CAPTER 18 ACID-BASE EQUILIBRIA FOLLOW UP PROBLEMS 18.1A Plan: Examine the formulas and classify each as an acid or base. Strong acids are the hydrohalic acids Cl, Br, and I, and oxoacids in which the number of O atoms exceeds the number of ionizable protons by at least two. Other acids are weak acids. Strong bases are soluble oxides or hydroxides of the Group 1A(1) metals and Ca, Sr, and Ba in Group A(). Other bases are weak bases. a) Chloric acid, ClO 3, is the stronger acid because acid strength increases as the number of O atoms in the acid increases. b) ydrochloric acid, Cl, is one of the strong hydrohalic acids whereas acetic acid, C 3 COO, is a weak carboxylic acid. c) Sodium hydroxide, NaO, is a strong base because Na is a Group 1A(1) metal. Methylamine, C 3 N, is an organic amine and, therefore, a weak base. 18.1B Plan: Examine the formulas and classify each as a strong acid, weak acid, strong base, or weak base. Strong acids are the hydrohalic acids Cl, Br, and I, and oxoacids in which the number of O atoms exceeds the number of ionizable protons by at least two. Other acids are weak acids. Strong bases are soluble oxides or hydroxides of the Group 1A(1) metals and Ca, Sr, and Ba in Group A(). Other bases are weak bases. a) (C 3 ) 3 N is a weak base. It contains a nitrogen atom with a lone pair of electrons, which classifies it as a base; however, it is not one of the strong bases. b) ydroiodic acid, I, is a strong acid (one of the strong acids listed above). c) BrO is a weak acid. It has an ionizable hydrogen, which makes it an acid. Specifically, it is an oxoacid, in which a polyatomic ion is the anion. In the case of this oxoacid, there is only one O atom for each ionizable hydrogen, so this is a weak acid. The reaction for the dissociation of this weak acid is: BrO(aq) O(l) BrO (aq) 3 O (aq). The corresponding equilibrium expression is: K a BrO- [ 3 O ] [BrO] d) Ca(O) is a strong base (one of the strong bases listed above). 18.A Plan: The product of [ 3 O ] and [O ] remains constant at 5 C because the value of K w is constant at a given temperature. Use K w [ 3 O ][O ] 1.0 x 10 14 to solve for [ 3 O ]. Calculating [ 3 O ]: 14 K [ 3 O w 1.0x10 ] 1.495x10 13 1.5x10 13 M [O ] 6.7x10 Since [O ] > [ 3 O ], the solution is basic. 18.B Plan: The product of [ 3 O ] and [O ] remains constant at 5 C because the value of K w is constant at a given temperature. Use K w [ 3 O ][O ] 1.0 x 10 14 to solve for [ 3 O ]. Calculating [O ]: 14 K [O w 1.0x10 ] 5.55555x10 5 5.6x10 5 M 10 [3O ] 1.8x10 Since [O ] > [ 3 O ], the solution is basic. 18.3A Plan: NaO is a strong base that dissociates completely in water. Subtract p from 14.00 to find the po, and calculate inverse logs of p and po to find [ 3 O ] and [O ], respectively. 18-1
p po 14.00 po 14.00 9.5 4.48 p log [ 3 O ] [ 3 O ] 10 p 10 9.5 3.01995x10 10 3.0x10 10 M po log [O ] [O ] 10 po 10 4.48 3.3113x10 5 3.3x10 5 M 18.3B Plan: Cl is a strong acid that dissociates completely in water. Subtract p from 14.00 to find the po, and calculate inverse logs of p and po to find [ 3 O ] and [O ], respectively. p po 14.00 po 14.00.8 11.7 p log [ 3 O ] [ 3 O ] 10 p 10.8 5.481x10 3 5.x10 3 M po log [O ] [O ] 10 po 10 11.7 1.9055x10 1 1.9x10 1 M 18.4A Plan: Identify the conjugate pairs by first identifying the species that donates (the acid) in either reaction direction. The other reactant accepts the and is the base. The acid has one more and 1 greater charge than its conjugate base. a) C 3 COO has one more than C 3 COO. 3 O has one more than O. Therefore, C 3 COO and 3 O are the acids, and C 3 COO and O are the bases. The conjugate acid/base pairs are C 3 COO/C 3 COO and 3 O / O. b) O donates a and acts as the acid. F accepts the and acts as the base. In the reverse direction, F acts as the acid and O acts as the base. The conjugate acid/base pairs are O/O and F/F. 18.4B Plan: To derive the formula of a conjugate base, remove one from the acid and decrease the charge by 1 (acids donate ). To derive the formula of a conjugate acid, add an and increase the charge by 1 (bases accept ). a) Adding a to SO 3 gives the formula of the conjugate acid: SO 3. b) Removing a from C 5 5 N gives the formula of the conjugate base: C 5 5 N c) Adding a to CO 3 gives the formula of the conjugate acid: CO 3. d) Removing a from CN gives the formula of the conjugate base: CN. 18.5A Plan: The two possible reactions involve reacting the acid from one conjugate pair with the base from the other conjugate pair. The reaction that favors the products (K c > 1) is the one in which the stronger acid produces the weaker acid. The reaction that favors reactants (K c < 1) is the reaction is which the weaker acid produces the stronger acid. a) The conjugate pairs are SO 3 (acid)/ SO 3 (base) and CO 3 (acid)/ CO 3 (base). Two reactions are possible: (1) SO 3 CO 3 SO 3 CO 3 and () SO 3 CO 3 SO 3 CO 3 The first reaction is the reverse of the second. Both acids are weak. Of the two, SO 3 is the stronger acid. Reaction (1) with the stronger acid producing the weaker acid favors products and K c > 1. Reaction () with the weaker acid forming the stronger acid favors the reactants and K c < 1. Therefore, reaction 1 is the reaction in which K c > 1. b) The conjugate pairs are F (acid)/f (base) and CN (acid)/cn (base). Two reactions are possible: (1) F CN F CN and () F CN F CN The first reaction is the reverse of the second. Both acids are weak. Of the two, F is the stronger acid. Reaction (1) with the stronger acid producing the weaker acid favors products and K c > 1. Reaction () with the weaker acid forming the stronger acid favors the reactants and K c < 1. Therefore, reaction is the reaction in which K c < 1. 18-
