. 3 3 a w b b NaOH(aq) Na (aq) OH (aq); NaOH is a strong base which completely dissociates into Na and OH. The initial concentration of NaOH will equal the concentration of OH donated by NaOH. a. [OH ] = 0.10 M; poh = log[oh ] = log(0.10) = 1.00 ph = poh = = Note that H2O is also present, but the amount of OH produced by H2O will be insignificant compared to the 0.10 M OH produced from the NaOH. 10 b. The [OH ] concentration donated by the NaOH is M. Water by itself donates

19 CHAPTER 14 ACIDS AND BASES M. In this problem, water is the major OH contributor and [OH ] = M. 7 poh = log ( ) = 7.00; ph = = 7.00 c. [OH ] = 2.0 M; poh = log (2.0) = 0.30; ph = (0.30) = a. Ca(OH) 2 Ca 2 OH ; Ca(OH) 2 is a strong base and dissociates completely. 4 [OH ] = 2( ) = M; poh = log [OH ] = 3.10; ph = poh = b. = 0.45 mol KOH/L KOH is a strong base, so [OH ] = 0.45 M; poh = log (0.45) = 0.35; ph = c. = M; NaOH is a strong base, so [OH ] = M. poh = log (3.750) = and ph = (0.5740) = Although we are justified in calculating the answer to four decimal places, in reality ph values are generally measured to two decimal places, sometimes three. 79. a. Major species: K, OH, H2O (KOH is a strong base.) [OH ] = M, poh = log (0.015) = 1.82; ph = poh = b. Major species: Ba, OH, H2O; Ba(OH) 2(aq) Ba (aq) 2 OH (aq); Since each mol of the strong base Ba(OH) 2 dissolves in water to produce two mol OH, then [OH ] = 2(0.015 M) = M. poh = log (0.030) = 1.52; ph = = a. Major species: Na, Li, OH, H O (NaOH and LiOH are both strong bases.) [OH ] = = M; poh = 1.000; ph = b. Major species: Ba, Rb, OH, H O; Both Ba(OH) and RbOH are strong bases and Ba(OH) donates 2 mol OH per mol Ba(OH) [OH ] = 2(0.0010) = M; poh = log (0.022) = 1.66; ph = ph = 10.50; poh = = 3.50; [OH ] = 10 = M 4 KOH(aq) K (aq) OH (aq); Since KOH is a strong base, a M KOH solution will produce a ph = solution ph = 10.50; poh = = 3.50; [OH ] = 10 = M 2 Sr(OH) 2(aq) Sr (aq) 2 OH (aq); Sr(OH) 2 donates two mol OH per mol Sr(OH) 2. [Sr(OH) ] = M OH = M Sr(OH)

20 364 CHAPTER 14 ACIDS AND BASES 4 A M Sr(OH) 2 solution will produce a ph = solution NH 3 is a weak base with K b = The major species present will be NH 3 and H2O (K b = Kw 14 = ). Since NH 3 has a much larger K b value compared to H2O, NH 3 is the stronger base present and will be the major producer of OH. To determine the amount of OH produced from NH 3, we must perform an equilibrium calculation. NH 3(aq) H2O(l) NH 4 (aq) OH (aq) Initial M 0 ~0 x mol/l NH 3 reacts with H2O to reach equilibrium Change x x x Equil x x x 5 K b = = (assuming x << 0.150) 3 x = [OH ] = M; Check assumptions: x is 1.1% of so the assumption x is valid by the 5% rule. Also, the contribution of OH from water will be insignificant (which 3 will usually be the case). Finishing the problem, poh = log [OH ] = log ( M) = 2.80; ph = poh = = Major species: H NNH (K = ) and H O (K = K = ); The weak base H NNH will dominate OH production. We must perform a weak base equilibrium calculation b 2 b w H2NNH 2 H2O H2NNH 3 OH K b = Initial 2.0 M 0 ~0 x mol/l H2NNH 2 reacts with H2O to reach equilibrium Change x x x Equil. 2.0 x x x 6 K b = = (assuming x << 2.0) 3 x = [OH ] = M; poh = 2.62; ph = Assumptions good (x is 0.12% of 2.0) [H2NNH 3 ] = M; [H2NNH 2] = 2.0 M; [H ] = 10 = M 85. These are solutions of weak bases in water. We must solve the equilibrium weak base problem. 4 a. (C2H 5) 3N H2O (C2H 5) 3NH OH K b = Initial 0.20 M 0 ~0 x mol/l of (C2H 5) 3N reacts with H2O to reach equilibrium Change x x x

