The following Assignment Keys have worked out solutions to the Assigned Homework Problems. They will be posted after quizzing on the assignments. The even-numbered problems in the Guided Reading portions of the assignments have answers provided in the back of the textbook. However the solutions are not worked out. If you do not understand problems in the Assignment, the time to get help is BEFORE the announced quiz. See me during office hours, or use email or phone (home). Do not wait until right before the quiz. ALSO be aware that there can be mistakes in the answer key (we all make them occasionally). If you think you spot one, let me know!
| ASSIGNMENT
1 ANSWER KEY HOMEWORK PROBLEMS: 9, 13, 17, 19, 25 |
| 9.
Molar mass of caffeine = 194.20g a. From mole fraction of solvent: 0.900 mol H2O x 18.02 g/mol = 16.22 g H2O. 0.100 mol caffeine x 194.20 g/mol = 19.42g caffeine molality: 0.100 mol caffeine/16.22 x 10-3 kg water = 6.17 m mass % of solvent: 16.22 g water/(16.22g + 19.42g) x 100% = 45.5% ppm solute: 19.42g/(16.22g + 19.42g) x 106 = 5.45 x 105 ppm b. mass % solvent: [(1,000,000 - 1269)/1,000,000] x 100% = 99.87% molality: Assuming 1 kg solvent, 1000g solvent = 0.9987x, where x = total mass. Total mass = 1001.3g; mass of solute is 1.3g. Then 1.3g x 1 mol/194.20g = 6.69 x 10-3 mol solute. 6.69 x 10-3 mol/1 kg = 6.69 x 10-3 m Xsolvent: (1000g/18.02 g/mol) / (1000g/18.02 g/mol + 6.69 x 10-3 mol) = 0.9999 c. Assume 100 g of solution. Then 85.5 g are solvent and 14.5 g are solute. molality: 14.5 g solute/194.20g/mol = 0.074665 mol solute. 0.074665 mol/0.0855 kg = 0.873 m ppm solute: 14.5 g/100g x 106 = 1.45 x 105 ppm Xsolvent: 85.5 g H2O x 1mol/18.02g = 4.7447mol; 4.7447mol/(4.7447mol + 0.074665mol) = 9845 d. mass % solvent: 0.2560 mol x 194.20 g/mol = 49.72 g; 1000 g/(1000g + 49.72g) x 100 = 95.26% ppm solute: Assume 100 g solution. (100-95.25 g)/100g) x 106 = 4.74 x 104 ppm Xsolvent: (1000/18.02 mol)/ (1000/18.02 mol + 0.2560 mol) = 0.995 |
| 13.a.
0.1885 mol/L x 0.225 L = 0.0424 mol b. [Al2(SO4)3] = 0.0424 mol/1.450 L = 0.0292 M; [Al3+] = 2 x 0.0292M = 0.0585 M; [SO42-] = 3 x 0.0292M = 0.0878 M 17. Molar mass of KOH = 56.105 g. 1.13 mol x 56.108 g/mol = 63.40 g KOH. a. Molality: 1.13 mol/(1.050kg - 0.063.4kg) = 1.15 m mass % solute: 63.40g/1050g x 100 = 6.04% b. 1 L = 1290g solution (from density). 0.30 x 1290g = 387 g solute in one liter. molarity: 387g/56.108 g/mol = 6.90 M molality: 6.90 mol/(1.290 - 0.387 kg) = 0.764 m c. 3.11 mol x 56.108 g/mol = 174.5 g KOH/kg solvent. mass % solute: 174.5g/1174.5 g x 100 = 14.86% molarity: 1 L weighs 1230g (density). 0.1486 x 1230g = 182.7 g KOH per liter. 182.7g/56.108 g/mol = 3.26 mol/L or 3.26 M |
| 19.
a. HOOH (contains OH covalent bonds and can hydrogen
bond to water molecules. Naphthalene cannot.) b. sodium hydroxide (KOH) SiO2 is a network covalent solid (See Ch. 9) and is insoluble in water because covalent bonds would have to be broken. KOH is an ionic solid that breaks up into ions that are attracted to the polar water molecules. c. HCl HCl reacts with water to form ions (H3O+ + -OH) which are attracted to polar water molecules. Although CHCl3 does have a dipole moment, it is still not very polar. d. CH3OH Methyl alcohol has an O—H covalent bond, and can hydrogen bond to water molecules. H3C—O—CH3 has a bent shape and a permanent dipole, but cannot make as many hydrogen bonds to water because it has no O—H covalent bond itself. It is therefore less polar than methyl alcohol and will not dissolve as readily in water. |
| 25. a.
C = kP = 0.0769 M/atm x 3.00 atm = 0.23 M b. C = kP = 0.0313 M/atm x 3.4 x 10-4 atm = 1.1 x 10-5 M |
| ASSIGNMENT
2 ANSWER KEY HOMEWORK PROBLEMS: 31, 33, 35, 47, 49 |
| 31.