18.5B Plan: For a), write the reaction that shows the reaction of ammonia with water; for b), write a reaction between ammonia and Cl; for c), write the reaction between the ammonium ion and NaO to produce ammonia. a) The following equation describes the dissolution of ammonia in water: N 3 (g) O(l) N 4 (aq) O (aq) weak base weak acid stronger acid strong base Ammonia is a known weak base, so it makes sense that it accepts a from O. The reaction arrow indicates that the equilibrium lies to the left because the question states, you smell ammonia (N 4 and O are odorless). N 4 and O are the stronger acid and base, so the reaction proceeds to the formation of the weaker acid and base. b) The addition of excess Cl results in the following equation: N 3 (g) 3 O (aq; from Cl) N 4 (aq) O(l) stronger base strong acid weak acid weak base Cl is a strong acid and is much stronger than N 4. Similarly, N 3 is a stronger base than O. The reaction proceeds to produce the weak acid and base, and thus the odor from N 3 disappears. c) The solution in (b) is mostly N 4 and O. The addition of excess NaO results in the following equation: N 4 (aq) O (aq; from NaO) N 3 (g) O(l) stronger acid strong base weak base weak acid N 4 and O are the stronger acid and base, respectively, and drive the reaction towards the formation of the weaker base and acid, N 3 (g) and O, respectively. The reaction direction explains the return of the ammonia odor. 18.6A Plan: If A is a stronger acid than B, K c > 1 and more A molecules will produce B molecules. If B is a stronger acid than A, K c < 1 and more B molecules will produce A molecules. There are more B molecules than there are A molecules, so the equilibrium lies to the right and K c > 1. A is the stronger acid. 18.6B Plan: Because D is a stronger acid than C, the reaction of D and C will have K c > 1, and there should be more C molecules than D molecules at equilibrium. There are more green/white acid molecules in the solution than black/white acid molecules. Therefore, the green/white acid molecules represent C, and the black/white acid molecules represent D. The green spheres represent C, and the black spheres represent D. Because the reaction of the stronger acid D with C will have K c > 1, the reverse reaction (C D ) will have K c < 1. 18.7A Plan: Write a balanced equation for the dissociation of N 4 in water. Using the given information, construct a reaction table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and make assumptions where possible to simplify the calculations. Since the p is known, [ 3 O ] can be found; that value can be substituted into the equilibrium expression. N 4 (aq) O(l) 3 O (aq) N 3 (g) Initial 0. M 0 0 Change x x x Equilibrium 0. x x x The initial concentration of N 4 0. M because each mole of N 4 Cl completely dissociates to form one mole of N 4. x [ 3 O ] [N 3 ] 10 p 10 5.0 1.0x10 5 M N33O K a 5 )( 5 x 1.0x10 1.0x10 ) N 4 ( 0. x ) 5 ( 0. 1.0x10 ) 5x10 10 18.7B Plan: Write a balanced equation for the dissociation of acrylic acid in water. Using the given information, construct a reaction table that describes the initial and equilibrium concentrations. Construct an equilibrium 18-3
expression. Since the p is known, [ 3 O ] can be found; that value can be used to find the equilibrium concentrations of all substances, which can then be substituted into the equilibrium expression to solve for the value of K a. CCCOO(aq) O(l) 3 O (aq) CCCOO (aq) Initial 0.3 M 0 0 Change x x x Equilibrium 0.3 x x x According to the information given in the problem, p at equilibrium.43. [ 3 O ] eq 10 p 10.43 3.7154x10 3 3.7x10 3 M x Thus, [ 3 O ] [ CCCOO ] 3.7x10 3 M [ CCCOO] (0.30 x) (0.30 3.7x10 3 ) M 0.963 M K a CCOO [ 3 O ] [ CCOO] K a (3.7x10 3 )(3.7x10 3 ) 4.603x10 5 4.6x10 5 (0.963) 18.8A Plan: Write a balanced equation for the dissociation of OCN in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and solve the quadratic expression for x, the concentration of 3 O. Use the concentration of the hydronium ion to solve for p. OCN(aq) O(l) 3 O (aq) OCN (aq) Initial 0.10 M 0 0 Change x x x Equilibrium 0.10 x x x OCN K a 3.5x10 4 3O x [ OCN] ( 0.10 x ) In this example, the dissociation of OCN is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression is solved using the quadratic formula. x 3.5x10 4 (0.10 x) x 3.5x10 5 3.5x10 4 x x 3.5x10 4 x 3.5x10 5 0 (ax bx c 0) a 1 b 3.5x10 4 c 3.5x10 5 x b ± b 4ac a ( ) ( )( ) 4 4 5 3.5x10 ± 3.5x10 4 1 3.5x10 x 1 ( ) x 5.7436675x10 3 5.7x10 3 M 3 O p log [ 3 O ] log [5.7436675x10 3 ].408.4 18.8B Plan: Write a balanced equation for the dissociation of C 6 5 COO in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Use pk a to solve for the value of K a. Construct an equilibrium expression, use simplifying assumptions when possible to solve for x, the concentration of 3 O. Use the concentration of the hydronium ion to solve for p. C 6 5 COO(aq) O(l) 3 O (aq) C 6 5 COO (aq) Initial 0.5 M 0 0 Change x x x Equilibrium 0.5 x x x 18-4
K a 10 pka 10 4.0 6.3096x10 5 6.3x10 5 K a 6.3x10 5 C 6 5 COO [ 3 O ] (x)(x) Assume x is negligible so 0.5 x 0.5 [C 6 5 COO] (0.5 x) (x)(x) (0.5) 6.3x10 5 x (6.3x10 5 ) (0.5); x 3.9686x10 3 4.0x10 3 Check the assumption by calculating the % error: 4.0x10 3 (100) 1.6% which is smaller than 5%, so the assumption is valid. 0.5 At equilibrium [ 3 O ] eq 4.0x10 3 M p log [ 3 O ] log [4.0x10 3 ].3979.40 18.9A Plan: Write the acid-dissociation reaction and the expression for K a. Set up a reaction table in which x the concentration of the dissociated acid and also [ 3 O ]. Use the expression for K a to solve for x, the concentration of cyanide ion at equilibrium. Then use the initial concentration of CN and the equilibrium concentration of CN to find % dissociation. Concentration CN(aq) O(l) 3 O (aq) CN (aq) Initial 0.75 0 0 Change x x x Equilibrium 0.75 x x x CN [ 3 O ] K a 6.x10 10 [CN] K a 6.x10 10 [x][x] Assume x is small compared to 0.75. [0.75 x] K a 6.x10 10 [x][x] [0.75] x.1564x10 5.x10 5 M Check the assumption by calculating the % error:.x10 5 (100) 0.009% which is smaller than 5%, so the assumption is valid. 0.75 Percent CN dissociated [CN] dissoc [CN] init (100) Percent CN dissociated (.x10 5 ) (100) 0.009% 0.75 18.9B Plan: Write the acid-dissociation reaction and the expression for K a. Percent dissociation refers to the amount of the initial concentration of the acid that dissociates into ions. Use the percent dissociation to find the concentration of acid dissociated, which also equals [ 3 O ]. A will be used as the formula of the acid. Set up a reaction table in which x the concentration of the dissociated acid and [ 3 O ]. Substitute [A], [A ], and [ 3 O ] into the expression for K a to find the value of K a. A(aq) O(l) 3 O (aq) A (aq) dissociated acid Percent A ( 100 ) initial acid x 3.16% 1.5 M (100) [Dissociated acid] x 0.047 M Concentration A(aq) O(l) 3 O (aq) A (aq) 18-5