21 CHAPTER 14 ACIDS AND BASES 365 Equil x x x K = = b, x = [OH ] = M Assumptions good (x is 4.5% of 0.20). [OH ] = M 12 [H ] = = M; ph = b. HONH 2 H2O HONH 3 OH K b = Initial 0.20 M 0 ~0 Equil x x x K = = b, x = [OH ] = M; Assumptions good [H ] = M; ph = These are solutions of weak bases in water. 10 a. C6H5NH 2 H2O C6H5NH 3 OH K b = Initial 0.20 M 0 ~0 x mol/l of C6H5NH 2 reacts with H2O to reach equilibrium Change x x x Equil x x x =, x = [OH ] = M; Assumptions good. 9 [H ] = K w/[oh ] = M; ph = b. C5H5N H2O C5H5NH OH K b = Initial 0.20 M 0 ~0 Equil x x x K = = b 9 5, x = M; Assumptions good.

22 366 CHAPTER 14 ACIDS AND BASES 5 10 [OH ] = M; [H ] = M; ph = This is a solution of a weak base in water. We must solve the weak base equilibrium problem. 4 C2H5NH 2 H2O C2H5NH 3 OH K b = Initial 0.20 M 0 ~0 x mol/l C2H5NH 2 reacts with H2O to reach equilibrium Change x x x Equil x x x K b = (assuming x << 0.20) 2 x = ; Checking assumption: 100 = 5.5% Assumption fails the 5% rule. We must solve exactly using either the quadratic equation or the method of successive approximations (see Appendix 1.4 of the text). Using successive approximations and carrying extra significant figures: 4 2 = , x = M (consistent answer) 2 12 x = [OH ] = M; [H ] = = M; ph = (C2H 5) 2NH H2O (C2H 5) 2NH 2 OH K b = Initial M 0 ~0 x mol/l (C2H 5) 2NH reacts with H2O to reach equilibrium Change x x x Equil x x x 3 K b = = (assuming x << 0.050) 3 x = ; Assumption is bad (x is 16% of 0.20). Using successive approximations: =, x = =, x = (consistent answer) 3 12 [OH ] = x = M; [H ] = K w/[oh ] = M; ph = 11.85

23 CHAPTER 14 ACIDS AND BASES To solve for percent ionization, just solve the weak base equilibrium problem. 5 a. NH 3 H2O NH 4 OH K b = Initial 0.10 M 0 ~0 Equil x x x K = = b, x = [OH ] = M; Assumptions good. 5 3 Percent ionization = 100 = 1.3% b. NH 3 H2O NH 4 OH Initial M 0 ~0 Equil x x x =, x = [OH ] = M; Assumptions good. Percent ionization = 100 = 4.2% Note: For the same base, the percent ionization increases as the initial concentration of base decreases a. HONH H O HONH OH K = b Initial 0.10 M 0 ~0 Equil x x x =, x = [OH ] = M; Assumptions good. Percent ionization = 100 = 0.033% 4 b. CH3NH 2 H2O CH3NH 3 OH K b = Initial 0.10 M 0 ~0 Equil x x x =, x = ; Assumption fails the 5% rule (x is 6.6% of 0.10). Using successive approximations and carrying extra significant figures: 4 3 = , x = (consistent answer) Percent ionization = 100 = 6.4%

24 368 CHAPTER 14 ACIDS AND BASES 91. Let cod = codeine, C18H21NO 3; using the K b reaction to solve: cod H2O codh OH Initial M 0 ~0 x mol/l codeine reacts with H2O to reach equilibrium Change x x x Equil x x x K b = ; Since ph = 9.59, then poh = = [OH ] = x = 10 = M; K b = = Using the K b reaction to solve where PY = pyrrolidine, C4H8NH: PY H2O PYH OH Initial M 0 ~0 Equil x x x K b = ; Since ph = 10.82: poh = 3.18 and [OH ] = x = 10 = M K b = = Polyprotic Acids 93. H2SO 3(aq) HSO 3(aq) H (aq) = 2 HSO 3(aq) SO 3 (aq) H (aq) = 94. H3C6H5O 7(aq) H2C6H5O 7(aq) H (aq) = 2 H2C6H5O 7(aq) HC6H5O 7 (aq) H (aq) = 2 3 HC6H5O 7 (aq) C6H5O 7 (aq) H (aq) =

