CHAPTER 11 ACIDS AND BASES

CHAPTER 11 ACIDS AND BASES 11.1 Table 11.5 of the text contains a list of important Brønsted acids and bases. (a) both (why?), (b) base, (c) acid, (d) base, (e) acid, (f) base, (g) base, (h) base, (i) acid, (j) acid. 11.2 Recall that the conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid. (a) (b) (c) (d) (e) nitrite ion: NO2 hydrogen sulfate ion (also called bisulfate ion): HSO4 hydrogen sulfide ion (also called bisulfide ion): HS cyanide ion: CN formate ion: HCOO 11.3 In general the components of the conjugate acidbase pair are on opposite sides of the reaction arrow. The base always has one fewer proton than the acid. (a) (b) (c) (d) (e) The conjugate acidbase pairs are (1) HCN (acid) and CN (base) and (2) CH3COO (base) and CH3COOH (acid). (1) HCO3 (acid) and CO3 2 (base) and (2) HCO3 (base) and H2CO3 (acid). (1) H2PO4 (acid) and HPO4 2 (base) and (2) NH3 (base) and NH4 (acid). (1) HClO (acid) and ClO (base) and (2) CH3NH2 (base) and CH3NH3 (acid). (1) H2O (acid) and OH (base) and (2) CO3 2 (base) and HCO3 (acid). 11.4 The conjugate acid of any base is just the base with a proton added. (a) H2S (b) H2CO3 (c) HCO3 (d) H3PO4 (e) H2PO4 (f) HPO4 2 (g) H2SO4 (h) HSO4 (i) HSO3 11.5 (a) The Lewis structures are - O O C O C O H and - O O C O C O - (b) H and C2H2O4 can act only as acids, C2HO4 can act as both an acid and a base, and C2O4 2 can act only as a base. 11.6 The conjugate base of any acid is simply the acid minus one proton. (a) CH2ClCOO (b) IO4 (c) H2PO4 (d) HPO4 2 (f) HSO4 (g) SO4 2 (h) IO3 (i) SO3 2 (k) HS (l) S 2 (m) OCl (e) PO4 3 (j) NH3

11.7 (a) Lewis acid; see the reaction with water shown in Section 11.1 of the text. (b) Lewis base; water can donate a pair of electrons to H to form H3O. (c) (d) Lewis base. Lewis acid; SO2 reacts with water to form H2SO3. Compare to CO2 above. Actually, SO2 can also act as a Lewis base under some circumstances. (e) (f) (g) (h) Lewis base; see the reaction with H to form ammonium ion. Lewis base; see the reaction with H to form water. Lewis acid; does H have any electron pairs to donate? Lewis acid; compare to the example of NH3 reacting with BF3. 11.8 AlCl3 is a Lewis acid with an incomplete octet of electrons and Cl is the Lewis base donating a pair of electrons. Cl Cl - Cl Al Cl + - Cl Cl Al Cl Cl 11.9 (a) Both molecules have the same acceptor atom (boron) and both have exactly the same structure (trigonal planar). Fluorine is more electronegative than chlorine so we would predict based on electronegativity arguments that boron trifluoride would have a greater affinity for unshared electron pairs than boron trichloride. (b) Since it has the larger positive charge, iron(iii) should be a stronger Lewis acid than iron(ii). 11.10 By definition Brønsted acids are proton donors, therefore such compounds must contain at least one hydrogen atom. In Problem 11.7, Lewis acids that do not contain hydrogen, and therefore are not Brønsted acids, are CO2, SO2, and BCl3. Can you name others? 11.11 No. The ph depends on the H + activity, which for aqueous solutions, depends on the autoionization of water. The autoionization constant of water (K W) is temperature dependent. At higher temperatures, K W gets larger as does the hydrogen ion concentration (and thus the hydrogen ion activity) of neutral water. Hence, the ph of neutral water decreases with increasing temperature. For example, the ph of neutral water is 6.51 at 60 C [see Table 11.2 in the text]. If the ph and temperature of a solution are known, it can be determined if it is an acidic or basic solution. 11.12 Yes. The ph is defined as the negative log of the hydrogen ion activity, which, for dilute solutions can be approximated by the concentration in M. Hence the ph for 1.0 M HCl(aq) can be approximated as 0 [ph = -log(1.0) = 0]; however, because 1.0 M is not very dilute the actual ph will differ slightly. If the hydrogen ion concentration is greater than about 1.0 M, then the ph can be negative. For example, 2.0 M HCl(aq) has a ph of approximately 0.3. 11.13 [H ] 1.4 10 3 M Assuming that the solution is dilute enough that hydrogen ion activities can be approximated by concentrations, we have

14 w 1.0 10 [OH ] K 7.1 10 3 [H ] 1.4 10 12 M 11.14 [OH ] 0.62 M Assuming that the solution is dilute enough that ion activities can be approximated by concentrations, we have 14 w 1.0 10 [H ] K 1.6 10 [OH ] 0.62 14 Note: The concentration of [OH ] is not very dilute, so the above answer only represents an approximation. 11.15 (a) HCl is a strong acid, so the concentration of hydrogen ion is also 0.0010 M. (What is the concentration of chloride ion?) We use the definition of ph. ph -log a(h + ) log[h ] log(0.0010) 3.00 M (b) KOH is an ionic compound and completely dissociates into ions. Assuming activities can be approximated by concentrations, we first find the concentration of hydrogen ion. 14 Kw 1.0 10 14 [H ] 1.3 10 M [OH ] 0.76 The ph is then found from its defining equation 11.16 (a) Ba(OH)2 is ionic and fully ionized in water. The concentration of the hydroxide ion is 5.6 10 4 M (Why? What is the concentration of Ba 2?) Assuming activities can be approximated by concentrations, we find the hydrogen ion concentration. 14 Kw 1.0 10 [H ] 1.8 10 4 [OH ] 5.6 10 11 The ph is then: ph log[h ] log(1.8 10 11 ) 10.74 M (b) Nitric acid is a strong acid, so the concentration of hydrogen ion is also 5.2 10 4 M. The ph is: 11.17 Since ph log[h ], we write [H ] 10 ph ph log[h ] log(5.2 10 4 ) 3.28 ph log[h ] log[1.3 10 14 ] 13.89 (a) [H ] 10 2.42 3.8 10 3 M (c) [H ] 10 6.96 1.1 10 7 M (b) [H ] 10 11.21 6.2 10 12 M (d) [H ] 10 15.00 1.0 10 15 M 11.18 Strategy: Here we are given the ph of a solution and asked to calculate [H ]. Because ph is defined as ph log[h ], we can solve for [H ] by taking the antilog of the ph; that is, [H ] 10 ph.