Initial: 1.5 0 0 Change: x x x Equilibrium: 1.5 x x x [Dissociated acid] x [A ] [ 3 O ] 0.047 M [A] 1.5 M 0.047 M 1.453 M Solving for K a. In the equilibrium expression, substitute the concentrations above and calculate K a. K a 3O A - [A] (0.047)(0.047) 1.503 x 10 3 1.5 x 10 3 (1.453) 18.10A Plan: Write the balanced equation and corresponding equilibrium expression for each dissociation reaction. Calculate the equilibrium concentrations of all species and convert [ 3 O ] to p. Find the equilibrium constant values from Appendix C, K a1 5.6x10 and K a 5.4x10 5. OOCCOO(aq) O(l) OOCCOO (aq) 3 O (aq) CO 4 3O K a1 5.6x10 [ CO 4] OOCCOO (aq) O(l) OOCCOO (aq) 3 O (aq) CO 4 O 3 K a CO 4 5.4x10 5 Assumptions: 1) Since K a1 >> K a, the first dissociation produces almost all of the 3 O, so [ 3 O ] eq [ 3 O ] from C O 4. ) Since K a1 (5.6x10 ) is fairly large, solve the first equilibrium expression using the quadratic equation. OOCCOO(aq) O(l) 3 O (aq) OOCCOO (aq) Initial 0.150 M 0 0 Change x x x Equilibrium 0.150 x x x CO 4 3O K a1 x [ CO 4] ( 0.150 x ) 5.6x10 x 5.6x10 x 8.4x10 3 0 (ax bx c 0) ( ) ( )( ) 3 5.6 x10 ± 5.6 x10 4 1 8.4 x10 x 1 ( ) x 0.067833 M 3 O Therefore, [ 3 O ] [C O 4 ] 0.068 M and p log (0.067833) 1.16856 1.17. The oxalic acid concentration at equilibrium is [ C O 4 ] init [ C O 4 ] dissoc 0.150 0.067833 0.8167 0.08 M. Solve for [C O 4 ] by rearranging the K a expression: CO 4 O 3 K a 5.4x10 5 CO 4 K a CO ( 5 4 5.4x10 ) CO 4 ( 0.067833) 5.4x10 O ( 0.067833) 5 M 3 18.10B Plan: Write the balanced equation and corresponding equilibrium expression for each dissociation reaction. Calculate the equilibrium concentrations of all species and convert [ 3 O ] to p. Find the equilibrium constant values from Appendix C, K a1 4.5x10 7 and K a 4.7x10 11. 18-6
CO 3 (aq) O(l) CO 3 (aq) 3 O (aq) K a1 3O CO 3 4.5x10 7 [ CO 3 ] CO 3 (aq) O(l) CO 3 (aq) 3 O (aq) K a 3O CO 3 4.7x10 11 - CO 3 Assumption: 1) Since K a1 >> K a, the first dissociation produces almost all of the 3 O, so [ 3 O ] eq [ 3 O ] from CO 3. ) Because K a1 (4.7x10 7 ) is fairly small, [ CO 3 ] init x [ CO 3 ] init. Thus, [ CO 3 ] 0.075 M x 0.075 M Solve the first equilibrium expression making the assumption that x is small. CO 3 (aq) O(l) 3 O (aq) CO 3 (aq) Initial 0.075 M 0 0 Change x x x Equilibrium 0.075 x x x K a1 3O CO 3 (x)(x) [ CO 3 ] (0.075 x) (x)(x) (0.075) 4.5x10 7 x (0.075)(4.5x10 7 ); x 1.8371x10-4 1.8x10-4 M Check the assumption by calculating the % error: 1.8x10 4 (100) 0.4% which is smaller than 5%, so the assumption is valid. 0.075 Therefore, [ 3 O ] [CO 3 ] 1.8x10-4 M and p log (1.8x10-4 ) 3.7447 3.74. The carbonic acid concentration at equilibrium is [ CO 3 ] init [ CO 3 ] dissoc 0.075 1.8x10-4 0.0748 0.075 M [ CO 3 ]. Solve for [CO 3 ] by rearranging the K a expression: K a 3O CO 3 4.7x10 11 CO 3 [CO 3 ] K a CO 3 3 O (4.7x10 11 )(1.8x10 4 ) (1.8x10 4 4.7x10 11 M ) 18.11A Plan: Pyridine contains a nitrogen atom that accepts from water to form O ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, convert pk b to K b, make simplifying assumptions (if valid), and solve for [O ]. Calculate [ 3 O ] using [ 3 O ][O ] 1.0x10 14 and convert to p. p K b 10 K b 10 8.77 1.6984x10 9 C 5 5 N(aq) O(l) C 5 5 N (aq) O (aq) Initial 0.10 M 0 0 Change x x x Equilibrium 0.10 x x x C5N5N O K b CNN 5 5 1.6984x10 9 Assume that 0.10 x 0.10. C5N5N O K b x CNN 5 5 ( 0.10 ) x 1.303165x10 5 1.3x10 5 M [O ] [C 5 5 N ] 5 [O ] 1.30365x10 Since ( 100) ( 100) 0.01313 which < 5%, the assumption that the dissociation of [C55N 5] 0.10 C 5 5 N 5 is small is valid. 18-7
14 K [ 3 O w 1.0x10 ] 7.6736x10 10 M 5 [O ] 1.303165x10 p log (7.6736x10 10 ) 9.1149995 9.11 (Since pyridine is a weak base, a p > 7 is expected.) 18.11B Plan: Amphetamine contains a nitrogen atom that accepts from water to form O ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, make simplifying assumptions (if valid), and solve for [O ]. Calculate [ 3 O ] using [ 3 O ][O ] 1.0x10 14 and convert to p. In the information below, the symbol B will be used to represent the formula of amphetamine. B(aq) O(l) B (aq) O (aq) Initial 0.075 M 0 0 Change x x x Equilibrium 0.075 x x x [B] O K b [B] 6.3x10 5 Assume that 0.075 x 0.075. [B] O x K b [B] (0.075) 6.3x10 5 x 0.001737.x10 3 M [O ] [B ] Check the assumption by calculating the % error:.x10 3 0.075 (100).9% which is smaller than 5%, so the assumption is valid. K [ 3 O w ] [O ] 1x10 14.x10 3 4.5x10 1 M p log (4.5x10 1 ) 11.344 11.35 Since amphetamine is a weak base, a p > 7 is expected. 18.1A Plan: The hypochlorite ion, ClO, acts as a weak base in water. Write a balanced equation and equilibrium expression for this reaction. The K b of ClO is calculated from the K a of its conjugate acid, hypochlorous acid, ClO (from Appendix C, K a.9x10 8 ). Make simplifying assumptions (if valid), solve for [O ], convert to [ 3 O ] and calculate p. ClO (aq) O(l) ClO(aq) O (aq) Initial 0.0 M 0 0 Change x x x Equilibrium 0.0 x x x O ClO 14 Kw K b K 1.0x10 3.44876x10 7 8 a.9x10 Since K b is very small, assume [ClO ] eq 0.0 x 0.. [ ClO] O K b ClO x.661x10 4 Therefore, [ClO] [O ].6x10 4 M. [ ClO] K b x ( 0.0 ) 3.44876x10 7 18-8
