1. One example for NH 3 as a conjugate acid: NH 2 − + H + ⟶ NH 3 ; NH 2 − + H + ⟶ NH 3 ; as a conjugate base: NH 4 + ( a q ) + OH − ( a q ) ⟶ NH 3 ( a q ) + H 2 O ( l ) NH 4 + ( a q ) + OH − ( a q ) ⟶ NH 3 ( a q ) + H 2 O ( l ) 3. (a) H 3 O + ( a q ) ⟶ H + ( a q ) + H 2 O ( l ) ; H 3 O + ( a q ) ⟶ H + ( a q ) + H 2 O ( l ) ; (b) HCl ( aq ) ⟶ H + ( a q ) + Cl − ( a q ) ; HCl ( aq ) ⟶ H + ( a q ) + Cl − ( a q ) ; (c) NH 3 ( a q ) ⟶ H + ( a q ) + NH 2 − ( a q ) ; NH 3 ( a q ) ⟶ H + ( a q ) + NH 2 − ( a q ) ; (d) CH 3 CO 2 H ( a q ) ⟶ H + ( a q ) + CH 3 CO 2 − ( a q ) ; CH 3 CO 2 H ( a q ) ⟶ H + ( a q ) + CH 3 CO 2 − ( a q ) ; (e) NH 4 + ( a q ) ⟶ H + ( a q ) + NH 3 ( a q ) ; NH 4 + ( a q ) ⟶ H + ( a q ) + NH 3 ( a q ) ; (f) HSO 4 − ( a q ) ⟶ H + ( a q ) + SO 4 2 − ( a q ) HSO 4 − ( a q ) ⟶ H + ( a q ) + SO 4 2 − ( a q ) 5. (a) H 2 O ( l ) + H + ( a q ) ⟶ H 3 O + ( a q ) ; H 2 O ( l ) + H + ( a q ) ⟶ H 3 O + ( a q ) ; (b) OH − ( a q ) + H + ( a q ) ⟶ H 2 O ( l ) ; OH − ( a q ) + H + ( a q ) ⟶ H 2 O ( l ) ; (c) NH 3 ( a q ) + H + ( a q ) ⟶ NH 4 + ( a q ) ; NH 3 ( a q ) + H + ( a q ) ⟶ NH 4 + ( a q ) ; (d) CN − ( a q ) + H + ( a q ) ⟶ HCN ( a q ) ; CN − ( a q ) + H + ( a q ) ⟶ HCN ( a q ) ; (e) S 2− ( a q ) + H + ( a q ) ⟶ HS − ( a q ) ; S 2− ( a q ) + H + ( a q ) ⟶ HS − ( a q ) ; (f) H 2 PO 4 − ( a q ) + H + ( a q ) ⟶ H 3 PO 4 ( a q ) H 2 PO 4 − ( a q ) + H + ( a q ) ⟶ H 3 PO 4 ( a q ) 7. (a) H 2 O, O 2− ; (b) H 3 O + , OH − ; (c) H 2 CO 3 , CO 3 2− ; CO 3 2− ; (d) NH 4 + , NH 4 + , NH 2 − ; NH 2 − ; (e) H 2 SO 4 , SO 4 2− ; SO 4 2− ; (f) H 3 O 2 + , H 3 O 2 + , HO 2 − ; HO 2 − ; (g) H 2 S; S 2− ; (h) H 6 N 2 2+ , H 6 N 2 2+ , H 4 N 2 9. The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO 3 (BA), H 2 O(BB), H 3 O + (CA), NO 3 − ( CB ) ; NO 3 − ( CB ) ; (b) CN − (BB), H 2 O(BA), HCN(CA), OH − (CB); (c) H 2 SO 4 (BA), Cl − (BB), HCl(CA), HSO 4 − ( CB ) ; HSO 4 − ( CB ) ; (d) HSO 4 − ( BA ) , HSO 4 − ( BA ) , OH − (BB), SO 4 2− SO 4 2− (CB), H 2 O(CA); (e) O 2− (BB), H 2 O(BA) OH − (CB and CA); (f) [Cu(H 2 O) 3 (OH)] + (BB), [Al(H 2 O) 6 ] 3+ (BA), [Cu(H 2 O) 4 ] 2+ (CA), [Al(H 2 O) 5 (OH)] 2+ (CB); (g) H 2 S(BA), NH 2 − ( BB ) , NH 2 − ( BB ) , HS − (CB), NH 3 (CA) 11. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H 2 O. As an acid: H 2 O ( a q ) + NH 3 ( a q ) ⇌ NH 4 + ( a q ) + OH − ( a q ) . H 2 O ( a q ) + NH 3 ( a q ) ⇌ NH 4 + ( a q ) + OH − ( a q ) . As a base: H 2 O ( a q ) + HCl ( a q ) ⇌ H 3 O + ( a q ) + Cl − ( a q ) H 2 O ( a q ) + HCl ( a q ) ⇌ H 3 O + ( a q ) + Cl − ( a q ) 13. amphiprotic: (a) NH 3 + H 3 O + ⟶ NH 4 OH + H 2 O , NH 3 + H 3 O + ⟶ NH 4 OH + H 2 O , NH 3 + OCH 3 − ⟶ NH 2 − + CH 3 OH ; NH 3 + OCH 3 − ⟶ NH 2 − + CH 3 OH ; (b) HPO 4 2− + OH − ⟶ PO 4 3− + H 2 O , HPO 4 2− + OH − ⟶ PO 4 3− + H 2 O , HPO 4 2− + HClO 4 ⟶ H 2 PO 4 − + ClO 4 − ; HPO 4 2− + HClO 4 ⟶ H 2 PO 4 − + ClO 4 − ; not amphiprotic: (c) Br − ; (d) NH 4 + ; NH 4 + ; (e) AsO 4 3− AsO 4 3− 15. In a neutral solution [H 3 O + ] = [OH − ]. At 40 °C, [H 3 O + ] = [OH − ] = (2.910 × 10 −14 ) 1/2 = 1.7 × × 10 −7 . 