Solution: From Equation 11.5 of the text: (a) ph log [H ] 5.20 log[h ] 5.20 To calculate [H ], we need to take the antilog of 5.20. [H ] 10 5.20 6.3 10 6 M Check: Because the ph is between 5 and 6, we can expect [H ] to be between 1 10 5 M and 1 10 6 M. Therefore, the answer is reasonable. (b) ph log [H ] 16.00 log[h ] 16.00 [H ] 10 16.00 1.0 10 16 M (c) Strategy: We are given the concentration of OH ions and asked to calculate [H ]. The relationship between [H ] and [OH ] in water or an aqueous solution is given by the ion-product of water, Kw [Equation (11.3) of the text]. Solution: The ion product of water is applicable to all aqueous solutions. At 25C, Kw 1.0 10 14 [H ][OH ] Rearranging the equation to solve for [H ], we write 14 14 1.0 10 1.0 10 6 [H ] 2.7 10 M 9 [OH ] 3.7 10 Check: Since the [OH ] < 1 10 7 M we expect the [H ] to be greater than 1 10 7 M. 11.19 The ph can be found by using Equation (11.8) of the text. ph 14.00 poh 14.00 9.40 4.60 The hydrogen ion concentration can be found as in Example 11.5 of the text. Taking the antilog of both sides: 4.60 log[h ] [H ] 2.5 10 5 M 11.20 1 L 0.360 mol 5.50 ml 1000 ml 1 L 3 1.98 10 mol KOH KOH is a strong base and therefore ionizes completely. The OH concentration equals the KOH concentration, because there is a 1:1 mole ratio between KOH and OH. [OH ] 0.360 M

poh log[oh ] 0.444 11.21 We can calculate the OH concentration from the poh. poh 14.00 ph 14.00 10.00 4.00 [OH ] 10 poh 1.0 10 4 M Since NaOH is a strong base, it ionizes completely. The OH concentration equals the initial concentration of NaOH. [NaOH] 1.0 10 4 mol L 1 So, we need to prepare 546 ml of 1.0 10 4 M NaOH. This is a dimensional analysis problem. We need to perform the following unit conversions. mol L 1 mol NaOH grams NaOH 546 ml 0.546 L 4 1.0 10 mol NaOH 40.00 g NaOH 3? g NaOH 546 ml 2.2 10 g NaOH 1000 ml soln 1 mol NaOH 11.22 Molarity of the HCl solution is: 1 mol HCl 18.4 g HCl 36.46 g HCl 3 662 10 L 0.762 M ph log(0.762) 0.118 11.23 The 0.1 M HNO 3 solution has the highest ionic concentration and thus would exhibit a greater deviation from ideal behavior, than the other two. 11.24 The 0.0560 M NaOH solution has the highest ionic concentration and thus would exhibit a greater deviation from ideal behavior, than the other two. 11.25 Using Equation 11.3 we can determine the H + concentration from the OH concentration and K w. Table 11.2 lists the value of K w as a function of temperature. At 60 C (333 K), K w equals 9.61 10 14. K W H OH 14 KW 9.61 10 OH 6.4 10 5 H 1.5 10 9 M 11.26 If we take the negative log of both sides of Equation 11.3 and then use the definitions of ph, poh, and pk W, we can derive an equation that relates ph, poh, and pk W. K W H OH

KW log log H OH KW log log H log OH KW Using the equation above, we can determine the ph given the poh and pk W. Table 11.2 lists the value of K w as a function of temperature. At 40 C (313 K), K w equals 2.92 10 14 and pk W = - log(k W) = -log(2.92 10-14 ) = 13.5. ph? pk poh = 13.5 4.5 9.0 W 11.27 A strong acid, such as HCl, will be completely ionized, choice (b). A weak acid will only ionize to a lesser extent compared to a strong acid, choice (c). A very weak acid will remain almost exclusively as the acid molecule in solution. Choice (d) is the best choice. 11.28 (1) The two steps in the ionization of a weak diprotic acid are: H2A(aq) H2O(l) H3O (aq) HA (aq) HA (aq) H2O(l) H3O (aq) A 2 (aq) The diagram that represents a weak diprotic acid is (c). In this diagram, we only see the first step of the ionization, because HA is a much weaker acid than H2A. (2) Both (b) and (d) are chemically implausible situations. Because HA is a much weaker acid than H2A, you would not see a higher concentration of A 2 compared to HA. 11.29 (a) strong acid, (b) weak acid, (c) strong acid (first stage of ionization), (d) weak acid, (e) weak acid, (f) weak acid, (g) strong acid, (h) weak acid, (i) weak acid. 11.30 (a) strong base (b) weak base (c) weak base (d) weak base (e) strong base 11.31 The maximum possible concentration of hydrogen ion in a 0.10 M solution of HA is 0.10 M. This is the case if HA is a strong acid. If HA is a weak acid, the hydrogen ion concentration is less than 0.10 M. The ph corresponding to 0.10 M [H ] is 1.00. (Why three digits?) For a smaller [H ] the ph is larger than 1.00 (why?). (a) false, the ph is greater than 1.00 (b) false, they are equal (c) true (d) false 11.32 (a) false, they are equal (b) true, find the value of log(1.00) on your calculator (c) true (d) false, if the acid is strong, [HA] 0.00 M 11.33 The direction should favor formation of F (aq) and H2O(l). Hydroxide ion is a stronger base than fluoride ion, and hydrofluoric acid is a stronger acid than water. 11.34 Cl is the conjugate base of the strong acid, HCl. It is a negligibly weak base and has no affinity for protons. Therefore, the reaction will not proceed from left to right to any measurable extent.

Another way to think about this problem is to consider the possible products of the reaction. CH3COOH(aq) Cl (aq) HCl(aq) CH3COO (aq) The favored reaction is the one that proceeds from right to left. HCl is a strong acid and will ionize completely, donating all its protons to the base, CH3COO. 11.35 If we assume that H and S do not change going from 25 C (298 K) to 60 C (333 K), we can use the van t Hoff equation [Equation (10.16) in the text]. ln K 2 Hū 1 1 K 1 R T 1 T 2 Plugging our numbers in to the above equation, we get the following: K 2 61530 J mol-1 1 ln 4 7.1 10 8.3145J mol -1 K -1 298 K 1 333K 2.61. We can solve for the equilibrium constant at 60 C (K 2) by raising both sides of the above equation to the power of e and then multiplying both sides by 7.1 10 4. K 2 7.1 10 4 e 2.61 5.2 10 5 Because the reaction is exothermic the equilibrium constant decreases with increasing temperature. 11.36 They are assumed to completely ionize. 11.37 Equilibrium constants, such as K a, tend to be temperature dependent so the temperature should be specified when K a constants are given. 11.38 Because all of the acids listed are monoprotic and the solutions have the same concentration, the solution of the weakest acid (smallest K a) will have the highest ph. HClO 4 is considered a strong acid and according to Table 11.3 formic acid (HCOOH) has a larger K a than acetic acid (CH 3COOH). Hence, the solution of acetic acid would have the largest ph. 11.39 We set up a table for the dissociation. C6H5COOH(aq) H (aq) C6H5COO (aq) Initial (M): 0.10 0.00 0.00 Change (M): x x x Equilibrium (M): (0.10 x) x x K a [H ][C H COO ] 6 5 6 5 [C H COOH] 6.5 10 5 2 x (0.10 x) x 2 (6.5 10 5 )x (6.5 10 6 ) 0 Solving the quadratic equation:

