//05 Chapter 16 Homework Solutions 6. a) H AsO b) CH 3 NH 3 + c) HSO d) H 3 PO 8. acid base conj. base conj. acid a) H O CHO OH CH O a) HSO HCO 3 SO H CO 3 b) H 3 O + HSO 3 H O H SO 3 10. a) H C 6 H 7 O 5 + H O H 3 C 6 H 7 O 5 + OH b) H C 6 H 7 O 5 + H O HC 6 H 7 O 5 + H 3 O + c) conjugate acid H 3 C 6 H 7 O 5 ; conjugate base HC 6 H 7 O 5 1. a) HC H 3 O, weak base c) OH, strong base e) NH +, weak base b) H CO 3, weak base d) HCl, negligible 16. a) HCl (aq) + H O (l) ; left b) NO (aq) + H 3 O + (aq), right c) HNO 3 (aq) + OH (aq), left 18. a) H O (l) H + (aq) + OH (aq) b) K w [H + [OH ; The concentration of water is constant at constant temperature. Therefore it is incorporated into K w. c) [OH > [H + 1 19. a) [H + 0.00005 x 10 10 M: basic 1 b) [H + 9 3. x 10 3.1 x 10 6 M: acidic c) Let x [H +, then [OH 100x 1 x(100x) x [H + 8 M: slightly basic. [D + [OD 8.9 x 10 16 let x [D + [OD x 8.9 x 10 16 x 3.0 x 10 8 M [D + [OD
. We need to begin by setting up the proper equations for comparison. ph B log[h + vs. ph A log(500[h + ) log(500) log[h + log[h +.70 ph B.70 The first solution will have a ph of.70 units lower than the second solution. 6. a) HNO 3, nitric acid, is an acid and will release H + ions. This will lower [OH and ph. 1 b) [H + 7.1 x 10 13 M 0.01 ph log(7.1 x 10 13 ) 1.15 Since ph > 7, the solution is basic. c) [H + antilog(6.6).5 x 10 7 M [OH 1 x 10 8 M 7.5 x 10 3. a) A strong base completely dissociates into ions in aqueous solution. It is a strong electrolyte. b) From Sr(OH) Sr + + OH 0.15 M Sr(OH) 0.50 M OH c) Mg(OH) is a strong base because each unit of Mg(OH) that dissolves dissociates to yield OH ions. It is not very soluble, so the concentration of dissolved units is not very high. 3. In each of these problems we are dealing with a strong acid. These dissociate completely, so hydrogen ion concentration equals initial acid concentration. a) [H + 0.0575 M; ph log(0.0575) 1. b) [H + HClO H+ 1 (0.73 g HClO ) 3.60 x 10 3 M; 100. g HClO HClO.00 L ph log(3.60 x 10 3 ). c) [H + 1.00 mol (0.00500 L soln1 ) Lsoln1 ph log(0.00667).18 d) total volume 0.175 L (175 ml) 0.000 mol mol H+ (0.0500 L) L HCl H+ HCl HCl H +.5 x 10 3 mol H+ 3 [H +.5 x 10 mol H + 0.018 M 0.175 L ph log(0.018) 1.89 HCl 1 0.750 L soln 0.00667 M; 0.0100 mol + (0.15 L) HI 1 mol H + L HI
36. a) [OH 0.0050mol L ph 1.00.30 11.70 b) [OH.055 g 0.500 L 56.11g 0.0050 M OH ; poh log(0.0050).30 7.35 x 10 M OH dsoln 0.0100 M OH OH OH poh log(7.35 x 10 ) 1.1; ph 1.00.1 1.86 c) [OH 0.50 mol Ca(OH) (0.0100 L csoln ) mol OHcsoln 1 mol 1 L Ca(OH) 0.5000 L poh log(0.0100).00; ph 1.00.00 1.00 d) V total 10.0 ml + 30.0 ml 0.0 ml (or 0.000 L) [OH 0.015 mol Ba(OH) mol (0.0100 L) OH 1 + L Ba(OH) 0.000 L 3 7.5 x 10 mol NaOH OH (0.0300 L) 1 0.013M OH L NaOH 0.000 L poh log(0.013) 1.88; ph 1.00 1.88 1.1 38. poh 1.00 1.00.00, [OH antilog(.00) M OH mol OH Ca(OH) [Ca(OH) 5.0 x 10 3 M Ca(OH) L mol OH. Begin by setting up the equilibrium and writing out the K a expression. HC 8 H 7 O H + + C 8 H 7 O K a + [H [C8H7O [HC H O Now we can get [H + from ph and we know that since both H + and C 8 H 7 O come only from the acid, their concentrations are equal. The equilibrium concentration of the acid equals its initial concentration less what ionizes. [H + antilog(.68).09 x 10 3 M [C 8 H 7 O [HC 8 H 7 O 0.085 M 0.0009 M 0.083 M K a (.09 x 10 0.083 3 ) 5.3 x 10 5 + 6. 11.0% [H (100%) 0.100 [H + 0.0110 M [CH ClCO (since CH ClCO H H + + CH ClCO ) [CH ClCO H 0.100 M 0.0110 M 0.089 M 8 7
+ [H [CH K a ClCO [CH ClCO H (0.0110) K a (0.100 0.0110) 1.36 x 10 3 50. HClO H + + ClO K a 3.0 x 10 8 [H [ClO [HClO init 0.0075 0 0 (all quantities are molar) change x +x +x equil 0.0075x +x +x x 3.0 x 10 8 (0.0075 x) since [HClO 100K a, ignore x in the subtraction 3.0 x 10 8 x 0.0075 x 1.5 x 10 5 M [H + [ClO [HClO 0.0075 M 0.0007 M 0.0075 M (assumption is ok) 5. All of these problems are set up just like in problem 50 above. a).1 b) 5.98 + 6. a) C 3 H 7 NH + H O C 3 H 7 NH 3 + + OH b) HPO + H O H PO + OH c) C 6 H 5 CO + H O C 6 H 5 CO H + OH [C3H 7NH3 [OH [C H NH [H [C 6 PO + 3 7 [HPO H 5 [C CO 6 H 5 [OH CO H[OH 66. H O + BrO HBrO + OH.0 x 10 6 [HBrO[OH [BrO init 1.15 0 0 (all quantities are molar) change x +x +x equil 1.15 x +x +x x.0 x 10 6 1.15 x since [OBr > 100K b, we may assume x is negligible