17. x = 3.051 × × 10 −7 M = [H 3 O + ] = [OH − ]; pH = −log3.051 × × 10 −7 = −(−6.5156) = 6.5156; pOH = pH = 6.5156 19. (a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4 21. [H 3 O + ] = 3.0 × × 10 −7 M ; [OH − ] = 3.3 × × 10 −8 M 23. [H 3 O + ] = 1 × × 10 −2 M ; [OH − ] = 1 × × 10 −12 M 25. [OH − ] = 3.1 × × 10 −12 M 27. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH − , which causes the solution to be basic. 29. [H 2 O] > [CH 3 CO 2 H] > [ H 3 O + ] [ H 3 O + ] ≈ [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] > [OH − ] 31. The oxidation state of the sulfur in H 2 SO 4 is greater than the oxidation state of the sulfur in H 2 SO 3 . 33. Mg ( OH ) 2 ( s ) + 2HCl ( a q ) ⟶ Mg 2+ ( a q ) + 2 Cl − ( a q ) + 2H 2 O ( l ) BB BA CB CA Mg ( OH ) 2 ( s ) + 2HCl ( a q ) ⟶ Mg 2+ ( a q ) + 2 Cl − ( a q ) + 2H 2 O ( l ) BB BA CB CA 35. K a = 2.3 × 10 − 11 K a = 2.3 × 10 −11 37. The stronger base or stronger acid is the one with the larger K b or K a , respectively. In these two examples, they are (CH 3 ) 2 NH and CH 3 NH 3 + . CH 3 NH 3 + . 39. triethylamine 41. (a) HSO 4 − ; HSO 4 − ; higher electronegativity of the central ion. (b) H 2 O; NH 3 is a base and water is neutral, or decide on the basis of K a values. (c) HI; PH 3 is weaker than HCl; HCl is weaker than HI. Thus, PH 3 is weaker than HI. (d) PH 3 ; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid. 43. (a) NaHSeO 3 < NaHSO 3 < NaHSO 4 ; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) ClO 2 − < BrO 2 − < IO 2 − ; ClO 2 − < BrO 2 − < IO 2 − ; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO 2 < HOClO 3 ; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) HTe − < HS − << PH 2 − < NH 2 − ; HTe − < HS − << PH 2 − < NH 2 − ; PH 2 − PH 2 − and NH 2 − NH 2 − are anions of weak bases, so they act as strong bases toward H + . HTe − HTe − and HS − are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) BrO 4 − < BrO 3 − < BrO 2 − < BrO − ; BrO 4 − < BrO 3 − < BrO 2 − < BrO − ; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic. 45. [ H 2 O ] > [ C 6 H 4 OH ( CO 2 H ) ] > [H + ] 0 > [C 6 H 4 OH ( CO 2 ) − ] ≫ [ C 6 H 4 O ( CO 2 H ) − ] > [ OH − ] [ H 2 O ] > [ C 6 H 4 OH ( CO 2 H ) ] > [H + ] 0 > [C 6 H 4 OH ( CO 2 ) − ] ≫ [ C 6 H 4 O ( CO 2 H ) − ] > [ OH − ] 47. 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H 3 O + . 48. (b) The addition of HCl 50. (a) Adding HCl will add H 3 O + ions, which will then react with the OH − ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO 2 , and decreasing the concentration of NO 2 − NO 2 − ions. (b) Adding HNO 2 increases the concentration of HNO 2 and shifts the equilibrium to the left, increasing the concentration of NO 2 − NO 2 − ions and decreasing the concentration of OH − ions. (c) Adding NaOH adds OH − ions, which shifts the equilibrium to the left, increasing the concentration of NO 2 − NO 2 − ions and decreasing the concentrations of HNO 2 . (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO 2 adds NO 2 − NO 2 − ions and shifts the equilibrium to the right, increasing the HNO 2 and OH − ion concentrations. 52. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO 2 H exists primarily as HCO 2 H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO 2 H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H 3 O + ] produced by the stronger acid. 54. (a) K b = 1.8 × 10 −5 ; K b = 1.8 × 10 −5 ; (b) K a = 4.5 × 10 −4 ; K a = 4.5 × 10 −4 ; (c) K b = 6.4 × 10 −5 ; K b = 6.4 × 10 −5 ; (d) K a = 5.6 × 10 −10 K a = 5.6 × 10 −10 56. K a = 1.2 × 10 −2 K a = 1.2 × 10 −2 58. (a) K b = 4.3 × 10 −12 K b = 4.3 × 10 −12 (b) K a = 1.6 × 10 10 K a = 1.6 × 10 10 (c) K b = 5.9 × 10 8 K b = 5.9 × 10 8 (d) K b = 4.2 × 10 −3 K b = 4.2 × 10 −3 (e) K b = 2.3 × 10 5 K b = 2.3 × 10 5 (f) K b = 6.3 × 10 −13 K b = 6.3 × 10 −13 60. (a) [ H 3 O + ] [ ClO − ] [ HClO ] = ( x ) ( x ) ( 0.0092 − x ) ≈ ( x ) ( x ) 0.0092 = 2.9 × 10 −8 [ H 3 O + ] [ ClO − ] [ HClO ] = ( x ) ( x ) ( 0.0092 − x ) ≈ ( x ) ( x ) 0.0092 = 2.9 × 10 −8 Solving for x gives 1.63 × × 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H 3 O + ] = [ClO – ] = 1.6 × × 10 −5 M [HClO – ] = 0.0092 M [OH − ] = 6.1 × × 10 −10 M ; (b) [ C 6 H 5 NH 3 + ] [ OH − ] [ C 6 H 5 NH 2 ] = ( x ) ( x ) ( 0.0784 − x ) ≈ ( x ) ( x ) 0.0784 = 4.3 × 10 −10 [ C 6 H 5 NH 3 + ] [ OH − ] [ C 6 H 5 NH 2 ] = ( x ) ( x ) ( 0.0784 − x ) ≈ ( x ) ( x ) 0.0784 = 4.3 × 10 −10 Solving for x gives 5.81 × × 10 −6 M . This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [ C 6 H 5 NH 3 + ] [ C 6 H 5 NH 3 + ] = [OH − ] = 5.8 × × 10 −6 M [C 6 H 5 NH 2 ] = 0.0784 M [H 3 O + ] = 1.7 × × 10 −9 M ; (c) [ H 3 O + ] [ CN − ] [ HCN ] = ( x ) ( x ) ( 0.0810 − x ) ≈ ( x ) ( x ) 0.0810 = 4.9 × 10 −10 [ H 3 O + ] [ CN − ] [ HCN ] = ( x ) ( x ) ( 0.0810 − x ) ≈ ( x ) ( x ) 0.0810 = 4.9 × 10 −10 Solving for x gives 6.30 × × 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H 3 O + ] = [CN − ] = 6.3 × × 10 −6 M [HCN] = 0.0810 M [OH − ] = 1.6 × × 10 −9 M ; (d) [ ( CH 3 ) 3 NH + ] [ OH − ] [ ( CH 3 ) 3 N ] = ( x ) ( x ) ( 0.11 − x ) ≈ ( x ) ( x ) 0.11 = 6.3 × 10 −5 [ ( CH 3 ) 3 NH + ] [ OH − ] [ ( CH 3 ) 3 N ] = ( x ) ( x ) ( 0.11 − x ) ≈ ( x ) ( x ) 0.11 = 6.3 × 10 −5 Solving for x gives 2.63 × × 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH 3 ) 3 NH + ] = [OH − ] = 2.6 × × 10 −3 M [(CH 3 ) 3 N] = 0.11 M [H 3 O + ] = 3.8 × × 10 −12 M ; (e) [ Fe ( H 2 O ) 5 ( OH ) + ] [ H 3 O + ] [ Fe ( H 2 O ) 6 2+ ] = ( x ) ( x ) ( 0.120 − x ) ≈ ( x ) ( x ) 0.120 = 1.6 × 10 −7 [ Fe ( H 2 O ) 5 ( OH ) + ] [ H 3 O + ] [ Fe ( H 2 O ) 6 2+ ] = ( x ) ( x ) ( 0.120 − x ) ≈ ( x ) ( x ) 0.120 = 1.6 × 10 −7 Solving for x gives 1.39 × × 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 × × 10 −4 M [ Fe ( H 2 O ) 6 2+ ] [ Fe ( H 2 O ) 6 2+ ] = 