25 CHAPTER 14 ACIDS AND BASES In both these polyprotic acid problems, the dominate equilibrium is the reaction. The amount of H produced from the subsequent K a reactions will be minimal since they are all have much smaller K values. a 3 a. H3PO 4 H H2PO 4 = Initial 0.10 M ~0 0 x mol/l H3PO 4 dissociates to reach equilibrium Change x x x Equil x x x 3 2 = =, x = Assumption is bad (x is 27% of 0.10). Using successive approximations: = , x = ; = , x = (consistent answer) 2 2 x = [H ] = M; ph = log ( ) = b. H2CO 3 H HCO 3 = Initial 0.10 M ~0 0 Equil x x x 7 = = 4 x = [H ] = M; ph = 3.68; Assumptions good. 96. The reactions are: 3 H3AsO 4 H H2AsO 4 = H2AsO 4 H HAsO 4 = HAsO 4 H AsO 4 = 6 10 We will deal with the reactions in order of importance, beginning with the largest K a,. 3 H3AsO 4 H H2AsO 4 = 5 10 = Initial 0.20 M ~0 0 Equil x x x =, x = = 3 10 M (By using successive approximations

26 370 CHAPTER 14 ACIDS AND BASES 2 [H ] = [H2AsO 4] = 3 10 M; [H3AsO 4] = = 0.17 M or the quadratic formula.) 8 Since = 8 10 is much smaller than the value, very little of 2 H2AsO 4 (and HAsO 4 ) dissociates compared to H3AsO 4. Therefore, [H ] and [H2AsO 4] will not change significantly by the reaction. Using the previously calculated concentrations of H and 2 H AsO to calculate the concentration of HAsO : =, [HAsO 4 ] = 8 10 M Assumption that the K reaction does not change [H ] and [ using to get [AsO ]. a2 3 4 ] is good. We repeat the process 10 = 6 10 = [AsO 4 ] = M Assumption good. So in 0.20 M analytical concentration of H3AsO 4: 2 [H3AsO 4] = 0.17 M; [H ] = [H2AsO 4] = 3 10 M [HAsO 4 ] = 8 10 M; [AsO 4 ] = 2 10 M 13 [OH ] = K w/[h ] = 3 10 M 97. The dominant H producer is the strong acid H SO. A 2.0 M H SO solution produces 2.0 M HSO and 2.0 M H. However, HSO is a weak acid which could also add H to the solution HSO 4 H SO 4 Initial 2.0 M 2.0 M 0 x mol/l HSO 4 dissociates to reach equilibrium Change x x x Equil. 2.0 x 2.0 x x 2 2 = =, x = Since x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H from

27 CHAPTER 14 ACIDS AND BASES HSO 4 is The total amount of H present is: 2 [H ] = = 2.0 M; ph = log (2.0) = 0.30 Note: In this problem, H from HSO 4 could have been ignored. However, this is not usually the case, especially in more dilute solutions of H SO For H2SO 4, the first dissociation occurs to completion. The hydrogen sulfate ion, HSO 4, is a weak 2 acid with = We will consider this equilibrium for additional H production: 2 HSO 4 H SO4 Initial M M 0 x mol/l HSO 4 dissociates to reach equilibrium Change x x x Equil x x x = = x, x = 0.012; Assumption is horrible (240% error). Using the quadratic formula: x = x x, x x = 0 x =, x = M [H ] = x = = M; ph = 2.10 Note: We had to consider both H2SO 4 and HSO 4 for H production in this problem. AcidBase Properties of Salts 99. One difficult aspect of acidbase chemistry is recognizing what types of species are present in solution, i.e., whether a species is a strong acid, strong base, weak acid, weak base or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a. Memorize the following strong acids: HCl, HBr, HI, HNO 3, HClO 4 and H2SO4 b. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH) 2, Sr(OH) 2 and Ba(OH) 2 c. All weak acids have a K a value less than 1 but greater than K w. Some weak acids are in Table 14.2 of the text. All weak bases have a K b value less than 1 but greater than K w. Some weak bases are in Table 14.3 of the text. d. All conjugate bases of weak acids are weak bases, i.e., all have a K b value less than 1 but greater than K w. Some examples of these are the conjugate bases of the weak acids in Table 14.2 of the text. e. All conjugate acids of weak bases are weak acids, i.e., all have a K a value less than 1 but greater than K. Some examples of these are the conjugate acids of the weak bases in Table 14.3 of the w