Molar mass of urea = 60.062g a. p = 0.217 M x 0.0821 L•atm/mol•K x 295 K = 5.26 atm b. Molarity of urea = (25.0 g/60.062g)/0.685 L = 0.6076 M p = 0.6076 M x 0.0821 L•atm/mol•K x 295 K = 14.7 atm c. 1 liter = 1120g (from density). 0.15 x 1120g = 168g urea. 168g/60.062g = 2.797 mol in a liter. p = 2.797 M x 0.0821 L•atm/mol•K x 295 K = 67.7 atm |
| 33. (2.5/760)atm
= (n x 0.0821 L•atm/mol•K x 298 K)/0.125 L; n =
1.68 x 10-5 mol Molar Mass = g/mol = 0.100 g/ 1.68 x 10-5 mol = 5.95 x 103 g/mol |
| 35.
a. Glycerine has molar mass of 92.094g. Assume 25.0 g
of glycerine and 75.0 g of water. Then m =
(25.0g/92.094g)/0.0750 kg = 3.62 m Delta Tf = 1.86°C/m x 3.62 m = 6.73°C, so Tf = - 6.73°C (original fp = 0°C) Delta Tb = 0.52°C/m x 3.62m = 1.88°C, so Tb = 101.88°C (original bp = 100°C) b. Propylene glycol has molar mass of 76.094g. m = (28.0g/76.094g)/0.325 kg = 1.13m. Delta Tf = 1.86°C/m x 1.13m = 2.10°C, so Tf = - 2.10°C. Delta Tb = 0.52°C/m x 1.13m = 0.59°C, so Tb = 101.59°C. c. 25.0 mL ethyl alcohol x 0.780g/mL x 1 mol/46.068g = 0.423 mol; m = 0.423 mol/0.735 kg = 0.576 m. Delta Tf = 1.86°C/m x 0.576 m = 1.07°C, so Tf = - 1.07°C. Delta Tb = 0.52°C/m x 0.576 m = 0.30°C, so Tb = 100.30°C. |
| 47. p/RT = [B] =
(4.18/760)atm/(0.0821 L•atm/mol•K x 298 K) =
2.248 x 10-4 M M = mol/L, so moles = M x L = 2.248 x 10-4 mol/L x 0.0500 L = 1.124 x 10-5 moles. Molar mass = g/mol = 0.225g/1.124 x 10-5 mol = 2.00 x 104 g/mol |
| 49.
The key here is to determine how many particles per
formula unit are formed when the solute is dissolved.
This number is i. a. i = 2; Delta Tf = 1.86°C/m x 0.25m x 2 = 0.93°C; Tf = - 0.93°C Delta Tb = 0.52°C/m x 0.25m x 2 = 0.26°C; Tb = 100.26°C. b. i = 4; Using the same k values, Tf = - 1.86°C and Tb = 100.52°C. c. i = 5; Using the same k values, Tf = - 2.32°C and Tb = 100.65°C |
| ASSIGNMENT
3 ANSWER KEY HOMEWORK PROBLEMS: 1, 3, 17, 21, 23 |
| 1. a. rate = — D[C2H6]/2Dt b. rate = D[CO2]/4Dt |
| 3. CO2 is being produced at 4 x 0.20 = 0.80 mol/L•s; H2O is being produced at 6 x 0.20 = 1.20 mol/L•s |
| 17.
a. k = rate/[NO]2[Br2] = 1.6
x 10-8 mol/L•min/(0.020M)2(0.030M)
= 1.33 x 10-3 L2/mol2•min b. [Br2] = rate/k[NO]2 = 3.5 x 10-7 mol/L•min/(1.33 x 10-3 L2/mol2•min)(0.043M)2 = 0.142 M c. 2.0 x 10-6 mol/L•min = (1.33 x 10-3 L2/mol2•min)(x)2(0.25x) Solve for x. 2.0 x 10-6/1.33 x 10-3 = 0.25x3, x3 = 6.0 x 10-8; x = 0.182 M |
| 21.
a. for [S2O82-]
Rate 3/Rate 2 = 2.22 x 10-4/1.85 x 10-4
= (0.0300/0.0250)m 1.2 = 1.2m, so m = 1. For [I-] Rate 4/Rate 3 = 3.06 x 10-4/2.22 x 10-4 = (0.0275/0.0200)n 1.38 = 1.38n, so n = 1. The overall order is 1 + 1 = 2 b. Rate = k[S2O82-][I-] c. k = rate/[S2O82-][I-] (use the numbers from any experiment in the table) k = 1.15 x 10-4/(0.0200)(0.0155) = 0.371 L/mol•min d. Rate = k[S2O82-][I-] = (0.371 L/mol•min)(0.105 mol/L)(0.0875 mol/L) = 3.41 x 10-3 mol/L•min |
| 23.