4 [O ].661x10 Since ( 100 ) ( 100) 0.1313 which < 5%, the assumption that the dissociation of ClO is [ClO ] 0.0 small is valid. [ 3 O 14 1.0x10 ] 3.8079x10 11 M 4.661x10 p log (3.8079x10 11 ) 10.4193 10.4 (Since hypochlorite ion is a weak base, a p > 7 is expected.) 18.1B Plan: The nitrite ion, NO, acts as a weak base in water. Write a balanced equation and equilibrium expression for this reaction. The K b of NO is calculated from the K a of its conjugate acid, nitrous acid, NO (from Appendix C, K a 7.1x10 4 ). Make simplifying assumptions (if valid), solve for [O ], convert to [ 3 O ] and calculate p. NO (aq) O(l) NO (aq) O (aq) Initial 0.80 M 0 0 Change x x x Equilibrium 0.80 x x x [ NO ] O K b NO 14 Kw K b K 1.0x10 1.4 x 10 11 4 a 7.1x10 Since K b is very small, assume [NO ] eq 0.80 x 0.8. [ NO ] O K b x NO (0.80) 1.4x10 11 x 3.3x10-6 M Check the assumption by calculating the % error: 3.3x10 6 (100) 0.00041% which is smaller than 5%, so the assumption is valid. 0.80 Therefore, [NO ] [O ] 3.3x10 6 M. [ 3 O ] 1x10 14 3.3x10 6 3.0x10 9 M p log (3.0x10 9 ) 8.59 8.5 Since nitrite ion is a weak base, a p > 7 is expected. 18.13A Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. a) The ions are K and ClO ; the K is from the strong base KO, and does not react with water. The ClO is from the weak acid ClO, so it reacts with water to produce O ions. Since the base is strong and the acid is weak, the salt derived from this combination will produce a basic solution. K does not react with water. ClO (aq) O(l) ClO (aq) O (aq) b) The ions are C 3 N 3 and NO 3 ; C 3 N 3 is derived from the weak base methylamine, C 3 N. Nitrate ion, NO 3, is derived from the strong acid NO 3 (nitric acid). A salt derived from a weak base and strong acid produces an acidic solution. NO 3 does not react with water. C 3 N 3 (aq) O(l) C 3 N (aq) 3 O (aq) c) The ions are Rb and Br. Rubidium ion is derived from rubidium hydroxide, RbO, which is a strong base because Rb is a Group 1A(1) metal. Bromide ion is derived from hydrobromic acid, Br, a strong hydrohalic 18-9
acid. Since both the base and acid are strong, the salt derived from this combination will produce a neutral solution. Neither Rb nor Br react with water. 18.13B Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. a) The ions are Fe 3 and Br. The Br is the anion of the strong acid Br, so it does not react with water. The Fe 3 ion is small and highly charged, so the hydrated ion, Fe( O) 6 3, reacts with water to produce 3 O. Since the base is weak and the acid is strong, the salt derived from this combination will produce an acidic solution. Br does not react with water. Fe( O) 6 3 (aq) O(l) Fe( O) 5 O (aq) 3 O (aq) b) The ions are Ca and NO ; the Ca is from the strong base Ca(O), and does not react with water. The NO is from the weak acid NO, so it reacts with water to produce O ions. Since the base is strong and the acid is weak, the salt derived from this combination will produce a basic solution. Ca does not react with water. NO (aq) O(l) NO (aq) O (aq) c) The ions are C 6 5 N 3 and I ; C 6 5 N 3 is derived from the weak base aniline, C 6 5 N. Iodide ion, I, is derived from the strong acid I (hydroiodic acid). A salt derived from a weak base and strong acid produces an acidic solution. I does not react with water. C 6 5 N 3 (aq) O(l) C 6 5 N (aq) 3 O (aq) 18.14A Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. a) The two ions that comprise this salt are cupric ion, Cu, and acetate ion, C 3 COO. Metal ions are acidic in water. Assume that the hydrated cation is Cu( O) 6. The K a is found in Appendix C. Cu( O) 6 (aq) O(l) Cu( O) 5 O (aq) 3 O (aq) K a 3x10 8 Acetate ion acts likes a base in water. The K b is calculated from the K a of acetic acid (1.8x10 5 ): 14 Kw K b K 1.0x10 5.6x10 10 5 a 1.8x10 C 3 COO (aq) O(l) C 3 COO(aq) O (aq) K b 5.6x10 10 Cu( O) 6 is a better proton donor than C 3 COO is a proton acceptor (i.e., K a > K b ), so a solution of Cu(C 3 COO) is acidic. b) The two ions that comprise this salt are ammonium ion, N 4, and fluoride ion, F. Ammonium ion is the acid 14 Kw of N 3 with K a K 1.0x10 5.7x10 10. 5 b 1.76x10 N 4 (aq) O(l) N 3 (aq) 3 O (aq) K a 5.7x10 10 14 Kw Fluoride ion is the base with K b K 1.0x10 1.5x10 11. 4 a 6.8x10 F (aq) O(l) F(aq) O (aq) K b 1.5x10 11 Since K a > K b, a solution of N 4 F is acidic. c) The ions are K and SO 3 ; the K is from the strong base KO, and does not react with water. The SO 3 can react as an acid: SO 3 (aq) O(l) SO 3 (aq) 3 O (aq) K a 6.5x10 8 SO 3 can also react as a base. Its K b value can be found by using the K a of its conjugate acid, SO 3. 18-10
SO 3 (aq) O(l) SO 3 (aq) O (aq) Since K a > K b, a solution of KSO 3 is acidic. K b K w K a 1.0x10 1.4x10 14 7.1x10 13 18.14B Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. a) The two ions that comprise this salt are sodium ion, Na, and bicarbonate ion, CO 3. The Na is from the strong base NaO, and does not react with water. The CO 3 can react as an acid: CO 3 (aq) O(l) CO 3 (aq) 3 O (aq) K a 4.7x10 11 CO 3 can also react as a base. Its K b value can be found by using the K a of its conjugate acid, CO 3. CO 3 (aq) O(l) CO 3 (aq) O (aq) K b K w 1.0x10 K 7 a 4.5x10.x10 8 Since K b > K a, a solution of NaCO 3 is basic. b) The two ions that comprise this salt are anilinium ion, C 6 5 N 3, and nitrite ion, NO. 14 Kw Anilinium ion is the acid of C 6 5 N with K a K 1.0x10.5x10 5. 10 b 4.0x10 C 6 5 N 3 (aq) O(l) C 6 5 N (aq) 3 O (aq) K a.5x10 5 14 Kw Nitrite ion is the base with K b K 1.0x10 1.4x10 11. 4 a 7.1x10 NO (aq) O(l) NO (aq) O (aq) K b 1.4x10 11 Since K a > K b, a solution of C 6 5 N 3 NO is acidic. c) The ions are Na and PO 4 ; the Na is from the strong base NaO, and does not react with water. The PO 4 can react as an acid: PO 4 (aq) O(l) PO 4 (aq) 3 O (aq) K a 6.3x10 8 PO 4 can also react as a base. Its K b value can be found by using the K a of its conjugate acid, 3 PO 4. PO 4 (aq) O(l) 3 PO 4 (aq) O (aq) Since K a > K b, a solution of Na PO 4 is acidic. K b K w K a 1.0x10 7.x10 18.15A Plan: A Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor. a) _ O _ O 14 14 3 1.4x10 1 O Al Al O O O O O trigonal planar tetrahedral ydroxide ion, O, donates an electron pair and is the Lewis base; Al(O) 3 accepts the electron pair and is the Lewis acid. Note the change in geometry caused by the formation of the adduct. b) 18-11