28 372 CHAPTER 14 ACIDS AND BASES text f. Alkali metal ions (Li, Na, K, Rb, Cs ) and heavier alkaline earth metal ions (Ca, Sr, Ba ) have no acidic or basic properties in water. g. All conjugate bases of strong acids (Cl, Br, I, NO 3, ClO 4, HSO 4) have no basic properties in water (K << K ) and only HSO has any acidic properties in water. b w 4 Lets apply these ideas to this problem to see what type of species are present. The letters in parenthesis is/are the generalization(s) above which identifies the species. KOH: strong base (b) KCl: neutral; K and Cl have no acidic/basic properties (f and g) KCN: CN is a weak base, K b = / = (c and d). Ignore K (f). 10 NH4Cl: NH 4 is a weak acid, K a = (c and e). Ignore Cl (g). HCl: strong acid (a) The most acidic solution will be the strong acid followed by the weak acid. The most basic solution will be the strong base followed by the weak base. The KCl solution will be between the acidic and basic solutions at ph = Most acidic most basic; HCl > NH4Cl > KCl > KCN > KOH 100. See Exercise for some generalizations on acidbase properties of salts. The letters in parenthesis is/are the generalization(s) listed in Exercise which identifies the species. 2 CaBr 2: neutral; Ca and Br have no acidic/basic properties (f and g) KNO 2: NO 2 is a weak base, K b = / = (c and d). Ignore K (f). HClO 4: strong acid (a) 4 HNO 2: weak acid, K a = (c) HONH3ClO 4: HONH 3 is a weak acid, K a = / = (c and e). Ignore ClO (g). 4 Using the information above (identity and K or K values), the ordering is: most acidic most basic: HClO 4 > HNO 2 > HONH3ClO 4 > CaBr 2 > KNO2 a 101. From the K a values, acetic acid is a stronger acid than hypochlorous acid. Conversely, the conjugate base of acetic acid, C2H3O 2, will be a weaker base than the conjugate base of hypochlorous acid, OCl. Thus, the hypochlorite ion, OCl, is a stronger base than the acetate ion, C2H3O 2. In general, the stronger the acid, the weaker the conjugate base. This statement comes from the relationship K w= Ka K, which holds for all conjugate acidbase pairs. b b 102. Since NH is a weaker base (smaller K value) than CH NH, the conjugate acid of NH will be a stronger acid than the conjugate acid of CH NH. Thus, NH is a stronger acid than CH NH. 3 b NaN 3 Na N 3; Azide, N 3, is a weak base since it is the conjugate base of a weak acid. All conjugate bases of weak acids are weak bases (K < K < 1). Ignore Na. w b 10 N 3 H2O HN 3 OH K b = = Initial M 0 ~0