a. For [H2], 0.0168/0.00671 =
(0.250/0.100)m; 2.5 = 2.5m; m = 1 For [Br2], 0.00671/0.00474 = (0.200/0.100)n; 1.416 = 2n; n = 1/2 (note: this is a fractional power; the square root of 2 gives the number on the left) The overall order is 1.5 L1/2/(mol1/2•s) b. Rate = k[H2][Br2]1/2 c. k = rate/[H2][Br2]1/2 = 0.00474/(0.100)(0.100)1/2 = 0.15 d. rate = 0.15 x 0.455 x 0.2151/2 = 3.16 x 10-2 molL•s |
| ASSIGNMENT
4 ANSWER KEY HOMEWORK PROBLEMS: 31, 33, 35, 39, 43, 45 |
| 31.
a. If you graph ln [(CH3)2N2]
on the y-axis vs. time (hours) on the x-axis, you will
get a straight line plot with a negative slope. This is
what you would expect for a first order reaction. b. You can use any two points from your straight line to determine the slope, —k. Take the negative of that and you have k. k = 0.150 h-1 c. Here, you can use either form of the equation relating concentration to time for a first order reaction and, after solving for [(CH3)2N2]0, put in your data and solve for t. [(CH3)2N2]0 = 1.05 M (take any set of data from the table and k, and solve for [A]0). Then ln ([A]0/[A]) = kt; ln (1.05/0.100) = 0.150 h-1 (t); t = 15.7 h. d. Rate = (0.150 h-1)(0.415 M) = 6.22 x 10-2 mol/L•h |
| 33.a.
ln(0.0200 M/x) = (3.42 x 10-4d-1)(60.0
days) = 0.02052; 0.200M/x = e0.02052 = 1.0207;
x = 0.0200M/1.0207 = 0.0196 M b. ln (0.0200M/0.00350M) = 3.42 x 10-4d-1)t; t = 1.743/3.42 x 10-4d-1 = 5.10 x 103 days, or 14.0 years. c. t1/2 = 0.693/3.42 x 10-4d-1 = 2.03 x 103 days |
| 35. a.
Here you can use masses in place of concentrations. ln (150 mg/43.2 mg) = k(0.750 h); k = 1.66 h-1 b. t1/2 = 0.693/1.66 h-1 = 0.417 h c. When 95% of the reactant has been consumed, there will be 5% left. 0.05 x 150 mg = 7.5 mg. ln(150 mg/7.5 mg) = 1.66 h-1 x t; t = 1.80 hours |
| 39.
Again, you may use masses in place of concentrations. ln (5.00 mg/x mg) = 0.0546 h-1)(8 hours) = 0.4368; 5.00 mg/x mg = e0.4368 = 1.55; x = 5.00/1.55 = 3.23 mg |
| 43. a.
[A]0 — [A] = kt; k = ([A]0
— [A])/t = (1.00 atm — 0.200 atm)/16.0 s = 0.0500
atm/s b. (0.150 atm — 0.0432 atm)/0.0500 atm s-1 = 2.14 s c. t1/2 = [A]0/2k = 0.500 atm/(2 x 0.0500 atm/s) = 5.00 s |
| 45.
a. t1/2 [A]0 = 1/k = (1.51
min)(0.250 M) = 0.3775 mol•min/L; k = 0.3775-1
= 2.65 L/mol•min b. 1/0.0915 — 1/0.187 = 2.65 t; 5.58 = 2.65 t; t = 2.11 min c. Use the rate expression for a second order reaction: rate = (2.65 L/mol•min)(0.335 mol/L)2 = 0.297 L/mol•min |
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ASSIGNMENT 5 ANSWER KEY HOMEWORK PROBLEMS: 49, 55, 59, 61, 63
49. Your plot should have ln k on the y-axis and 1/T on the x-axis. If you have done it correctly, you will get a straight line with a negative slope. The slope is —2.35 x 104 K. Since the slope is — Ea/R, you can solve for the activation energy by multiplying the slope by —R. 2.35 x 104 K x 8.31 J/mol•K = 1.95 x 105 J/mol or 195 kJ/mol. 55. a. ln (22.8/0.066) = (Ea/8.31 J/mol•K)[1/838 — 1/1101]; Ea = 2.50 x 105 J/mol or 250 kJ/mol b. ln (0.066/k) = (2.50 x 105 J/mol / 8.31 J/mol•K)[1/758 — 1/838] = 3.7889; 0.066/k = e3.7889 = 44.21; k = 0.066/44.21 = 1.49 x 10-3 L/mol•min
c. ln (22.8/11.6) = (2.50 x 105 J/mol / 8.31 J/mol•K)[1/T1 — 1/1001]; 2.245 x 10-5 = 1/T1 — 9.99 x 10-4; 1.02 x 10-3 = 1/T1; T1 = 979 K (or 706 °C)
59. a. Rate = k1[NO3][CO] b. Rate = k2[I2]
c. Rate = k3[NO][O2]
61. Rate of slow step = k2[H2][I]2; [I] is an intermediate and cannot appear in the rate expression. Rate of fast step = k1[I2] = k-1[I]2
Solving for [I]2 in terms of [I2], [I]2 = k1/k-1 [I2].