O O S O S O O O O O Sulfur trioxide accepts the electron pair and is the Lewis acid. Water donates an electron pair and is the Lewis base. c) 3 N 3 N N 3 Co 6 Co 3 N N 3 N 3 Co 3 accepts six electron pairs and is the Lewis acid. Ammonia donates an electron pair and is the Lewis base. 18.15B Plan: A Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor. a) B(O) 3 is the Lewis acid because it is accepting electron pairs from water, the Lewis base. b) Cd accepts four electron pairs and is the Lewis acid. Each iodide ion donates an electron pair and is the Lewis base. c) Each fluoride ion donates an electron pair to form a bond with boron in SiF 6. The fluoride ion is the Lewis base and the boron tetrafluoride is the Lewis acid. N 3 3 END OF CAPTER PROBLEMS 18.1 The Arrhenius definition classifies substances as being acids or bases by their behavior in the solvent water. 18. All Arrhenius acids contain hydrogen and produce hydronium ion ( 3 O ) in aqueous solution. All Arrhenius bases contain an O group and produce hydroxide ion (O ) in aqueous solution. Neutralization occurs when each 3 O molecule combines with an O molecule to form two molecules of O. Chemists found that the rxn was independent of the combination of strong acid with strong base. In other words, the reaction of any strong base with any strong acid always produced 56 kj/mol ( 56 kj/mol). This was consistent with Arrhenius s hypothesis describing neutralization, because all other counter ions (those present from the dissociation of the strong acid and base) were spectators and did not participate in the overall reaction. 18.3 The Arrhenius acid-base definition is limited by the fact that it only classifies substances as an acid or base when dissolved in the single solvent water. The anhydrous neutralization of N 3 (g) and Cl(g) would not be included in the Arrhenius acid-base concept. In addition, it limits a base to a substance that contains O in its formula. N 3 does not contain O in its formula but produces O ions in O. 18.4 Strong acids and bases dissociate completely into their ions when dissolved in water. Weak acids only partially dissociate. The characteristic property of all weak acids is that a significant number of the acid molecules are not dissociated. For a strong acid, the concentration of hydronium ions produced by dissolving the acid is equal to the initial concentration of the undissociated acid. For a weak acid, the concentration of hydronium ions produced when the acid dissolves is less than the initial concentration of the acid. 18.5 Plan: Recall that an Arrhenius acid contains hydrogen and produces hydronium ion ( 3 O ) in aqueous solution. a) Water, O, is an Arrhenius acid because it produces 3 O ion in aqueous solution. Water is also an Arrhenius base because it produces the O ion as well. 18-1
b) Calcium hydroxide, Ca(O) is a base, not an acid. c) Phosphorous acid, 3 PO 3, is a weak Arrhenius acid. It is weak because the number of O atoms equals the number of ionizable atoms. d) ydroiodic acid, I, is a strong Arrhenius acid. 18.6 Only (a) NaSO 4 18.7 Plan: All Arrhenius bases contain an O group and produce hydroxide ion (O ) in aqueous solution. Barium hydroxide, Ba(O), and potassium hydroxide, KO, (b and d) are Arrhenius bases because they contain hydroxide ions and form O when dissolved in water. 3 AsO 4 and ClO, (a) and (c), are Arrhenius acids, not bases. 18.8 (b) O is a very weak Arrhenius base. 18.9 Plan: K a is the equilibrium constant for an acid dissociation which has the generic equation O A(aq) O(l) 3 O (aq) A 3 A (aq). The K a expression is. [ O] is treated as a constant [ A] and omitted from the expression. Write the acid-dissociation reaction for each acid, following the generic equation, and then write the K a expression. a) CN(aq) O(l) 3 O (aq) CN (aq) 3O CN K a [ CN] b) CO 3 (aq) O(l) 3 O (aq) CO 3 (aq) 3O CO 3 K a CO 3 c) COO(aq) O(l) 3 O (aq) COO (aq) 3O COO K a COO 18.10 a) C 3 N 3 (aq) O(l) 3 O (aq) C 3 N (aq) 3O C3N K a C3N 3 b) ClO(aq) O(l) 3 O (aq) ClO (aq) 3O ClO K a [ ClO] c) S(aq) O(l) 3 O (aq) S (aq) 3O S K a S 18-13
18.11 Plan: K a is the equilibrium constant for an acid dissociation which has the generic equation O A(aq) O(l) 3 O (aq) A 3 A (aq). The K a expression is. [ O] is treated as a constant [ A] and omitted from the expression. Write the acid-dissociation reaction for each acid, following the generic equation, and then write the K a expression. a) NO (aq) O(l) 3 O (aq) NO (aq) 3O NO K a [ NO ] b) C 3 COO(aq) O(l) 3 O (aq) C 3 COO (aq) 3O C3COO K a C 3COO c) BrO (aq) O(l) 3 O (aq) BrO (aq) 3O BrO K a [ BrO ] 18.1 a) PO 4 (aq) O(l) 3 O (aq) PO 4 (aq) 3O PO 4 K a PO 4 b) 3 PO (aq) O(l) 3 O (aq) PO (aq) 3O PO K a 3PO c) SO 4 (aq) O(l) 3 O (aq) SO 4 (aq) 3O SO 4 K a SO 4 18.13 Plan: K a values are listed in the Appendix. The larger the K a value, the stronger the acid. The K a value for hydroiodic acid, I, is not shown because K a approaches infinity for strong acids and is not meaningful. I is the strongest acid (it is one of the six strong acids), and acetic acid, C 3 COO, is the weakest: C 3 COO < F < IO 3 < I 18.14 Cl > NO > ClO > CN 18.15 Plan: Strong acids are the hydrohalic acids Cl, Br, I, and oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more; these include NO 3, SO 4, and ClO 4. All other acids are weak acids. Strong bases are metal hydroxides (or oxides) in which the metal is a Group 1A(1) metal or Ca, Sr, or Ba in Group A(). Weak bases are N 3 and amines. a) Arsenic acid, 3 AsO 4, is a weak acid. The number of O atoms is four, which exceeds the number of ionizable atoms, three, by one. This identifies 3 AsO 4 as a weak acid. b) Strontium hydroxide, Sr(O), is a strong base. Soluble compounds containing O ions are strong bases. Sr is a Group metal. c) IO is a weak acid. The number of O atoms is one, which is equal to the number of ionizable atoms identifying IO as a weak acid. d) Perchloric acid, ClO 4, is a strong acid. ClO 4 is one example of the type of strong acid in which the number of O atoms exceeds the number of ionizable atoms by more than two. 18-14