.0 x 10 6 x 1.15 x.1 x 10 3 M [OH (1.15 M 0.00001 M 1.15 M so the assumption is ok) poh log(.1 x 10 3 ).67; ph 1.00.67 11.33 68. codeine + H O Hcodeine + + OH + [Hcod [OH [codeine poh 1.00 9.95.05; [OH antilog(.05) 8.9 x 10 5 M Since the only source of hydroxide ion is the hydrolysis of water by codeine, [Hcod + [OH and [codeine e [codeine i [OH e 5.0 x 10 3 M 8.9 x 10 5 M.9 x 10 3 M. 5 (8.9 x 10 ) 3.9 x 10 1.6 x 10 6 7. a) Fluoride ion is a weak base, that is the conjugate of a known weak acid. Therefore you must calculate the K b. 1 6.8 x 10 1.5 x 10 11 Now the problem is worked exactly like problem 66 above. H O + F HF + OH 1.5 x 10 11 [HF[OH [F init 0.036 0 0 (all quantities are molar) change x +x +x equil 0.036 x +x +x 1.5 x 10 11 x 0.036 x since [F > 100K b, we may assume x is negligible 1.5 x 10 11 x 0.036 x 7.35 x 10 7 M [OH (1.15 M 0.00001 M 1.15 M so the assumption is ok) poh log(7.35 x 10 7 ) 6.13; ph 1.00 6.13 7.87 1 b) 1 x 10 5 19 1x 10 This value is so large that it means that S is a strong base. You could have arrived at the same conclusion if you happened to remember that H S is a weak acid and so it s conjugate base HS is a weak base. Thus, deprotonating HS would lead to a strong base. Na S + H O NaOH + NaHS [OH 0.17 mol NaS L ph log(0.17) 13.10 NaOH NaS OH NaOH 0.17 M
You can ignore HS as a base because of Le Châtelier s principle. The presence of so much hydroxide ion from the reaction between S and water suppresses the hydrolysis of additional water by HS. 1 c) 5.6 x 10 10 5 1.8 x 10 OAc [OAc 0.035 mol NaOAc 0.055 mol OAc Ba(OAc) + mol L 1 mol NaOAc L At this point the problem is solved just as in part (a). ph 8.95 76. a) neutral b) acidic c) basic d) acidic e) acidic 80. [PO 3 (50.0 g TSP ) TSP 1 0.305 M 163.9 gtsp 1.00 L PO 3 + H O HPO + OH K w. x 10 K a [HPO 3 [PO [OH [OH [HPO x, [PO 3 0.305 x. x 10 x (0.305 x) since 0.305 < 100K b use the quadratic equation x 0.07 M [OH poh log(0.07) 1.13, ph 1.00 1.13 1.87 Ba(OAc) 0.15 M 8. To answer these questions you must compare bond strengths (Table 8.3, p. 80) and bond polarity (Figure 8.7, p. 67). a) The HS bond (339 kj/mol) is weaker than the HCl bond (31 kj/mol), so bond the fact that the HCl bond is more polar than the HS bond must account for the difference in acidities. b) The HS bond (339 kj/mol) is weaker than the HO bond (63 kj/mol), while the HO bond is more polar than the HS bond. The weaker bond strength accounts for the more acidic H S. It may not seem like a large difference (< 10%), but remember that bond polarity controls acidity only when the bond energies lie within a certain energy separation. c) Since both hydrogens are bound to oxygen, it bond strengths should be exceedingly close together. In this situation, bond polarity will almost certainly be the deciding factor, and in the first acid the 3 chlorines will pull electron density towards themselves quite strongly. This will make release of a proton more likely. 86. a) NO ; one fewer oxygen on NO stabilizes the negative charge less, making it more likely to pickup a proton. b) AsO 3 ; phosphorus is more electronegative. This pulls some of the negative charge off
the oxygen atoms making them less attractive to the hydrogen ions. c) CO 3 ; the higher negative charge creates a stronger interaction with approaching hydrogen ions. 9. acid/base a) HNO /OH b) FeBr 3 /Br c) Zn + /NH 3 d) SO /H O 93. a) Cu + b) Fe 3+ c) Al 3+ In each case, the metal ion with the larger charge or smaller size (charges equal) will yield the more acidic solution.