x 2.5 10 3 M [H ] ph log(2.5 10 3 ) 2.60 This problem could be solved more easily if we could assume that (0.10 x) 0.10. If the assumption were mathematically valid, then it would not be necessary to solve a quadratic equation, as we did above. Re-solve the problem above, making the assumption. Was the assumption valid? What is our criterion for deciding? 11.40 Strategy: Recall that a weak acid only partially ionizes in water. We are given the initial quantity of a weak acid (CH3COOH) and asked to calculate the concentrations of H, CH3COO, and CH3COOH at equilibrium. First, we need to calculate the initial concentration of CH3COOH. In determining the H concentration, we ignore the ionization of H2O as a source of H, so the major source of H ions is the acid. We follow the procedure outlined in Section 11.4 of the text. Solution: Step 1: Calculate the concentration of acetic acid before ionization. 1 mol acetic acid 4 0.0560 g acetic acid 9.33 10 mol acetic acid 60.05 g acetic acid 4 9.33 10 mol 0.0500 L soln 0.0187 M acetic acid Step 2: We ignore water's contribution to [H ]. We consider CH3COOH as the only source of H ions. Step 3: Letting x be the equilibrium concentration of H and CH3COO ions in mol L 1, we summarize: CH3COOH(aq) H (aq) CH3COO (aq) Initial (M): 0.0187 0 0 Change (M): x x x Equilibrium (M): 0.0187 x x x Step 4: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x. You can look up the Ka value in Table 11.3 of the text. K a [H ][CH 3 COO ] [CH 3 COOH] 5 ( x)( x) 1.8 10 (0.0187 x) At this point, we can make an assumption that x is very small compared to 0.0187. Hence, 0.0187 x 0.0187 5 ( x)( x) 1.8 10 0.0187

x 5.8 10 4 M [H ] [CH3COO ] [CH3COOH] (0.0187 5.8 10 4 )M 0.0181 M Check: Testing the validity of the assumption, 4 5.8 10 0.0187 The assumption is valid. 100% 3.1% 5% 11.41 First we find the hydrogen ion concentration. [H ] 10 ph 10 6.20 6.3 10 7 M If the concentration of [H ] is 6.3 10 7 M, that means that 6.3 10 7 M of the weak acid, HA, ionized because of the 1:1 mole ratio between HA and H. Setting up a table: HA(aq) H (aq) A (aq) Initial (M): 0.010 0 0 Change (M): 6.3 10 7 6.3 10 7 6.3 10 7 Equilibrium (M): 0.010 6.3 10 7 6.3 10 7 Substituting into the acid ionization constant expression: K a [H ][A ] [HA] (6.3 107 )(6.3 10 7 ) 0.010 4.0 10 11 We have omitted the contribution to [H ] due to water. 11.42 A ph of 3.26 corresponds to a H activity of 5.5 10 4, which will give an approximate H + concentration of 5.5 10 4 M,. Let the original concentration of formic acid be I. If the concentration of [H ] is 5.5 10 4 M, that means that 5.5 10 4 M of HCOOH ionized because of the 1:1 mole ratio between HCOOH and H. HCOOH(aq) H (aq) HCOO (aq) Initial (M): I 0 0 Change (M): 5.5 10 4 5.5 10 4 5.5 10 4 Equilibrium (M): I (5.5 10 4 ) 5.5 10 4 5.5 10 4 Substitute Ka and the equilibrium concentrations into the ionization constant expression (assuming dilute solution behavior) to solve for I. K a [H ][HCOO ] [HCOOH] 1.7 10 4 4 2 (5.5 10 ) 4 x (5.5 10 ) I [HCOOH] 2.3 10 3 M 11.43 (a) Set up a table showing initial and equilibrium concentrations.

C6H5COOH(aq) H (aq) C6H5COO (aq) Initial (M): 0.20 0.00 0.00 Change (M): x x x Equilibrium (M): (0.20 x) x x Using the value of Ka from Table 11.3 of the text: K a [H ][C 6 H 5 COO ] [C 6 H 5 COOH] 6.5 10 5 2 x (0.20 x) We assume that x is small so (0.20 x) 0.20 2 5 x 6.5 10 0.20 x 3.6 10 3 M [H ] [C6H5COO ] 3 3.6 10 M Percent ionization 100% 1.8% 0.20 M (b) Set up a table as above. C6H5COOH(aq) H (aq) C6H5COO (aq) Initial (M): 0.00020 0.00000 0.00000 Change (M): x x x Equilibrium (M): (0.00020 x) x x Using the value of Ka from Table 11.3 of the text: K a [H ][C 6 H 5 COO ] [C 6 H 5 COOH] 5 6.5 10 2 x (0.00020 x) In this case we cannot apply the approximation that (0.00020 x) 0.00020 (see the discussion in Example 11.11of the text). We obtain the quadratic equation: x 2 (6.5 10 5 )x (1.3 10 8 ) 0 The positive root of the equation is x 8.6 10 5 M. (Is this less than 5% of the original concentration, 0.00020 M? That is, is the acid more than 5% ionized?) The percent ionization is then: 5 8.6 10 M Percent ionization 100% 43% 0.00020 M

Note that the extent to which a weak acid ionizes depends on the initial concentration of the acid. The more dilute the solution, the greater the percent ionization (see Figure 11.8 of the text). 11.44 Percent ionization is defined as: ionized acid concentration at equilibrium percent ionization 100% initial concentration of acid For a monoprotic acid, HA, the concentration of acid that undergoes ionization is equal to the concentration of H ions or the concentration of A ions at equilibrium. Thus, we can write: [H ] percent ionization 100% [HA] 0 (a) First, recognize that hydrofluoric acid is a weak acid. It is not one of the six strong acids, so it must be a weak acid. Step 1: Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x, that represents the change in concentration. Let (x) be the depletion in concentration ( mol L 1 ) of HF. From the stoichiometry of the reaction, it follows that the increase in concentration for both H and F must be x. Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations. HF(aq) H (aq) F (aq) Initial (M): 0.60 0 0 Change (M): x x x Equilibrium (M): 0.60 x x x Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x. K a [H ][F ] [HF] You can look up the Ka value for hydrofluoric acid in Table 11.3 of your text. 4 ( x)( x) 7.1 10 (0.60 x) At this point, we can make an assumption that x is very small compared to 0.60. Hence, 0.60 x 0.60 Oftentimes, assumptions such as these are valid if K is very small. A very small value of K means that a very small amount of reactants go to products. Hence, x is small. If we did not make this assumption, we would have to solve a quadratic equation. Solving for x. 4 ( )( ) 7.1 10 x x 0.60 x 0.021 M [H ]

Step 3: Having solved for the [H ], calculate the percent ionization. percent ionization [H ] [HF] 0 100% 0.021 M 100% 3.5% 0.60 M (b) (c) are worked in a similar manner to part (a). However, as the initial concentration of HF becomes smaller, the assumption that x is very small compared to this concentration will no longer be valid. You must solve a quadratic equation. (b) K a [H ][F ] [HF] x 2 (0.0046 x) 7.1 104 x 2 (7.1 10 4 )x (3.3 10 6 ) 0 x 1.5 10 3 M 3 1.5 10 M Percent ionization 100% 33% 0.0046 M (c) K a [H ][F ] [HF] x 2 (0.00028 x) 7.1 104 x 2 (7.1 10 4 )x (2.0 10 7 ) 0 x 2.2 10 4 M 4 2.2 10 M Percent ionization 100% 79% 0.00028 M As the solution becomes more dilute, the percent ionization increases. 11.45 Given 14% ionization, the concentrations must be: [H ] [A ] 0.14 0.040 M 0.0056 M [HA] (0.040 0.0056) M 0.034 M The value of Ka can be found by substitution. 11.46 The equilibrium is: [H ][A ] (0.0056) K a 9.2 10 [HA] 0.034 2 4 C9H8O4(aq) H (aq) C9H7O4 (aq) Initial (M): 0.20 0 0 Change (M): x x x Equilibrium (M): 0.20 x x x