Substituting in this value for the intermediate [I]2, Rate = (k2k1/k-1)[H2][I2], which fits the experimentally observed rate expression.
63. Both c and d are consistent with the observed second order rate law. Neither a nor b are consistent, since [CO] appears in the rate-determining slow step for these proposed mechanisms, but is not observed experimentally.
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ASSIGNMENT 6 ANSWER KEY HOMEWORK PROBLEMS: 1, 5, 7, 11, 13, 15
1. (a) It takes somewhere between 60 and 80 seconds. There is no change in reactant and product concentrations after 80 seconds. (b) After 30 seconds the rate of the forward reaction is slowing down but is greater than the rate of the reverse reaction. After 90 seconds, both rates are the same.
5. (a) K = (PIF5)2/(PI2)(PF2)5 (b) K = 1/(PCO)(PH2)5 (c) K = (PSO2)2/(PH2S)2(PO2)3 (d) K = 1/(PH2)2
7. (a) K = (PNO)2 [Cu2+]/[NO3-]2[H+]8 (b) K = (PSO2)2/(PO2)3 (c) K = 1/[Ca2+][CO32-] 11. (a) C3H8(g) + 5 O2(g) —> 3 CO2(g) + 4 H2O(g) (b) 2 C5H6(g) —> C10H12(g)
(c) H3PO4(aq) —> 3 H+(aq) + PO43-(aq)
(d) 2 H+(aq) + CO32-(aq) —> CO2(g) + H2O(g)
13. (a) K' = K2 = 4.8 x 10-6 (b) K" = 1/K2 = 2.1 x 105 15. To obtain the equation of interest, keep the first reaction equation as is, and reverse the second one. K = K1 x 1/K2 = 1.6 x 1/0.67 = 2.4
| ASSIGNMENT 7 ANSWER KEY HOMEWORK PROBLEMS: 19, 23, 25, 29, 31, 35, 41 |
| 19. K = (PCH4)(PH2O))/(PCO)(PH2)3 = (0.0391)(0.0124)/(0.921)(1.21)3 = 2.97 x 10-4 |
| 23. Initial
pressures: 0.692 atm NH3 and 0.0532 atm H2S Changes in pressure: (0.692 x 0.124) = + 0.086 atm NH3 and + 0.086 atm H2S Equilibrium pressures: 0.778 atm NH3 and 0.139 atm H2S K = (0.778)(0.139) = 0.108 |
| 25. Q =
(0.30)(0.16)/(0.50) = 0.096; Q > K (a) System is not at equilibrium because Q > K. (b) System will move toward reactants to reach equilibrium. |
| 29. 0.0049 = (0.213)2(0.883)6/(PNH3)4(0.255)3;
(PNH3)4 = (0.213)2(0.883)6/(0.0049)(0.255)3
= 385 Take the 4th root of 385; PNH3 = 4.43 atm |
| 31. 1.3 x 105
= (PH2S)2/(0.103)2(0.417);
(PH2S)2 = (1.3 x 105)(0.103)2(0.417)
= 5.75 x 102; PH2S = 24 atm |
| 35. Initial
pressures: PCO2 = PCF4 = 0.713 atm,
PCOF2 = 0.00 atm Changes in pressure: PCO2 = PCF4 = -x atm; PCOF2 = +2x atm Equilibrium pressures: PCO2 = PCF4 = 0.713 - x atm; PCOF2 = 2x atm K = 0.50 = (2x)2/(0.713 - x)2. Take square root of both sides: 0.707 = 2x/(0.713 - x); Solve for x: x = 0.186 atm; Equilibrium pressures: PCO2 = PCF4 = 0.713 - x atm = 0.527 atm; PCOF2 = 2x atm = 0.372 atm |
| 41. Initial
pressures: PHCN = 0.45 atm; PC2N2 =
PH2 = 0.32 atm Changes in pressure: PHCN = -2x atm; PC2N2 = PH2 = +x atm Equilibrium pressures: PHCN = 0.45 - 2x atm; PC2N2 = PH2 = 0.32 +x atm 0.17 = (0.32 +x atm)2/(0.45 - 2x atm)2; Take sq. rt. of both sides: 0.412 = (0.32 + x)/(0.45 - 2x) Solve for x: x = - 0.074 atm. So at equilibrium, PHCN = 0.45 - 2x atm = 0.60 atm and PC2N2 = PH2 = 0.32 +x atm = 0.25 atm. NOTE: Q is greater than K at initial pressures (check this). Therefore the reaction will go in the direction toward reactants to come to equilibrium. That is why x is a negative quantity. [If you had checked Q first, you might have reversed the signs of the change to begin with. Then x would be a positive quantity.] |
| ASSIGNMENT 8 ANSWER KEY |
| Chapter 12: 43, 47, 49 |
| 43.
(a) (1) reverse (2) reverse (3) neither (4) forward (5)
reverse (b) Decreasing T will decrease K. None of the above factors will increase K (although increasing T will increase K in this case). |
| 47.