29 CHAPTER 14 ACIDS AND BASES 373 x mol/l of N 3 reacts with H2O to reach equilibrium Change x x x Equil x x x 10 K b = = = (assuming x << 0.010) 6 9 x = [OH ] = M; [H ] = = M Assumptions good. 6 6 [HN 3] = [OH ] = M; [Na ] = M; [N 3] = = M 104. C2H5NH3Cl C2H5NH 3 Cl ; C2H5NH 3 is the conjugate acid of the weak base C2H5NH 2 4 (K b = ). As is true for all conjugate acids of weak bases, C2H5NH 3 is a weak acid. Cl has no basic (or acidic) properties. Ignore Cl. Solving the weak acid problem: 4 11 C2H5NH 3 C2H5NH 2 H K a = K w/ = Initial 0.25 M 0 ~0 x mol/l C2H5NH 3 dissociates to reach equilibrium Change x x x Equil x x x 11 K a = = (assuming x << 0.25) 6 x = [H ] = M; ph = 5.68; Assumptions good. 6 [C2H5NH 2] = [H ] = M; [C2H5NH 3 ] = 0.25 M; [Cl ] = 0.25 M 9 [OH ] = K w/[h ] = M 105. a. CH3NH3Cl CH3NH 3 Cl : CH3NH 3 is a weak acid. Cl is the conjugate base of a strong acid. Cl has no basic (or acidic) properties. 11 CH3NH 3 CH3NH 2 H K a = = CH3NH 3 CH3NH 2 H Initial 0.10 M 0 ~0 x mol/l CH3NH 3 dissociates to reach equilibrium Change x x x Equil x x x 11 K a = = (assuming x << 0.10)

30 374 CHAPTER 14 ACIDS AND BASES 6 x = [H ] = M; ph = 5.82 Assumptions good. b. NaCN Na CN : CN is a weak base. Na has no acidic (or basic) properties. 5 CN H2O HCN OH K b = = Initial M 0 ~0 x mol/l CN reacts with H2O to reach equilibrium Change x x x Equil x x x 5 K b = = 4 x = [OH ] = M; poh = 3.05; ph = Assumptions good a. KNO 2 K NO 2: NO 2 is a weak base. Ignore K. 11 NO 2 H2O HNO 2 OH K b = = Initial 0.12 M 0 ~0 Equil x x x 11 K b = = 6 x = [OH ] = M; poh = 5.77; ph = 8.23 Assumptions good. b. NaOCl Na OCl : OCl is a weak base. Ignore Na. 7 OCl H2O HOCl OH K b = = Initial 0.45 M 0 ~0 Equil x x x 7 K = = b 4 x = [OH ] = M; poh = 3.44; ph = Assumptions good. c. NH ClO NH ClO : NH is a weak acid. ClO is the conjugate base of a strong acid. ClO has no basic (or acidic) properties NH 4 NH 3 H K a = = Initial 0.40 M 0 ~0

31 CHAPTER 14 ACIDS AND BASES 375 Equil x x x 10 K a = = 5 x = [H ] = M; ph = 4.82; Assumptions good All these salts contain Na, which has no acidic/basic properties, and a conjugate base of a weak acid (except for NaCl where Cl is a neutral species.). All conjugate bases of weak acids are weak bases since K b for these species is between K w and 1. To identify the species, we will use the data given to determine the K b value for the weak conjugate base. From the K b value and data in Table 14.2 of the text, we can identify the conjugate base present by calculating the K a value for the weak acid. We will use A as an abbreviation for the weak conjugate base. A H2O HA OH Initial mol/1.00 L 0 ~0 x mol/l A reacts with H2O to reach equilibrium Change x x x Equil x x x K b = ; From the problem, ph = 8.07: poh = = 5.93; [OH ] = x = 10 = M 11 K = = = K value for the conjugate base of a weak acid. b b The K a value for the weak acid equals K w/k b: K a = = From Table 14.2 of the text, this K a value is closest to HF. Therefore, the unknown salt is NaF BHCl BH Cl ; Cl is the conjugate base of the strong acid HCl, so Cl has no acidic/basic properties. BH is a weak acid since it is the conjugate acid of a weak base, B. Determining the K a value for BH : BH B H Initial 0.10 M 0 ~0 x mol/l BH dissociates to reach equilibrium Change x x x Equil x x x