(a) K a [H ][C 9 H 7 O 4 ] [C 9 H 8 O 4 ] Assuming (0.20 x) 0.20 x [H ] 7.7 10 3 M 3 7.7 10 Percent ionization x M 100% 100% 3.9% 0.20 0.20 M (b) At ph 1.00 the concentration of hydrogen ion is 0.10 M ([H ] 10 ph ). The extra hydrogen ions will tend to suppress the ionization of the weak acid (LeChâtelier's principle, Section 10.4 of the text). The position of equilibrium is shifted in the direction of the unionized acid. Let's set up a table of concentrations with the initial concentration of H equal to 0.10 M. C9H8O4(aq) H (aq) C9H7O4 (aq) Initial (M): 0.20 0.10 0 Change (M): x x x Equilibrium (M): 0.20 x 0.10 x x K a 9 7 4 [H ][C H O ] [C H O ] 9 8 4 4 x(0.10 x) 3.0 10 (0.20 x) Assuming (0.20 x) 0.20 and (0.10 x) 0.10 x 6.0 10 4 M 6.0 10 Percent ionization x M 100% 100% 0.30% 0.20 0.20 M The high acidity of the gastric juices appears to enhance the rate of absorption of unionized aspirin molecules through the stomach lining. In some cases this can irritate these tissues and cause bleeding. 11.47 (a) We construct the usual table. 4 NH3(aq) H2O(l) NH4 (aq) OH (aq) Initial (M): 0.10 0.00 0.00 Change (M): x x x Equilibrium (M): (0.10 x) x x K b [NH 4 ][OH ] [NH 3 ] 5 1.8 10 2 x (0.10 x)

Assuming (0.10 x) 0.10, we have: 2 5 x 1.8 10 0.10 x 1.3 10 3 M [OH ] poh log(1.3 10 3 ) 2.89 ph 14.00 2.89 11.11 By following the identical procedure, we can show: (b) ph 8.96. 11.48 Strategy: Weak bases only partially ionize in water. B(aq) H2O(l) BH (aq) OH (aq) Note that the concentration of the weak base given refers to the initial concentration before ionization has started. The ph of the solution, on the other hand, refers to the situation at equilibrium. To calculate Kb, we need to know the concentrations of all three species, [B], [BH ], and [OH ] at equilibrium. We ignore the ionization of water as a source of OH ions. Solution: We proceed as follows. Step 1: The major species in solution are B, OH, and the conjugate acid BH. Step 2: First, we need to calculate the hydroxide ion concentration from the ph value. Calculate the poh from the ph. Then, calculate the OH concentration from the poh. poh 14.00 ph 14.00 10.66 3.34 poh log[oh ] poh log[oh ] Taking the antilog of both sides of the equation, 10 poh [OH ] [OH ] 10 3.34 4.6 10 4 M Step 3: If the concentration of OH is 4.6 10 4 M at equilibrium, that must mean that 4.6 10 4 M of the base ionized. We summarize the changes. B(aq) H2O(l) BH (aq) OH (aq) Initial (M): 0.30 0 0 Change (M): 4.6 10 4 4.6 10 4 4.6 10 4 Equilibrium (M): 0.30 (4.6 10 4 ) 4.6 10 4 4.6 10 4 Step 4: Substitute the equilibrium concentrations into the ionization constant expression to solve for Kb.

K b [BH ][OH ] [B] 4 2 (4.6 10 ) K b 7.1 10 7 (0.30) 11.49 A ph of 11.22 corresponds to a [H ] of 6.03 10 12 M and a [OH ] of 1.66 10 3 M. Setting up a table: NH3(aq) H2O(l) NH4 (aq) OH (aq) Initial (M): I 0.00 0.00 Change (M): 1.66 10 3 1.66 10 3 1.66 10 3 Equilibrium (M): I (1.66 10 3 ) 1.66 10 3 1.66 10 3 K b [NH 4 ][OH ] [NH 3 ] 1.8 10 5 3 3 (1.66 10 )(1.66 10 ) 3 I (1.66 10 ) Assuming 1.66 10 3 is small relative to x, then 11.50 The reaction is: x 0.15 M [NH3] NH3(aq) H2O(l) NH4 (aq) OH (aq) Initial (M): 0.080 0 0 Change (M): x x x Equilibrium (M): 0.080 x x x At equilibrium we have: K a [NH 4 ][OH ] [NH 3 ] 2 2 5 x x 1.8 10 (0.080 x) 0.080 x 1.2 10 3 M 1.2 10 Percent NH 3 present as NH4 100% 1.5% 0.080 11.51 Phosphoric acid is a weak acid ( K a1 = 7.5 10 3 ) and its ionization constants decrease markedly for the second ( K a2 = 6.2 10 8 ) and third stages ( K a3 3 = 4.8 10 13 ) [see Table 11.6 in the text]. Thus, in a solution containing phosphoric acid, the concentration of the nonionized acid (H 3PO 4) is the highest, and the only other species present in significant concentrations are H + and H 2PO 4 ions. H 3PO 4 acts as a Brønsted acid, H 2PO 4 can act either as a Brønsted acid or base.

11.52 (a) The ph of a 0.040 M HCl solution (strong acid) is: ph log(0.040) 1.40. (b) Follow the procedure for calculating the ph of a diprotic acid to calculate the ph of the sulfuric acid solution. Strategy: Determining the ph of a diprotic acid in aqueous solution is more involved than for a monoprotic acid. The first stage of ionization for H2SO4 goes to completion. We follow the procedure for determining the ph of a strong acid for this stage. The conjugate base produced in the first ionization (HSO4 ) is a weak acid. We follow the procedure for determining the ph of a weak acid for this stage. Solution: We proceed according to the following steps. Step 1: H2SO4 is a strong acid. The first ionization stage goes to completion. The ionization of H2SO4 is H2SO4(aq) H (aq) HSO4 (aq) The concentrations of all the species (H2SO4, H, and HSO4 ) before and after ionization can be represented as follows. H2SO4(aq) H (aq) HSO4 (aq) Initial (M): 0.040 0 0 Change (M): 0.040 0.040 0.040 Final (M): 0 0.040 0.040 Step 2: Now, consider the second stage of ionization. HSO4 is a weak acid. Set up a table showing the concentrations for the second ionization stage. Let x be the change in concentration. Note that the initial concentration of H is 0.040 M from the first ionization. HSO4 (aq) H (aq) SO4 2 (aq) Initial (M): 0.040 0.040 0 Change (M): x x x Equilibrium (M): 0.040 x 0.040 x x Write the ionization constant expression for Ka. Then, solve for x. You can find the Ka value in Table 11.6 of the text. K a [H ][SO 4 2 ] [HSO 4 ] 2 (0.040 x)( x) 1.3 10 (0.040 x) Since Ka is quite large, we cannot make the assumptions that 0.040 x 0.040 and 0.040 x 0.040 Therefore, we must solve a quadratic equation. x 2 0.053x (5.2 10 4 ) 0