(a) K = (0.16)(0.27)/(0.43)2 = 0.23 (b) At equilibrium, 0.23 = (0.10 + x)(0.27 + x)/(0.43 - 2x)2; Working out the details, you should get a quadratic equation: 0.08x2 + 0.766x - 0.0155 = 0. The meaningful solution is: x = 1.9 x 10-2 atm. At equilibrium, PICl = 0.39 atm; PI2 = 0.12 atm, and PCl2 = 0.29 atm. |
| 49. Let K1 = 62.5 and T1 = 800 K; K2 = ? and T2 = 606 K. Substitute into the van't Hoff equation and solve for K2. Remember to convert the enthalpy change to joules, -9400 J, to match the units of R. Remember that 1/800 - 1/606 gives you a negative number. This times another negative number gives you a positive number: +0.4524. ln(K2/62.5) = +0.4524. K2 = e0.4524 x 62.5 = 98.3 |
| Chapter 13: 1-11 odd, 15, 17, 19, 25acd, 27, 33, 35 |
| 1.
In each pair, the acid is given first and its conjugate
base is given second. (a) H3O+ and H2O; HCN and CN- (b) HNO2 and NO2-; H2O and -OH (c) HCHO2 and CHO2-; H3O+ and H2O |
| 3. (a) acid (b) acid (c) base |
| 5. (a) H2O (b) H2PO4- (c) NH4+ (d) HF (e) Zn(H2O)3(OH)+ |
| 7. (a) pH = — 0.78; pOH = 14.78; solution is acidic (b) pH = 0.48; pOH = 13.52; solution is acidic (c) pH = 7.34; pOH = 6.66; solution is basic (d) pH = 13.14; pOH = 0.86; solution is basic |
| 9.
(a) [H+] = 1.0 x 10-4; [-OH]
= 1.0 x 10-10 (b) [H+] = 3.0 x 10-9;
[-OH] = 3.3 x 10-6 (c) [H+] = 1; [-OH] = 1.0 x 10-14 (d) [H+] = 2.5 x 10-13; [-OH] = 4.0 x 10-2 |
| 11. Solution 1 is more acidic. Solution 2 has the higher pH. |
| 15. Ratio of [H+] in acid rain to [H+] in normal rain: 1 x 10-3/3.162 x 10-6 = 316:1 |
| 17.
(a) [0.25g Ba(OH)2 x (1 mol/171.3g)]/0.655 L =
2.23 x 10-3 M Ba(OH)2; [-OH]
= 2 times this, or 4.46 x 10-3 M; pOH = 2.35;
pH = 11.65. (b)The pH of the original KOH solution is 13.17. The new molarity is (0.300L/3.00L) x 0.149 M, or 0.0149 M; The pH of this solution is 12.17. Making a 10-fold dilution decreases the pH by one pH unit. |
| 19. [(0.295 L x 0.786 M HNO3) + (5.00g/127.9 g per mol HI)]/ 0.295 L = 0.9185 M H+; pH = 0.037. The key to this problem is to recognize that both acids are strong and will dissociate completely in water. Therefore moles of HNO3 + moles of HI = total moles of H+ in 295 mL of water. |
| 25.
The equations should show the original species + a water
molecule as reactants, followed by a double arrow,
followed by products. Using the given species as the
reactant, the products should be as follows: (a) Ni(H2O)4(OH)2 + H3O+ (c) H3O+ + HS- (d) H3O+ + PO43- |
| 27.
For the following equations, the arrow should be a double
one to indicate an equilibrium. (a) PH4+(aq) —> H+ (aq) + PH3(aq); Ka = [H+][PH3]/[ PH4+] (b) HS-(aq) —> H+ (aq) + S2-(aq); Ka = [H+][S2-]/[ HS-] (c) HBrO2(aq) —> H+ (aq) + BrO2-(aq); Ka = [H+][BrO2-]/[HBrO2] |
| 33.
Initial concentration of HC6H11O2
= 0.450 mol/0.200L = 0.225 M. At equilibrium, [H+] = [C6H11O2-] = 1.7 x 10-3 M. At equilibrium, [HC6H11O2] = 0.225 - 1.7 x 10-3 = 0.2233M Ka = (1.7 x 10-3)2/0.2233 = 1.3 x 10-5 |
| 35.
Initial concentration of HC6H7O6
= (2.00g/176.13 g/mol)/0.100 L = 0.114 M At equil: [H+] = [C6H7O6-] = 10-2.54 = 2.884 x 10-3. [HC6H7O6] = 0.114 - 2.884 x 10-3 = 0.111 M. Ka = (2.884 x 10-3)2/0.111 = 7.5 x 10-5 |
| ASSIGNMENT 9 ANSWER KEY |
| CHAPTER 13: 39, 41, 47, 49, 51, 53, 55, 59 |
| 39.