32 376 CHAPTER 14 ACIDS AND BASES K a = ; From the problem, ph = 5.82: [H ] = x = 10 = M; K a = = K b for the base, B = K w\k a = / = From Table 14.3 of the text, this K bvalue is closest to CH3NH 2so the unknown salt is CH3NH3Cl Major species present: Al(H O) (K = ), NO (neutral) and H O (K = ); Al(H O) is a stronger acid than water so it will be the dominant H producer a 3 2 w Al(H2O) 6 Al(H2O) 5(OH) H Initial M 0 ~0 3 x mol/l Al(H2O) 6 dissociates to reach equilibrium Change x x x Equil x x x 5 K a = = 4 4 x = M = [H ]; ph = log ( ) = 3.08; Assumptions good Major species: Co(H2O) 6 (K a = ), Cl (neutral) and H2O (K w = ); Co(H2O) 6 will determine the ph since it is a stronger acid than water. Solving the weak acid problem in the usual manner: Co(H2O) 6 Co(H2O) 5(OH) H K a = Initial 0.10 M 0 ~0 Equil x x x K = = a, x = [H ] = M ph = log ( ) = 3.00; Assumptions good Reference Table 14.6 of the text and the solution to Exercise for some generalizations on acidbase properties of salts. a. NaNO 3 Na NO 3 neutral; Neither species has any acidic/basic properties. b. NaNO 2 Na NO 2 basic; NO 2 is a weak base and Na has no effect on ph.

33 CHAPTER 14 ACIDS AND BASES NO 2 H2O HNO 2 OH K b = = c. C5H5NHClO 4 C5H5NH C1O 4 acidic; C5H5NH is a weak acid and ClO 4 has no effect on ph. 6 CHNH 5 5 H C5H5N K a = = d. NH NO NH NO acidic; NH is a weak acid (K = ) and NO is a weak base (K = ). Since >, then the solution is acidic a 2 11 b NH 4 H NH3 K a = ; NO 2 H2O HNO 2 OH K b = e. KOCl K OCl basic; OCl is a weak base and K has no effect on ph. 7 OCl H2O HOCl OH K b = = f. NH4OCl NH 4 OCl basic; NH 4 is a weak acid and OCl is a weak base. Since, then the solution is basic NH 4 NH 3 H K a = ; OCl H2O HOCl OH K b = a. KCl K Cl neutral; K and Cl have no effect on ph. b. NHCHO NH 4 C2H3O 2 neutral; NH 4 is a weak acid and C2H3O 2 is a weak base. Since, ph = NH 4 NH 3 H K a = = CHO HO 2 HC2H3O 2 OH K b = = c. CH3NH3Cl CH3NH 3 Cl acidic; CH3NH 3 is a weak acid and Cl has no effect on ph. 11 CH3NH 3 H CH3NH 2 K a = = d. KF K F basic; F is a weak base and K has no effect on ph. 11 F H2O HF OH K b = =

34 378 CHAPTER 14 ACIDS AND BASES e. NH4F NH 4 F acidic; NH 4 is a weak acid and F is a weak base. Since, then the solution is acidic NH 4 H NH 3 K a = ; F H2O HF OH K b = f. CH3NH3CN CH3NH 3 CN basic; CH3NH 3 is a weak acid and CN is a weak base. Since, the solution is basic. 11 CH3NH 3 H CH3NH 2 K a = CN H2O HCN OH K b = = Relationships Between Structure and Strengths of Acids and Bases 113. a. HIO 3 < HBrO 3; As the electronegativity of the central atom increases, acid strength increases. b. HNO 2 < HNO 3; As the number of oxygen atoms attached to the central nitrogen atom increases, acid strength increases. c. HOI < HOCl; Same reasoning as in a. d. H3PO 3< H3PO 4; Same reasoning as in b a. BrO 3 < IO 3; These are the conjugate bases of the acids in Exercise a. Since HBrO 3is the stronger acid, the conjugate base of HBrO 3 (BrO 3) will be the weaker base. IO 3 will be the stronger base since HIO is the weaker acid. 3 b. NO 3 < NO 2; These are the conjugate bases of the acids in Exercise b. Conjugate base strength is inversely related to acid strength. c. OCl < OI. These are the conjugate bases of the acids in Exercise c a. H2O < H2S < H2Se; As the strength of the H S X bond decreases, acid strength increases. b. CH3CO2H < FCH2CO2H < F2CHCO2H < F3CCO2H; As the electronegativity of neighboring atoms increases, acid strength increases. c. NH 4 < HONH 3 ; Same reason as in b. d. NH 4 < PH 4 ; Same reason as in a In general, the stronger the acid, the weaker the conjugate base. a. SeH < SH < OH ; These are the conjugate bases of the acids in Exercise a. The ordering of the base strength is the opposite of the acids.