x 2 4 0.053 (0.053) 4(1)( 5.2 10 ) 2(1) 0.053 0.070 x 2 x 8.5 10 3 M or x 0.062 M The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer. Step 3: Having solved for x, we can calculate the H concentration at equilibrium. We can then calculate the ph from the H concentration. [H ] 0.040 M x [0.040 (8.5 10 3 )]M 0.049 M ph log(0.049) 1.31 Without doing any calculations, could you have known that the ph of the sulfuric acid would be lower (more acidic) than that of the hydrochloric acid? The concentrations of HSO 4, SO 4 2 and H + are 0.039 M, 8.5 10 3 M and 0.49 M, respectively, 11.53 For the first stage of ionization: H2CO3(aq) H (aq) HCO3 (aq) Initial (M): 0.025 0.00 0.00 Change (M): x x x Equilibrium (M): (0.025 x) x x K a 1 [H ][HCO 3 ] [H 2 CO 3 ] 2 2 7 x x 4.2 10 (0.025 x) 0.025 x 1.0 10 4 M For the second ionization, HCO3 (aq) H (aq) CO3 2 (aq) Initial (M): 1.0 10 4 1.0 10 4 0.00 Change (M): x x x Equilibrium (M): (1.0 10 4 ) x (1.0 10 4 ) x x K a2 [H ][CO 3 2 ] [HCO 3 ] 4 4 11 [(1.0 10 ) x]( x) (1.0 10 )( x) 4.8 10 4 4 (1.0 10 ) x (1.0 10 ) x 4.8 10 11 M

Since HCO3 is a very weak acid, there is little ionization at this stage. Therefore we have: [H ] [HCO3 ] 1.0 10 4 M and [CO3 2 ] x 4.8 10 11 M 11.54 Strategy We can follow Example 11.14 to solve this problem. The initial concentration of HCl is less than 10 6 M; therefore, the autoionization of water cannot be ignored. Because this is a strong acid, we can use Equation 11.26 to calculate the ph. Solution Using the initial concentration of HCl given, Equation 11.26 gives x 2 (2.0 10 7 ) x (1.0 10 14 ) = 0 This is a quadratic equation, so the solution can be determined using the quadratic formula: x?? 2? 4 2.010 (2.010 )? (1)(?.010 ) 2 x = 2.4 10 7 M or 4.1 10 8 M The second solution is physically impossible, so x = [H + ] = 2.4 10 7 M, which gives a ph of 6.62. 11.55 To answer this question we can follow the procedure in Example 11.13. The K a of formic acid is 1.7 10 4. Plugging the appropriate values in to Equation (11.25) we get the following: 4 14 4 7 M 4 14 M 3 2???????????????????????????????????.7 10 1.0 10? 0 3 4 2 11 18 H 1.7 10 H 3.4 10 H 1.7 10 0 The roots of the equation above can be solved by hand using algebra, solved by using computer software, or by plotting the above equation and determining for what values of [H + ] the lefthand side of the above equation equals zero. Using the latter method it was found that 2.4 10-7 M, was one of the roots of the above equation. Hence, the ph of a 2.0 10-7 M solution of formic acid is 6.6. Notice that both the autoionization of water and the dissociation of formic acid had to be considered for this problem. 11.56 Whenever K a2 is small enough that it is valid to assume that the second dissociation does not change the concentrations of HA and H +, then [A 2 ] = K a2. 11.57 For the first stage of ionization: H3PO4(aq) H (aq) H 2PO4 (aq) Initial (M): 0.0015 0.00 0.00 Change (M): x x x Equilibrium (M): (0.0015 x) x x

K a 1 [H ][HCO 3 ] [H 2 CO 3 ] 7.5 10 3 x 2 (0.0015 x) We can not use an approximation to solve for x. Because the initial concentration is so low, it is not valid to assume that x is negligible compared to the initial concentration. So, we have to solve the following quadratic equation: x 2 7.5 10 3 x 1.125 10 5 0 Solve this quadratic equation, we get x = 1.28 10 3 M For the second ionization, H 2PO4 (aq) H (aq) HPO4 2 (aq) Initial (M): 1.28 10 3 1.28 10 3 0.00 Change (M): x x x Equilibrium (M): (1.28 10 3 ) x (1.28 10 3 ) x x K a2 [H ][HPO 4 2 ] [H 2 PO 4 ] 6.2 10 8 [(1.28 103 ) x](x) (1.28 10 3 ) x (1.28 103 )(x) (1.28 10 3 ) x 6.2 10 8 M Since H 2PO4 is a very weak acid, there is little ionization at this stage. Hence, the hydrogen ion concentration and ph are not changed by the second dissociation step. ph log(1.28 10 3 ) = 2.89 11.58 The strength of the HX bond is the dominant factor in determining the strengths of binary acids. As with the hydrogen halides (see Section 11.5 of the text), the HX bond strength decreases going down the column in Group 6A. The compound with the weakest HX bond will be the strongest binary acid: H2Se > H2S > H2O. 11.59 All the listed pairs are oxoacids that contain different central atoms whose elements are in the same group of the periodic table and have the same oxidation number. In this situation the acid with the most electronegative central atom will be the strongest. (a) (b) H2SO4 > H2SeO4. H3PO4 > H3AsO4 11.60 The CHCl2COOH is a stronger acid than CH2ClCOOH. Having two electronegative chlorine atoms compared to one, will draw more electron density toward itself, making the OH bond

more polar. The hydrogen atom in CHCl2COOH is more easily ionized compared to the hydrogen atom in CH2ClCOOH. 11.61 The conjugate bases are C6H5O for phenol and CH3O for methanol. The C6H5O is stabilized by resonance: O O O O The CH3O ion has no such resonance stabilization. A more stable conjugate base means an increase in the strength of the acid. 11.62 (a) The K cation does not hydrolyze. The Br anion is the conjugate base of the strong acid HBr. Therefore, Br will not hydrolyze either, and the solution is neutral, ph 7. (b) (c) (d) Al 3 is a small metal cation with a high charge, which hydrolyzes to produce H ions. The NO3 anion does not hydrolyze. It is the conjugate base of the strong acid, HNO3. The solution will be acidic, ph < 7. The Ba cation does not hydrolyze. The Cl anion is the conjugate base of the strong acid HCl. Therefore, Cl will not hydrolyze either, and the solution is neutral, ph 7. Bi 3 is a small metal cation with a high charge, which hydrolyzes to produce H ions. The NO3 anion does not hydrolyze. It is the conjugate base of the strong acid, HNO3. The solution will be acidic, ph < 7. 11.63 Strategy: In deciding whether a salt will undergo hydrolysis, ask yourself the following questions: Is the cation a highly charged metal ion or an ammonium ion? Is the anion the conjugate base of a weak acid? If yes to either question, then hydrolysis will occur. In cases where both the cation and the anion react with water, the ph of the solution will depend on the relative magnitudes of Ka for the cation and Kb for the anion (see Table 11.8 of the text). Solution: We first break up the salt into its cation and anion components and then examine the possible reaction of each ion with water. (a) (b) (c) (d) The Na cation does not hydrolyze. The Br anion is the conjugate base of the strong acid HBr. Therefore, Br will not hydrolyze either, and the solution is neutral. The K cation does not hydrolyze. The SO3 2 anion is the conjugate base of the weak acid HSO3 and will hydrolyze to give HSO3 and OH. The solution will be basic. Both the NH4 and NO2 ions will hydrolyze. NH4 is the conjugate acid of the weak base NH3, and NO2 is the conjugate base of the weak acid HNO2. We know that the Ka of NH4 (5.6 10 10 ) is greater than the Kb of NO2 (2.2 10 11 ). Therefore, the solution will be acidic. Cr 3 is a small metal cation with a high charge, which hydrolyzes to produce H ions. The NO3 anion does not hydrolyze. It is the conjugate base of the strong acid, HNO3. The solution will be acidic.