(a) 5.1 x 10-6 = x2/(0.894-x) or
approx. x2/0.894. x = 2.1 x 10-3 =
[H+] (b) [-OH] = 1.0 x 10-14/2.1 x 10-3 = 4.8 x 10-12 (c) pH = - log 2.1 x 10-3 = 2.68 (d) % ionization = (2.1 x 10-3/0.894) x 100% = 0.24% |
| 41.
Ka = 1.1 x 10-10 = x2/(0.173-x)
or approx. x2/0.173. x = 4.36 x 10-6
= [H+]. pH = 5.36 % ionization = (4.36 x 10-6 /0.173) x 100% = 0.0025% |
| 47.
(a) 0.0012 = x2/(2.9 - x) or approx. x2/2.9.
x = 5.90 x 10-2 . pH = 1.23 (b) Since [H+] = [HC8H4O4-] from the first ionization, we have Ka2 = [H+][C8H4O42-]/[HC8H4O4-], or 1.6 x 10-12 = (5.90 x 10-2)[C8H4O42-]/(5.90 x 10-2), so [C8H4O42-] = 1.6 x 10-12 = Ka2. (The second ionization constant is so much smaller than the first, that we can assume that most of the H+ ions will come from the first ionization. This is why the estimate of the concentration of the second anion is the same as the second ionization constant. See Example 13.9.) |
| 49.
(a)NH3(aq) + H2O(l)
–> NH4+(aq) + -OH(aq)
NOTE: all of the arrows should be double arrows to
indicate reversible reactions. (d) CO32-(aq) + H2O(l) –> HCO3-(aq) + -OH(aq) (e) F-(aq) + H2O(l) –> HF(aq) + -OH(aq) |
| 51. (b) < (c) < (a) < (d) |
| 53.
(a) picrate ion: Kb = 1.0 x 10-14/0.16
= 6.25 x 10-14 (b) trichloroacetate ion: Kb = 1.0 x 10-14/0.20 = 5.0 x 10-14 |
| 55.
(a) Cod(aq) + H2O(l) –>
HCod+(aq) + -OH(aq)
NOTE: the arrow should be a double arrow to indicate a
reversible reaction. (b) Kb = 1.0 x 10-14/1.2 x 10-8 = 8.3 x 10-7 (c) At equilibrium, [Cod] = 0.0020 - x, [HCod+] = [-OH] = x 8/3 x 1--7 = x2/(0.0020 - x) or approx. x2/0.0020. X = 4.1 x 10-5 = [-OH]. pOH = 4.39; pH = 9.61 |
| 59. (a) basic (b) acidic (c) acidic (d) basic (e) neutral |
| CHAPTER 14: 9 |
| 9. For
this problem, Ka of lactic acid is 1.4
x 10-4. The ratio of lactic acid to lactate
times Ka gives the hydrogen ion
concentration, from which you can calculate the pH of the
solution. (a) [H+] = (0.250/0.250) 1.4 x 10-4 = 1.4 x 10-4 M; pH = 3.85 (b) [H+] = (0.250/1.25) 1.4 x 10-4 = 2.8 x 10-5 M; pH = 4.55 (c) [H+] = (0.250/0.0800) 1.4 x 10-4 = 4.4 x 10-4 M; pH = 3.36 (d) [H+] = (0.250/0.0500) 1.4 x 10-4 = 7.0 x 10-4 M; pH = 3.15 |
| ASSIGNMENT
10 ANSWER KEY CHAPTER 14: 11, 13, 15, 19, 25, 39, 43, 45 |
| 11.
Note: all of the buffers in problem 9 can handle 0.250
moles of strong base per liter of buffer solution. Since
the concentration of the weak base varies, it is the
capacity to handle strong acid that varies. Listed below
is the number of moles of strong acid a liter of each of
these buffers can handle. a. 0.250 moles of strong acid b. 0.125 moles of strong acid c. 0.0800 moles of strong acid d. 0.0500 moles of strong acid |
| 13.
Note that you have to get the Ka of HNO2
from Table 13.2 (p. 359). [H+] = Ka ([HNO2]/[NO2-]) = 6.0 x 10-4 (0.0410/0.0250) = 9.84 x 10-4 M; pH = 3.01. |
| 15.
The pKa of the acid component of the
buffer should be close to the desired pH of the buffer. (a) HF/F (pKa = 3.2); (b) H2CO3/HCO3- (pKa = 6.35); (c) HPO42-/PO43- (pKa= 12.3) |
| 19.
Moles NaC2H3O2 = 0.1828;
moles HC2H3O2 = 0.2081 (a) In a buffer situation, we can use moles instead of molarities: [H+] = 1.8 x 10-5 (0.2081/0.1828) = 2.05 x 10-5 M; pH = 4.69. (b) pH = 4.69. The volume units cancel out in this equation, indicating that dilution with water does not affect the pH of the buffer. |
| 25.