18.16 a) weak base b) strong base c) strong acid d) weak acid 18.17 Plan: Strong acids are the hydrohalic acids Cl, Br, I, and oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more; these include NO 3, SO 4, and ClO 4. All other acids are weak acids. Strong bases are metal hydroxides (or oxides) in which the metal is a Group 1A(1) metal or Ca, Sr, or Ba in Group A(). Weak bases are N 3 and amines. a) Rubidium hydroxide, RbO, is a strong base because Rb is a Group 1A(1) metal. b) ydrobromic acid, Br, is a strong acid, because it is one of the listed hydrohalic acids. c) ydrogen telluride, Te, is a weak acid, because is not bonded to an oxygen or halide. d) ypochlorous acid, ClO, is a weak acid. The number of O atoms is one, which is equal to the number of ionizable atoms identifying ClO as a weak acid. 18.18 a) weak base b) strong acid c) weak acid d) weak acid 18.19 Autoionization reactions occur when a proton (or, less frequently, another ion) is transferred from one molecule of the substance to another molecule of the same substance. O(l) O(l) 3 O (aq) O (aq) SO 4 (l) SO 4 (l) 3 SO 4 (solvated) SO 4 (solvated) 18.0 O(l) O(l) 3 O (aq) O (aq) 3O O K c [ O ] [ O] is a constant and is included with the value of K c to obtain K w : K w [ O] x K c [ 3 O ][O ] 18.1 a) p increases by a value of 1. b) [ 3 O ] increases by a factor of 1000. 18. Plan: The lower the concentration of hydronium ( 3 O ) ions, the higher the p. p increases as K a or the molarity of acid decreases. Recall that pk a log K a. a) At equal concentrations, the acid with the larger K a will ionize to produce more hydronium ions than the acid with the smaller K a. The solution of an acid with the smaller K a 4x10 5 has a lower [ 3 O ] and higher p. b) pk a is equal to log K a. The smaller the K a, the larger the pk a is. So the acid with the larger pk a, 3.5, has a lower [ 3 O ] and higher p. c) Lower concentration of the same acid means lower concentration of hydronium ions produced. The 0.01 M solution has a lower [ 3 O ] and higher p. d) At the same concentration, strong acids dissociate to produce more hydronium ions than weak acids. The 0.1 M solution of a weak acid has a lower [ 3 O ] and higher p. e) Bases produce O ions in solution, so the concentration of hydronium ion for a solution of a base solution is lower than that for a solution of an acid. The 0.01 M base solution has the higher p. f) po equals log [O ]. At 5 C, the equilibrium constant for water ionization, K w, equals 1x10 14 so 14 p po. As po decreases, p increases. The solution of po 6.0 has the higher p. 18.3 Plan: Part a) can be approached two ways. Because NaO is a strong base, the [O ] eq [NaO] init. One method involves calculating [ 3 O ] using K w [ 3 O ][O ], then calculating p from the relationship p log [ 3 O ]. The other method involves calculating po and then using p po 14.00 to calculate p. Part b) also has two acceptable methods analogous to those in part a); only one method will be shown. a) First method: K w [ 3 O ][O ] [ 3 O ] Kw [O ] 14 1.0x10 0.0111 9.0090x10 13 M 18-15
p log [ 3 O ] log (9.0090x10 13 ) 1.0453 1.05 Second method: po log [O ] log (0.0111) 1.954677 p 14.00 po 14.00 1.954677 1.0453 1.05 With a p > 7, the solution is basic. b) For a strong acid such as Cl: [ 3 O ] [Cl] 1.35x10 3 M p log (1.35x10 3 ).869666 po 14.00.869666 11.130334 11.13 With a p < 7, the solution is acidic. 18.4 a) p log (0.0333) 1.47756 1.478; acidic b) po log (0.0347) 1.45967 1.460; basic 18.5 Plan: I is a strong acid, so [ 3 O ] [I] and the p can be calculated from the relationship p log [ 3 O ]. Ba(O) is a strong base, so [O ] x [Ba(O) ] and po log [O ]. a) [ 3 O ] [I] 6.14x10 3 M. p log (6.14x10 3 ).1183.1. Solution is acidic. b) [O ] x [Ba(O) ] (.55 M) 5.10 M po log (5.10) 0.70757 0.708. Solution is basic. 18.6 a) po log (7.5x10 4 ) 3.1378 p 14.00 3.1378 10.876 10.88 basic b) p log (1.59x10 3 ).79860 po 14.00.79860 11.0140 11.0 acidic 18.7 Plan: The relationships are: p log [ 3 O ] and [ 3 O ] 10 p ; po log [O ] and [O ] 10 po ; and 14 p po. a) [ 3 O ] 10 p 10 9.85 1.415375x10 10 1.4x10 10 M 3 O po 14.00 p 14.00 9.85 4.15 [O ] 10 po 10 4.15 7.0794578x10 5 7.1x10 5 M O b) p 14.00 po 14.00 9.43 4.57 [ 3 O ] 10 p 10 4.57.691535x10 5.7x10 5 M 3 O [O ] 10 po 10 9.43 3.715353x10 10 3.7x10 10 M O 18.8 a) [ 3 O ] 10 p 10 3.47 3.38844x10 4 3.4x10 4 M 3 O po 14.00 p 14.00 3.47 10.53 [O ] 10 po 10 10.53.95109x10 11 3.0x10 11 M O b) p 14.00 po 14.00 4.33 9.67 [ 3 O ] 10 p 10 9.67.13796x10 10.1x10 10 M 3 O [O ] 10 po 10 4.33 4.67735x10 5 4.7x10 5 M O 18.9 Plan: The relationships are: p log [ 3 O ] and [ 3 O ] 10 p ; po log [O ] and [O ] 10 po ; and 14 p po. a) [ 3 O ] 10 p 10 4.77 1.6984x10 5 1.7x10-5 M 3 O po 14.00 p 14.00 4.77 9.3 [O ] 10 po 10 9.3 5.8884x10 10 5.9x10 10 M O b) p 14.00 po 14.00 5.65 8.35 [ 3 O ] 10 p 10 8.35 4.46684x10 9 4.5x10 9 M 3 O [O ] 10 po 10 5.65.387x10 6.x10 6 M O 18.30 a) [ 3 O ] 10 p 10 8.97 1.071519x10 9 1.1x10 9 M 3 O 18-16
po 14.00 p 14.00 8.97 5.03 [O ] 10 po 10 5.03 9.335x10 6 9.3x10 6 M O b) p 14.00 po 14.00 11.7.73 [ 3 O ] 10 p 10.73 1.86087x10 3 1.9x10 3 M 3 O [O ] 10 po 10 11.7 5.3703x10 1 5.4x10 1 M O 18.31 Plan: The p is increasing, so the solution is becoming more basic. Therefore, O ion is added to increase the p. Since one mole of 3 O will react with one mole of O, the difference in [ 3 O ] would be equal to the [O ] added. Use the relationship [ 3 O ] 10 p to find [ 3 O ] at each p. [ 3 O ] 10 p 10 3.15 7.07946x10 4 M 3 O [ 3 O ] 10 p 10 3.65.387x10 4 M 3 O Add (7.07946x10 4 M.387x10 4 M) 4.84074x10 4 4.8x10 4 mol of O per liter 18.3 The p is decreasing so the solution is becoming more acidic. Therefore, 3 O ion is added to decrease the p. [ 3 O ] 10 p 10 9.33 4.67735x10 10 M 3 O [ 3 O ] 10 p 10 9.07 8.51138x10 10 M 3 O Add (8.51138x10 10 M 4.67735x10 10 M) 3.83403x10 10 3.8x10 10 mol of 3 O per liter 18.33 Plan: The p is increasing, so the solution is becoming more basic. Therefore, O ion is added to increase the p. Since one mole of 3 O will react with one mole of O, the difference in [ 3 O ] would be equal to the [O ] added. Use the relationship [ 3 O ] 10 p to find [ 3 O ] at each p. [ 3 O ] 10 p 10 4.5 3.01995x10 5 M 3 O [ 3 O ] 10 p 10 5.5 5.63413x10 6 M 3 O 3.01995x10 5 M 5.63413x10 6 M.4576x10 5 M O must be added. Moles of O 5.4576x10 mol ( 5.6 L) 1.3763x10 4 1.4x10 4 mol of O L 18.34 The p is decreasing so the solution is becoming more acidic. Therefore, 3 O ion is added to decrease the p. [ 3 O ] 10 p 10 8.9 1.06x10 9 M 3 O [ 3 O ] 10 p 10 6.33 4.67735x10 7 M 3 O Add (4.67735x10 7 M 1.06x10 9 M)(87.5 ml)(10 3 L/1 ml) 4.0816x10 8 4.1x10 8 mol of 3 O 18.35 Scene A has a p of 4.8. [ 3 O ] 10 p 10 4.8 1.58489x10 5 M 3 O Scene B: [ 3 O ] ( 5 5 spheres 1.58489x10 M 3O ) spheres p log [ 3 O ] log [1.98x10 4 ] 3.7 1.98x10 4 M 3 O 18.36 Plan: Apply Le Chatelier s principle in part a). In part b), given that the p is 6.80, [ 3 O ] can be calculated by using the relationship [ 3 O ] 10 p. The problem specifies that the solution is neutral (pure water), meaning [ 3 O ] [O ]. A new K w can then be calculated. a) eat is absorbed in an endothermic process: O(l) heat 3 O (aq) O (aq). As the temperature increases, the reaction shifts to the formation of products. Since the products are in the numerator of the K w expression, rising temperature increases the value of K w. b) [ 3 O ] 10 p 10 6.80 1.58489x10 7 M 3 O 1.6x10 7 M [ 3 O ] [O ] K w [ 3 O ][O ] (1.58489x10 7 )(1.58489x10 7 ).511876x10 14.5x10 14 For a neutral solution: p po 6.80 18-17