11.64 There are two possibilities: (i) MX is the salt of a strong acid and a strong base so that neither the cation nor the anion react with water to alter the ph and (ii) MX is the salt of a weak acid and a weak base with Ka for the acid equal to Kb for the base. The hydrolysis of one would be exactly offset by the hydrolysis of the other. 11.65 There is an inverse relationship between acid strength and conjugate base strength. As acid strength decreases, the proton accepting power of the conjugate base increases. In general the weaker the acid, the stronger the conjugate base. All three of the potassium salts ionize completely to form the conjugate base of the respective acid. The greater the ph, the stronger the conjugate base, and therefore, the weaker the acid. The order of increasing acid strength is HZ < HY < HX. 11.66 The salt, sodium acetate, completely dissociates upon dissolution, producing 0.36 M [Na ] and 0.36 M [CH3COO ] ions. The [CH3COO ] ions will undergo hydrolysis because they are a weak base. CH3COO (aq) H2O(l) CH3COOH(aq) OH (aq) Initial (M): 0.36 0.00 0.00 Change (M): x x x Equilibrium (M): (0.36 x) x x K b [CH 3 COOH][OH ] [CH 3 COO ] 10 5.6 10 2 x (0.36 x) Assuming (0.36 x) 0.36, then x [OH ] 1.4 10 5 poh log(1.4 10 5 ) 4.85 ph 14.00 4.85 9.15 11.67 The salt ammonium chloride completely ionizes upon dissolution, producing 0.42 M [NH4 ] and 0.42 M [Cl ] ions. NH4 will undergo hydrolysis because it is a weak acid (NH4 is the conjugate acid of the weak base, NH3). Step 1: Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x, that represents the change in concentration. Let (x) be the depletion in concentration ( mol L 1 ) of NH4. From the stoichiometry of the reaction, it follows that the increase in concentration for both H3O and NH3 must be x. Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations. NH4 (aq) H2O(l) NH3(aq) H3O (aq) Initial (M): 0.42 0.00 0.00 Change (M): x x x Equilibrium (M): (0.42 x) x x

is Step 2: You can calculate the Ka value for NH4 from the Kb value of NH3. The relationship or Ka Kb Kw K 14 Kw 1.0 10 a K 5 b 1.8 10 5.6 10 10 Step 3: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x. K a [NH 3 ][H 3 O ] [NH 4 ] 2 2 10 x x 5.6 10 0.42 x 0.42 x [H ] 1.5 10 5 M ph log(1.5 10 5 ) 4.82 Since NH4Cl is the salt of a weak base (aqueous ammonia) and a strong acid (HCl), we expect the solution to be slightly acidic, which is confirmed by the calculation. 11.68 HCO3 H CO3 2 Ka 4.8 10 11 HCO3 H2O H2CO3 OH K 14 Kw 1.0 10 b K 7 a 4.2 10 2.4 10 8 HCO3 has a greater tendency to hydrolyze than to ionize (Kb > Ka). The solution will be basic (ph > 7). 11.69 The acid and base reactions are: acid: base: HPO4 2 (aq) H (aq) PO4 3 (aq) HPO4 2 (aq) H2O(l) H2PO4 (aq) OH (aq) Ka for HPO4 2 is 4.8 10 13. Note that HPO4 2 is the conjugate base of H2PO4, so Kb is 1.6 10 7. Comparing the two K's, we conclude that the monohydrogen phosphate ion is a much stronger proton acceptor (base) than a proton donor (acid). The solution will be basic. 11.70 Strategy Determining the equilibrium concentrations of the species of a diprotic acid in aqueous solution is more involved than for a monoprotic acid. We can follow the same procedure as that used as in Example 11.15. Solution We proceed as follows: First ionization: The first ionization of oxalic acid is given by C 2H 2O 4(aq) H+ (aq) + C 2H 2O 4 (aq) K a1 = 6.5 10 2

Letting x be the equilibrium concentration of H + and C 2H 2O 4, Equation 11.19 gives K a1 x 2 [C 2 HO 4 ] 0 Š x 6.5 10 Š2 x 2 0.015 Š x Because the equilibrium constant is not that small we can not use an approximation and must calculate x using the quadratic equation: x 2 + (6.5 10 2 ) x 9.75 10 4 = 0 The result is x = 1.3 10-2 M. The second ionization constant is significantly smaller than the first (6.1 10 5 versus 6.5 10 2 ), so the second ionization will make very little additional contribution to the H + ion concentration. Therefore, we can calculate the ph of the solution at this point: ph = log 10(1.3 10-2 ) = 1.9 11.71 The most basic oxides occur with metal ions having the lowest positive charges (or lowest oxidation numbers). (a) Al2O3 < BaO < K2O (b) CrO3 < Cr2O3 < CrO 11.72 (a) 2HCl(aq) Zn(OH)2(s) ZnCl2(aq) 2H2O(l) (b) 2OH (aq) Zn(OH)2(s) Zn(OH)4 2 (aq) 11.73 Al(OH)3 is an amphoteric hydroxide. The reaction is: Al(OH)3(s) OH (aq) Al(OH)4 (aq) This is a Lewis acid-base reaction. Can you identify the acid and base? 11.74 The answer to this depends upon the enthalpy of reaction for the acid dissociation process: HA(aq) A (aq) + H + (aq) If H o is positive (endothermic) then increasing the temperature will shift the equilibrium to the right, according to LeChâtelier s principle, corresponding to an increase in the K a, giving a stronger acid. By the same argument, if H o is negative (exothermic), increasing T will decrease K a indicating a weaker acid. For formic acid we have HCOOH(aq) HCOO (aq) + H + (aq) So, H o o = H f [HCOO o (aq)] H f [HCOO (aq)] H f o[hcooh(aq)] From a library search, we find the enthalpy of formation in aqueous solution for formic acid, the formate ion and the H + ion are and 425.68, 425.76 and 0.00 kj mol 1 (by definition), respectively. This gives H o = 425.76 kj mol 1 + 0.00 kj mol 1 ( 425.68 kj mol 1 ) = 0.08 kj mol 1 This predicts that the dissociation of formic acid is very slightly exothermic, so, by LeChâtelier s principle, we would expect a slight decrease in acid strength with increasing temperature; however, H o is very small and given the fact that enthalpy of reaction values are