This buffer contains 0.15 mol KH2PO4
and 0.0951 mol K2HPO4. (a) [H+] = 6.2 x 10-8 (0.15/0.0951) = 9.78 x 10-8; pH = 7.01 (b) The moles of strong acid added must be subtracted from moles of weak base and added to moles of weak acid. So [H+] = 6.2 x 10-8 (0.200/0.0405) = 3.06 x 10-7 M; pH = 6.51 (c) Moles of strong base added must be subtracted from moles of weak acid and added to moles of weak base. So [H+] = 6.2 x 10-8 (0.100/0.1451) = 4.27 x 10-8 M; pH = 7.37 |
| 39.
The indicator should have its endpoint pH near the
expected pH of the solution at the equivalence point. (a) methyl orange; (b) phenolphthalein; (c) all three will work for a strong acid/strong base titration; (d) methyl orange. |
| 43.
(a) H+(aq) + -OH(aq)
—> H2O(l) (b) Ba2+, NO3-, H2O (c) 0.0500 L x 0.237 mol/L x (2 mol -OH/1 mol Ba(OH)2) = 0.0237 moles -OH. Therefore 0.0237 moles of HNO3 are needed to get to the equivalence point. 0.0237 moles/0.4000 M HNO3 = 0.05925 L or 59.25 mL is the volume needed. (d) pOH = —log (0.237 x 2) = —log 0.474 = 0.324; pH = 14.00 — 0.324 = 13.7 (e) At the halfway point, we have 0.01185 moles -OH in 0.079625 L of solution. [-OH] = 0.1488 M; pOH = 0.827; pH = 13.2 (f) pH = 7 (true for all strong acid-strong base titrations) |
| 45.
(a) HBut(aq) + -OH(aq)
—> But-(aq) + H2O(l) (b) But-, Na+, H2O; because But- is a weak base, there will also be some HBut and -OH (c) 0.0500 L x 0.350 M = 0.0175 mol HBur = 0.0175 mol -OH needed. 0.0175 mol/0.225 M = 0.0778 L, or 77.8 mL needed to get to equivalence point (d) [H+]2 = 1.5 x 10-5 x 0.350 = 5.25 x 10-6; [H+] = 2.29 x 10-3; pH = 2.64. (e) pH = pKa = 4.82 (f) [-OH]2 = (1.0 x 10-14/1.5 x 10-5)(0.0175 mol/0.1278 L) = 9.13 x 10-11 [-OH] = 9.55 x 10-6; pOH = 5.02; pH = 8.98 |
| ASSIGNMENT
11 ANSWER KEY CHAPTER 16: 1, 3, 5, 9, 11, 15, 17, 19 |
| 1.
For the equations, the solid should be on the left of the
double arrow and the aqueous ions on the right. For each
compound, the ions and the Ksp
expression are given: (a) 2 Co3+(aq), 3 S2-(aq); Ksp = [Co3+]2[S2-]3 (b) Pb2+(aq), 2 Cl-(aq); Ksp = [Pb2+][ Cl-]2 (c) 2 Zn2+(aq), P2O74-(aq); Ksp = [Zn2]2[P2O74-] (d) Sc3+(aq), 3 -OH; Ksp = [Sc3+][-OH] 3 |
| 3.
Note: all the arrows should be double arrows to indicate
an equilibrium. (a) Hg2Cl2(s) —> Hg22+(aq) + 2 Cl-(aq) (b) PbCrO4(s) —> Pb2+(aq) + CrO42-(aq) (c) MnO2(s) —> Mn4+(aq) + 2 O2-(aq) (d) Al2S3(s) —> 2 Al3+(aq) + 3 S2-(aq) |
| 5.
(a) Ksp = [Cd2+][-OH]2;
[-OH]2 = 2.5 x 10-14/1.5
x 10-6 = 1.7 x 10-8; [-OH]
= 1.3 x 10-4 M (b) Ksp = [Cu2+]3[AsO43-]2; [Cu2+]3 = 7.6 x 10-36/(2.4 x 10-4)2 = 1.3 x 10-28; [Cu2+] = 5.1 x 10-10 M (c) Ksp = [Zn2+][C2O42-]; [Zn2+] = 2.7 x 10-8/8.8 x 10-3 = 3.1 x 10-6 M |
| 9.
(a) BaF2(s) —> Ba2+(aq)
+ 2 F-(aq); Ksp = [Ba2+][F-]2;
[Ba2+] = 1.8 x 10-7/(0.025)2
= 2.9 x 10-4 M (b) [F-]2 = 1.8 x 10-7/0.0045 = 4.0 x 10-5; [F-] = 6.3 x 10-3 M This is the amount of F- in solution. (0.025 — 6.3x 10-3)/0.025 = 0.748; 0.748 x 100% = 74.8% This is the percentage that has precipitated out. |
| 11.