18.37 The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors, while the Arrhenius definition looks at acids as containing ionizable atoms and at bases as containing hydroxide ions. In both definitions, an acid produces hydronium ions and a base produces hydroxide ions when added to water. Ammonia, N 3, and carbonate ion, CO 3, are two Brønsted-Lowry bases that are not Arrhenius bases because they do not contain hydroxide ions. Brønsted-Lowry acids must contain an ionizable atom in order to be proton donors, so a Brønsted-Lowry acid that is not an Arrhenius acid cannot be identified. (Other examples are also acceptable.) 18.38 Every acid has a conjugate base, and every base has a conjugate acid. The acid has one more and one more positive charge than the base from which it was formed. 18.39 a) Acid-base reactions are proton transfer processes. Thus, the proton will be transferred from the stronger acid to the stronger base to form the weaker acid and weaker base. b) B(aq) A (aq) A(aq) B (aq) The spontaneous direction of a Brønsted-Lowry acid-base reaction is that the stronger acid will transfer a proton to the stronger base to produce the weaker acid and base. Thus at equilibrium there should be relatively more of weaker acid and base present than there will be of the stronger acid and base. Since there is more A and B in sample and less B and A, B must be the stronger acid and A must be the stronger base. 18.40 An amphoteric substance can act as either an acid or a base. In the presence of a strong base (O ), the dihydrogen phosphate ion acts like an acid by donating hydrogen: PO 4 (aq) O (aq) O(aq) PO 4 (aq) In the presence of a strong acid (Cl), the dihydrogen phosphate ion acts like a base by accepting hydrogen: PO 4 (aq) Cl(aq) 3 PO 4 (aq) Cl (aq) 18.41 Plan: K a is the equilibrium constant for an acid dissociation which has the generic equation O A(aq) O(l) 3 O (aq) A 3 A (aq). The K a expression is. [ O] is treated as a constant [ A] and omitted from the expression. Write the acid-dissociation reaction for each acid, following the generic equation, and then write the K a expression. a) When phosphoric acid is dissolved in water, a proton is donated to the water and dihydrogen phosphate ions are generated. 3 PO 4 (aq) O(l) PO 4 (aq) 3 O (aq) 3O PO 4 K a 3PO4 b) Benzoic acid is an organic acid and has only one proton to donate from the carboxylic acid group. The atoms bonded to the benzene ring are not acidic hydrogens. C 6 5 COO(aq) O(l) C 6 5 COO (aq) 3 O (aq) 3O C65COO K a C65COO c) ydrogen sulfate ion donates a proton to water and forms the sulfate ion. SO 4 (aq) O(l) SO 4 (aq) 3 O (aq) 3O SO 4 K a SO 4 18.4 a) Formic acid, an organic acid, has only one proton to donate from the carboxylic acid group. The remaining atom, bonded to the carbon, is not an acidic hydrogen. COO(aq) O(l) COO (aq) 3 O (aq) 18-18
K a 3O COO [ COO] b) When chloric acid is dissolved in water, a proton is donated to the water and chlorate ions are generated. ClO 3 (aq) O(l) ClO 3 (aq) 3 O (aq) 3O ClO 3 K a ClO3 c) The dihydrogen arsenate ion donates a proton to water and forms the hydrogen arsenate ion. AsO 4 (aq) O(l) AsO 4 (aq) 3 O (aq) 3O AsO 4 K a AsO 4 18.43 Plan: To derive the conjugate base, remove one from the acid and decrease the charge by 1 (acids donate ). Since each formula in this problem is neutral, the conjugate base will have a charge of 1. a) Cl b) CO 3 c) O 18.44 a) PO 4 3 b) N 3 c) S 18.45 Plan: To derive the conjugate acid, add an and increase the charge by 1 (bases accept ). a) N 4 b) N 3 c) C 10 14 N 18.46 a) O b) SO 4 c) 3 O 18.47 Plan: The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid. a) Cl O Cl 3 O acid base conjugate base conjugate acid Conjugate acid-base pairs: Cl/Cl and 3 O / O b) ClO 4 SO 4 ClO 4 3 SO 4 acid base conjugate base conjugate acid Conjugate acid-base pairs: ClO 4 /ClO 4 and 3 SO 4 / SO 4 Note: Perchloric acid is able to protonate another strong acid, SO 4, because perchloric acid is a stronger acid. (ClO 4 s oxygen atoms exceed its hydrogen atoms by one more than SO 4.) c) PO 4 SO 4 PO 4 SO 4 base acid conjugate acid conjugate base Conjugate acid-base pairs: SO 4 /SO 4 and PO 4 /PO 4 18.48 a) N 3 NO 3 N 4 NO 3 base acid conjugate acid conjugate base Conjugate pairs: NO 3 /NO 3 ; N 4 /N 3 b) O O O O base acid conjugate acid conjugate base Conjugate pairs: O /O ; O/O c) N 4 BrO 3 N 3 BrO 3 acid base conjugate base conjugate acid Conjugate pairs: N 4 /N 3 ; BrO 3 /BrO 3 18.49 Plan: The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid. 18-19