rarely accurate to high precision and that enthalpies are themselves temperature dependent, all that we can really conclude is that the temperature dependence of acidity is likely to be relatively small, but that the exact direcetion of that dependence cannot be reliably concluded from the data. 11.75 We first find the number of moles of CO2 produced in the reaction: 1 mol NaHCO 1 mol CO 0.350 g NaHCO 4.17 10 mol CO 3 2 3 3 2 84.01 g NaHCO3 1 mol NaHCO3 V CO2 n CO2 RT P (4.17 103 mol)(0.0831 L bar K 1 mol 1 )(37.0 273)K (1.00 bar) 0.107 L 11.76 Choice (c) because 0.70 M KOH has a higher ph than 0.60 M NaOH. Adding an equal volume of 0.60 M NaOH lowers the [OH ] to 0.65 M, hence lowering the ph. 11.77 If we assume that the unknown monoprotic acid is a strong acid that is 100% ionized, then the [H ] concentration will be 0.0642 M. ph log (0.0642) 1.19 Since the actual ph of the solution is higher, the acid must be a weak acid. 11.78 (a) For the forward reaction one of the NH3 acts as the acid and the other acts as the base. For the reverse reaction NH4 and NH2 are the acid and base, respectively. (b) NH4 corresponds to H ; NH2 corresponds to OH. For the neutral solution, [NH4 ] [NH2 ]. 11.79 The reaction of a weak acid with a strong base is driven to completion by the formation of water. Irrespective of whether the strong base is reacting with a strong monoprotic acid or a weak monoprotic acid, the same number of moles of acid is required to react with a constant number of moles of base. Therefore the volume of base required to react with the same concentration of acid solutions (either both weak, both strong, or one strong and one weak) will be the same. 11.80 K a [H ][A ] [HA] [HA] 0.1 M [A ] 0.1 M Therefore, K a [H ] K w [OH ] [ OH ] K K w a 11.81 High oxidation state leads to covalent compounds and low oxidation state leads to ionic compounds. Therefore, CrO is ionic and basic and CrO3 is covalent and acidic. 11.82 We can calculate H, S, and G for the autoionization of water using the values given in Appendix 2 of the text.

H o H OH H2O 1 1 1 mol H H H o o o 1 S o H OH H2O 1 1 1 1 1 1 S S S o o o 1 1 G o H OH H2O 1 1 1 l G G G o o o 1 Notice that we can also calculate G using G = H TS, and that we get consistent results. G o o o 1 1 1 1 H TS We can use the following equation to calculate the equilibrium constant (K) from G, where R is the gas constant and T is the temperature in units Kelvin. K e G RT At 25 C we get the following: G RT 4 1 1 1 K e e 1.01 10 14 Notice that the units for G and R have to be consistent. For the 60 C calculations, we will assume that the temperature dependences of H and S are negligible in this small temperature range. First we will calculate G at 60 C and then use it to calculate K. G o o o 1 1 1 1 H TS G RT 4 1 1 1 K e e 1.07 10 13 This calculated value is slightly different than that listed in Table 11.2 in the text (K w =9.61 10 14 at 60ºC). The discrepancy is due to the fact that we have assumed that the enthalpy and entropy of reaction are temperature independent. If you include the temperature dependence of these quantities, using the techniques outlined in Chapter 7 and 8, the calculated value and that in the table will agree quite closely. 11.83 HCOOH HCOO H K = Ka 1.7 10 4 H OH H2O K = K W 1 1.0 10 14 HCOOH OH HCOO H2O

The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions. K K a K w 1 (1.7 10 4 )(1.0 10 14 ) 1.7 10 10 11.84 We can write two equilibria that add up to the equilibrium in the problem. CH3COOH(aq) H (aq) CH3COO (aq) H (aq) NO2 (aq) HNO2(aq) K = K a 1.8 10 5 K = K 1 1 a 2.2 103 4 4.5 10 CH3COOH(aq) NO2 (aq) CH3COO (aq) HNO2(aq) The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions. K (1.8 10 5 )(2.2 10 3 ) 4.0 10 2 11.85 H H2O OH H2 base1 acid2 base2 acid1 Water is acting as a Brønsted acid because it is donating a hydrogen ion (H + ). The hydride ion (H ) is acting as a Brønsted base because it is accepting the hydrogen ion (H + ). 11.86 We can calculate H, S, and G for the base dissociation of ammonia using the values given in Appendix 2 of the text. NH 3(aq) H 2O(l) NH 4 + (aq) OH (aq) H o NH4 OH NH3 H2O 1 1 1 1 H H H H o o o o 1 S o NH4 OH NH3 H2O S S S S o o o o 1 1 1 1 1 1 1 1 1 1 G o NH4 OH NH3 H2O 1 1 1 1 G G G G o o o o 1 Notice that we can also calculate G using G = H TS, and that we get consistent results. G o o o 1 1 1 1 H TS We can use the following equation to calculate the equilibrium constant (K) from G, where R is the gas constant and T is the temperature in units Kelvin.

K e G RT At 25 C we get the following: G RT 4 1 1 1 K e e 1.9 10 5 Which is close to the accepted value of 1.8 10-5. For the 50 C calculations, we will assume that the temperature dependences of H and S are negligible in this small temperature range. First we will calculate G at 50 C and then use it to calculate K. H TS o o o 1 1 1 1 G RT 4 1 1 1 K b e e 2.1 10 5 We could have also just used the van t Hoff equation [Equation (10.16) in the text] to solve this problem. 11.87 In this specific case the Ka of ammonium ion is the same as the Kb of acetate ion [Ka(NH4 ) 5.6 10 10, Kb(CH3COO ) 5.6 10 10, see Table 11.5 in the text]. The two are of exactly (to two significant figures) equal strength. The solution will have ph 7.00. What would the ph be if the concentration were 0.1 M in ammonium acetate? 0.4 M? 11.88 Kb 8.91 10 6 K a Kw 1.1 10 K b 9 ph 7.40 [H ] 10 7.40 3.98 10 8 K a [H ][conjugate base] [acid] Therefore, 9 [conjugate base] Ka 1.1 10 [acid] 8 [H ] 3.98 10 0.028 11.89 The fact that fluorine attracts electrons in a molecule more strongly than hydrogen should cause NF3 to be a poor electron pair donor and a poor base. NH3 is the stronger base. 11.90 Because the PH bond is weaker than the N-H bond, there is a greater tendency for PH4 to ionize than NH 4 +. Because PH4 is a stronger acid than NH 4 +, the conjugate base of PH4 (PH 3) is a weaker base than the conjugate base of NH 4 + (NH 3).