0.120 M [OH − ] = 7.2 × × 10 −11 M 62. pH = 2.41 64. [C 10 H 14 N 2 ] = 0.049 M ; [C 10 H 14 N 2 H + ] = 1.9 × × 10 −4 M ; [ C 10 H 14 N 2 H 2 2+ ] [ C 10 H 14 N 2 H 2 2+ ] = 1.4 × × 10 −11 M ; [OH − ] = 1.9 × × 10 −4 M ; [H 3 O + ] = 5.3 × × 10 −11 M 66. K a = 1.2 × 10 −2 K a = 1.2 × 10 −2 68. K b = 1.77 × 10 −5 K b = 1.77 × 10 −5 70. (a) acidic; (b) basic; (c) acidic; (d) neutral 72. [H 3 O + ] and [ HCO 3 − ] [ HCO 3 − ] are practically equal 74. [C 6 H 4 (CO 2 H) 2 ] 7.2 × × 10 −3 M , [C 6 H 4 (CO 2 H)(CO 2 ) − ] = [H 3 O + ] 2.8 × × 10 −3 M , [ C 6 H 4 ( CO 2 ) 2 2− ] [ C 6 H 4 ( CO 2 ) 2 2− ] 3.9 × × 10 −6 M , [OH − ] 3.6 × × 10 −12 M 76. (a) K a2 = 1.5 × 10 −11 ; K a2 = 1.5 × 10 −11 ; (b) K b = 4.3 × 10 −12 ; K b = 4.3 × 10 −12 ; (c) [ Te 2− ] [ H 3 O + ] [ HTe − ] = ( x ) ( 0.0141 + x ) ( 0.0141 − x ) ≈ ( x ) ( 0.0141 ) 0.0141 = 1.5 × 10 −11 [ Te 2− ] [ H 3 O + ] [ HTe − ] = ( x ) ( 0.0141 + x ) ( 0.0141 − x ) ≈ ( x ) ( 0.0141 ) 0.0141 = 1.5 × 10 −11 Solving for x gives 1.5 × × 10 −11 M . Therefore, compared with 0.014 M , this value is negligible (1.1 × × 10 −7 %). 78. Excess H 3 O + is removed primarily by the reaction: H 3 O + ( a q ) + H 2 PO 4 − ( a q ) ⟶ H 3 PO 4 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + H 2 PO 4 − ( a q ) ⟶ H 3 PO 4 ( a q ) + H 2 O ( l ) Excess base is removed by the reaction: OH − ( a q ) + H 3 PO 4 ( a q ) ⟶ H 2 PO 4 − ( a q ) + H 2 O ( l ) OH − ( a q ) + H 3 PO 4 ( a q ) ⟶ H 2 PO 4 − ( a q ) + H 2 O ( l ) 80. [H 3 O + ] = 1.5 × × 10 −4 M 82. [OH − ] = 4.2 × × 10 −4 M 84. (a) The added HCl will increase the concentration of H 3 O + slightly, which will react with CH 3 CO 2 − CH 3 CO 2 − and produce CH 3 CO 2 H in the process. Thus, [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] decreases and [CH 3 CO 2 H] increases. (b) The added KCH 3 CO 2 will increase the concentration of [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] which will react with H 3 O + and produce CH 3 CO 2 H in the process. Thus, [H 3 O + ] decreases slightly and [CH 3 CO 2 H] increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce OH − ions, which will react with the H 3 O + , thus reducing [H 3 O + ]. Some additional CH 3 CO 2 H will dissociate, producing [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] ions in the process. Thus, [CH 3 CO 2 H] decreases slightly and [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] increases. (e) The added CH 3 CO 2 H will increase its concentration, causing more of it to dissociate and producing more [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] and H 3 O + in the process. Thus, [H 3 O + ] increases slightly and [ CH 3 CO 2 − ] [ CH 3 CO 2 − ] increases. 86. pH = 8.95 88. 37 g (0.27 mol) 90. (a) pH = 5.222; (b) The solution is acidic. (c) pH = 5.220 92. At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example. 94. (a) pH = 2.50; (b) pH = 4.01; (c) pH = 5.60; (d) pH = 8.35; (e) pH = 11.08