a) N 3 3 PO 4 N 4 PO 4 base acid conjugate acid conjugate base Conjugate acid-base pairs: 3 PO 4 / PO 4 ; N 4 /N 3 b) C 3 O N 3 C 3 O N base acid conjugate acid conjugate base Conjugate acid-base pairs: N 3 /N ; C 3 O/C 3 O c) PO 4 SO 4 PO 4 SO 4 base acid conjugate acid conjugate base Conjugate acid-base pairs: SO 4 /SO 4 ; PO 4 /PO 4 18.50 a) N 4 CN N 3 CN acid base conjugate base conjugate acid Conjugate acid-base pairs: N 4 /N 3 ; CN/CN b) O S O S acid base conjugate base conjugate acid Conjugate acid-base pairs: O/O ; S/S c) SO 3 C 3 N SO 3 C 3 N 3 acid base conjugate base conjugate acid Conjugate acid-base pairs: SO 3 /SO 3 ; C 3 N 3 /C 3 N 18.51 Plan: Write total ionic equations (show all soluble ionic substances as dissociated into ions) and then remove the spectator ions to write the net ionic equations. The (aq) subscript denotes that each species is soluble and dissociates in water. The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid. a) Na (aq) O (aq) Na (aq) PO 4 (aq) O(l) Na (aq) PO 4 (aq) Net: O (aq) PO 4 (aq) O(l) PO 4 (aq) base acid conjugate acid conjugate base Conjugate acid-base pairs: PO 4 / PO 4 and O/O b) K (aq) SO 4 (aq) K (aq) CO 3 (aq) K (aq) SO 4 (aq) K (aq) CO 3 (aq) Net: SO 4 (aq) CO 3 (aq) SO 4 (aq) CO 3 (aq) acid base conjugate base conjugate acid Conjugate acid-base pairs: SO 4 /SO 4 and CO 3 /CO 3 18.5 a) 3 O (aq) CO 3 (aq) CO 3 (aq) O(l) acid base conjugate acid conjugate base Conjugate acid-base pairs: 3 O / O; CO 3 /CO 3 b) N 4 (aq) O (aq) N 3 (aq) O(l) acid base conjugate base conjugate acid Conjugate acid-base pairs: N 4 /N 3 ; O/O 18.53 Plan: The two possible reactions involve reacting the acid from one conjugate pair with the base from the other conjugate pair. The reaction that favors the products (K c > 1) is the one in which the stronger acid produces the weaker acid. The reaction that favors reactants (K c < 1) is the reaction is which the weaker acid produces the stronger acid. The conjugate pairs are S (acid)/s (base) and Cl (acid)/cl (base). Two reactions are possible: (1) S Cl S Cl and () S Cl S Cl The first reaction is the reverse of the second. Cl is a strong acid and S a weak acid. Reaction (1) with the stronger acid producing the weaker acid favors products and K c > 1. Reaction () with the weaker acid forming the stronger acid favors the reactants and K c < 1. 18.54 K c > 1: NO 3 F NO 3 F K c < 1: NO 3 F NO 3 F 18-0
18.55 Plan: An acid-base reaction that favors the products (K c > 1) is one in which the stronger acid produces the weaker acid. Use the figure to decide which of the two acids is the stronger acid. a) Cl N 3 N 4 Cl strong acid stronger base weak acid weaker base Cl is ranked above N 4 in the list of conjugate acid-base pair strength and is the stronger acid. N 3 is ranked above Cl and is the stronger base. N 3 is shown as a stronger base because it is stronger than Cl, but is not considered a strong base. The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and K c > 1. The stronger acid is more likely to donate a proton than the weaker acid. b) SO 3 N 3 SO 3 N 4 stronger acid stronger base weaker base weaker acid SO 3 is ranked above N 4 and is the stronger acid. N 3 is a stronger base than SO 3. The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and K c > 1. 18.56 Neither a or b have K c > 1. 18.57 Plan: An acid-base reaction that favors the reactants (K c < 1) is one in which the weaker acid produces the stronger acid. Use the figure to decide which of the two acids is the weaker acid. a) N 4 PO 4 N 3 PO 4 weaker acid weaker base stronger base stronger acid K c < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left. b) SO 3 S SO 3 S - weaker base weaker acid stronger acid stronger base K c < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left. 18.58 a) K c < 1 b) K c > 1 18.59 a) The concentration of a strong acid is very different before and after dissociation since a strong acid exhibits 100% dissociation. After dissociation, the concentration of the strong acid approaches 0, or [A] 0. b) A weak acid dissociates to a very small extent (<<100%), so the acid concentration after dissociation is nearly the same as before dissociation. c) Same as b), but the percent, or extent, of dissociation is greater than in b). d) Same as a) 18.60 No, Cl and C 3 COO are never of equal strength because Cl is a strong acid with K a > 1 and C 3 COO is a weak acid with K a < 1. The K a of the acid, not the concentration of 3 O in a solution of the acid, determines the strength of the acid. 18.61 Plan: We are given the percent dissociation of the original A solution (33%), and we know that the percent dissociation increases as the acid is diluted. Thus, we calculate the percent dissocation of each diluted sample and see which is greater than 33%. To determine percent dissociation, we use the following formula: Percent A dissociated ([A] dissoc /[A] init ) X 100, with [A] dissoc equal to the number of 3 O (or A ) ions and [A] init equal to the number of A plus the number of 3 O (or A ) Calculating the percent dissociation of each diluted solution: Solution 1. Percent dissociation [4/(54)] X 100 44% Solution. Percent dissociation [/(7 )] X 100 % Solution 3. Percent dissociation [3/(6 3)] X 100 33% Therefore, scene 1 represents the diluted solution. 18-1
18.6 Water will add approximately 10 7 M to the 3 O concentration. (The value will be slightly lower than for pure water.) a) C 3 COO(aq) O(l) 3 O (aq) C 3 COO (aq) 0.10 x x x K a 1.8x10 5 3O C3COO C3COO ( K a 1.8x10 5 x)( x) Assume x is small compared to 0.1 so 0.1 x 0.1. 0.1 x ( ) K a 1.8x10 5 ( x )( x ) ( 0.1) x 1.3416x10 3 M Since the 3 O concentration from C 3 COO is many times greater than that from O, [ 3 O ] [C 3 COO ]. b) The extremely low C 3 COO concentration means the 3 O concentration from C 3 COO is near that from O. Thus [ 3 O ] [C 3 COO ]. c) C 3 COO(aq) O(l) 3 O (aq) C 3 COO (aq) C 3 COONa(aq) C 3 COO (aq) Na (aq) K a 1.8x10 5 ( x )( 0.1 x ) Assume x is small compared to 0.1. ( 0.1 x) x [ 3 O ] 1.8x10 5 [C 3 COO ] 0.1 x 0.1 M Thus, [C 3 COO ] > [ 3 O ] 18.63 The higher the negative charge on a species, the more difficult it is to remove a positively charged ion. 18.64 Plan: Write the acid-dissociation reaction and the expression for K a. Set up a reaction table and substitute the given value of [ 3 O ] for x; solve for K a. Butanoic acid dissociates according to the following equation: C 3 C C COO(aq) O(l) 3 O (aq) C 3 C C COO (aq) Initial: 0.15 M 0 0 Change: x x x Equilibrium: 0.15 x x x According to the information given in the problem, [ 3 O ] eq 1.51x10 3 M x Thus, [ 3 O ] [C 3 C C COO ] 1.51x10 3 M [C 3 C C COO] (0.15 x) (0.15 1.51x10 3 ) M 0.14849 M 3O C3CCCOO K a C3CCCOO ( 3 )( 3 1.51x10 1.51x10 ) K a 0.14849 1.5355x10 5 1.5x10 5 ( ) 18.65 Any weak acid dissociates according to the following equation: A(aq) O(l) 3 O (aq) A (aq) [ 3 O ] 10 p 10 4.88 1.3186x10 5 M Thus, [ 3 O ] [A ] 1.3186x10 5 M, and [A] (0.035 1.3186x10 5 ) 0.03499 M 3O A K a A 18-