11.91 The autoionization for deuterium-substituted water is: D2O D OD [D ][OD ] 1.35 10 15 (1) (a) The definition of pd is: 15 pd log[d ] log 1.35 10 7.43 (b) To be acidic, the pd must be < 7.43. (c) Taking log of both sides of equation (1) above: log[d ] log[od ] log(1.35 10 15 ) pd pod 14.87 11.92 Because the deuterium (D, also symbolized using 2 H 1 ) is heavier than protium ( vibrations of the D 2O molecule are lower in frequency than those of H 2O and the zero point energy of D 2O is lower than that of H 2O making it a little more stable. 1 H 1 ) the 11.93 (a) HNO2 (b) HF (c) BF3 (d) NH3 (e) H2SO3 (f) HCO3 and CO3 2 The reactions for (f) are: HCO3 (aq) H (aq) CO2(g) H2O(l) CO3 2 (aq) 2H (aq) CO2(g) H2O(l) 11.94 First we must calculate the molarity of the trifluoromethane sulfonic acid. (Molar mass 150.1 g mol 1 ) 1 mol 0.616 g 150.1 g Molarity 0.0164 M 0.250 L Since trifluoromethane sulfonic acid is a strong acid and is 100% ionized, the [H ] is 0.0165 M. ph log(0.0164) 1.79 11.95 (a) The Lewis structure of H3O is: H O H + H Note that this structure is very similar to the Lewis structure of NH3. The geometry is trigonal pyramidal. (b) H4O 2 does not exist because the positively charged H3O has no affinity to accept the positive H ion. If H4O 2 existed, it would have a tetrahedral geometry.

11.96 The reactions are HF H F (1) F HF HF2 (2) Note that for equation (2), the equilibrium constant is relatively large with a value of 5.2. This means that the equilibrium lies to the right. Applying Le Châtelier s principle, as HF ionizes in the first step, the F that is produced is partially removed in the second step. More HF must ionize to compensate for the removal of the F, at the same time producing more H. 11.97 The equations are: Cl2(g) H2O(l) HCl(aq) HClO(aq) HCl(aq) AgNO3(aq) AgCl(s) HNO3(aq) In the presence of OH ions, the first equation is shifted to the right: H (from HCl) OH H2O Therefore, the concentration of HClO increases. (The bleaching action is due to ClO ions.) 11.98 Given the equation: HbH O2 HbO2 H (a) (b) (c) From the equilibrium equation, high oxygen concentration puts stress on the left side of the equilibrium and thus shifts the concentrations to the right to compensate. HbO2 is favored. High acid, H concentration, places stress on the right side of the equation forcing concentrations on the left side to increase, thus releasing oxygen and increasing the concentration of HbH. Removal of CO2 decreases H (in the form of carbonic acid), thus shifting the reaction to the right. More HbO2 will form. Breathing into a paper bag increases the concentration of CO2 (re-breathing the exhaled CO2), thus causing more O2 to be released as explained above. 11.99 The solution for the first step is standard: H3PO4(aq) H (aq) H2PO4 (aq) Initial (M): 0.100 0.000 0.000 Change (M): x x x Equil. (M): (0.100 x) x x K a1 [H ][H 2 PO 4 ] [H 3 PO 4 ] 3 7.5 10 2 x (0.100 x) In this case we probably cannot say that (0.100 x) 0.100 due to the magnitude of Ka. We obtain the quadratic equation: x 2 (7.5 10 3 )x (7.5 10 4 ) 0

The positive root is x 0.0239 M. We have: For the second ionization: [H ] [H2PO4 ] 0.0239 M [H3PO4] (0.100 0.0239) M 0.076 M H2PO4 (aq) H (aq) HPO4 2 (aq) Initial (M): 0.0239 0.0239 0.000 Change (M): y y y Equil (M): (0.0239 y) (0.0239 y) y K a2 [H ][HPO 4 2 ] [H 2 PO 4 ] 8 (0.0239 y)( y) (0.0239)( y) 6.2 10 (0.0239 y) (0.0239) Thus, y 6.2 10 8 M. [H ] [H2PO4 ] 0.0239 M 2 8 4 y [HPO ] 6.2 10 We set up the problem for the third ionization in the same manner. HPO4 2 (aq) H (aq) PO4 3 (aq) Initial (M): 6.2 10 8 0.0239 0 Change (M): z z z Equil. (M): (6.2 10 8 ) z 0.0239 z z K a3 [H ][PO 4 3 ] [HPO 4 2 ] 13 (0.0239 z)( z) (0.239)( z) 4.8 10 8 8 (6.2 10 ) z (6.2 10 ) z 1.2 10 18 M The equilibrium concentrations are: M [H ] [H2PO4 ] 0.0239 M [H3PO4] 0.076 M [HPO4 2 ] 6.2 10 8 M [PO4 3 ] 1.2 10 18 M 11.100 (a) We carry an additional significant figure throughout this calculation to minimize rounding errors.

Number of moles NaOH M vol (L) 0.0568 M 0.0138 L 7.838 10 4 mol If the acid were all dimer, then: 4 mol NaOH 7.838 10 mol 4 mol of dimer 3.919 10 mol 2 2 If the acetic acid were all dimer, the pressure that would be exerted would be: P nrt V (3.919 104 mol)(0.08314 L bar K 1 mol 1 )(324 K) 0.360 L 0.02932 bar However, the actual pressure is 0.0346 bar. If mol of dimer dissociates to monomers, then 2 monomer forms. (CH3COOH)2 2CH3COOH 1 2 The total moles of acetic acid is: moles dimer monomer (1 ) 2 1 Using partial pressures: Pobserved P(1 ) 0.0346 bar (0.02932 bar)(1 ) 0.180 (b) The equilibrium constant is: 2 2 2 2 P Pobserved 2 CH3COOH 1 4 Pobserved K p 4.63 10 P 2 CH3COOH 1 1 2 Pobserved 1 3 11.101 0.100 M Na2CO3 0.200 M Na 0.100 M CO3 2 First stage: CO3 2 (aq) H2O(l ) HCO3 (aq) OH (aq) Initial (M): 0.100 0 0 Change (M): x x x Equilibrium (M): 0.100 x x x K 14 Kw 1.0 10 1 K 11 2 4.8 10 2.1 10 K 1 [HCO 3 ][OH ] [CO 3 2 ] 4

Second stage: 2 2 4 x x 2.1 10 0.100 x 0.100 x 4.6 10 3 M [HCO3 ] [OH ] HCO3 (aq) H2O(l) H2CO3(aq) OH (aq) Initial (M): 4.6 10 3 0 4.6 10 3 Change (M): y y y Equilibrium (M): (4.6 10 3 ) y y (4.6 10 3 ) y K 2 [H 2 CO 3 ][OH ] [HCO 3 ] 3 3 8 y[(4.6 10 ) y] ( y)(4.6 10 ) 2.4 10 3 3 (4.6 10 ) y (4.6 10 ) At equilibrium: y 2.4 10 8 M [Na ] 0.200 M [HCO3 ] (4.6 10 3 ) M (2.4 10 8 ) M 4.6 10 3 M [H2CO3] 2.4 10 8 M [OH ] (4.6 10 3 ) M (2.4 10 8 ] M 4.6 10 3 M 14 1.0 10 12 [H ] 2.2 10 M 3 4.6 10 11.102 The four equations that we were to start with are listed below with their corresponding numbers. K w H + OH (11.3) K a H A (11.11) HA HA A HA 0 (11.23) OH A H (11.24) There are many ways of deriving Equation (11.25) from these equations. Step-by-step the different variables have to be removed until the equation contains just [H + ] and constants. We can start by solving for [HA] in Equations (11.11) and (11.23) and then setting them equal to each other.