[Ca2+] = 3.0 x 10-3 g/40.08 g per
mol = 7.485 x 10-5 M 0.00050 g Na2SO4 x 1 mol/142.5 g = 3.52 x 10-6 moles SO42-; Ksp of CaSO4 = 7.1 x 10-5 P = (7.485 x 10-3)(3.52 x 10-6) = 2.63 x 10-10; No ppt will form. To make a ppt form, the minimum amount of sulfate ion needed is: [SO42-] = 7.1 x 10-5/7.485 x 10-5 = 0.95 M |
| 15. 0.025 g Pb(N3)2 x 1 mol/291.26 g = 8.58 x 10-5 moles; [Pb(N3)2] = 8.58 x 10-5 moles/0.100 L = 8.58 x 10-4 M; Ksp = [Pb2+][N3]2 = (s)(2s)2 = (8.58 x 10-4)(1.72 x 10-3)2 = 3.0 x 10-9 |
| 17.
(a) Ksp = [Ag+][Cl-]
= (s)(s) = s2 = 1.8 x 10-10; s =
1.3 x 10-5 mol/L (b) 0.025 M BaCl2 = 0.050 M Cl-; 1.8 x 10-10 = (s)(0.050 + s) or (s)(0.050) s = 1.8 x 10-10/(0.050) = 3.6 x 10-9 mol/L (c) 1.8 x 10-10 = (0.17 + s)(s) or about (0.17)(s); s = 1.8 x 10-10/0.17 = 1.1 x 10-9 mol/L |
| 19.
1 g PbCl2 x 1 mol/278.1 g = 3.6 x 10-3
mol; in 1.0 L, the molarity of PbCl2 is 3.6 x
10-3 M. Will there be a precipitate? Ksp = 1.7 x 10-5 = (s)(2s)2 ; s3 = 1.7 x 10-5 /4 = 4.25 x 10-6; s = 1.6 x 10-2 mol/L. There will be no precipitate, because the solubility is greater than the molarity of the solute. Another way to look at this is to calculate P, which turns out to be 4.7 x 10-8, which is less than Ksp. |
| ASSIGNMENT
12 ANSWER KEY CHAPTER 17: 1, 5, 7, 11, 15, 17, 21, 25, 27, 31 |
| 1. (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous |
| 5. (a) + (b) + (c) + (d) + |
| 7. (a) — (b) — (c) — (d) + |
| 11.
(a) DS°
= +126.8 J/K — (+197.6 J/K + (2)(+130.6 J/K) =
—332 J/K (b) DS° = 2(+210.7 J/K) — (+191.5 J/K + 205.0 J/K) = +24.9 J/K (c) DS° = (+70.4 J/K + 213.6 J/K) — 112.1 J/K = +171.9 J/K (d) DS° = [(2)(+51.5 J/K) + 223.0 J/K] — [(2)(+72.1 J/K) + 202.7 J/K] = — 20.9 J/K |
| 15.
(a) DS°
= [(2)(+188.7 J/K) + (2)(+248.1 J/K)] — [(2)(+205.7
J/K) + (3)(+205.0 J/K)] = —152.8 J/K (b) DS° = (+72.7 J/K + 69.9 J/K + 210.7 J/K) — (+42.6 J/K + 2(0.0 J/K) + 146.4 J/K) = +164.3 J/K (c) DS° = (—128.9 J/K + 248.1 J/K + (2)(+69.9 J/K)) — (+20.1 J/K + 4(0.0 J/K) + 29.9 J/K = +209 J/K |
| 17.
(a) DG°
= —362 kJ — (318 K)(0.0192 kJ/K) = — 368
kJ (b) DG° = +745 kJ — (318K)(0.113 kJ/K) = +709 kJ (c) DG° = —22.5 kJ — (318K)(—0.0922 kJ/K) = +6.82 kJ |
| 21.
(a) DG°
= (0.0 kJ + 2(—55.2 kJ)) — (2(—167.2 kJ) +
0.0 kJ) = + 224 kJ (b) DG° = [—77.6 kJ — 300.2 kJ —237.2 kJ(2)] — [—744.5 kJ + 4(0.0 kJ) + 0.0 kJ] = —107.7 kJ (c) DG° = [0.0 kJ + 0.0 kJ + 2(—157.2 kJ)] — [2(—104.0 kJ) + 2(—237.2 kJ)] = +368 kJ |
| 25.
For this problem, you can either use free energies of
formation or you can use the Gibbs-Helmholtz equation.
Below is the solution for use of free energies of
formation. The reaction equation is: 2 CH3OH(l)
—> 2 CH4(g) + O2(g). DG° = [—50.7 kJ(2) + 0.0 kJ] — (—166.3)(2) = +231.3 kJ NO, this is not a thermodynamically feasible reaction at 25°C and 1 atm. |
| 27.
(a) You can use the Gibbs-Helmholtz equation to solve for
DS°. DS° = (DH° — DG°)/T = (+135.6 kJ — 34.6 kJ)/298 K = +0.339 kJ/K. The sign is reasonable since the reaction produces a solid, a liquid and a gas from a solid. (b) DG° = +135.6 kJ — (1000K)(0.3389 kJ/K) = —203.3 kJ |
| 31.
(a) DS°
= (DH°
— DG°)/T = (—353.2 kJ — (—452.4
kJ))/298 K = +0.3329 kJ/K. Yes, the sign is reasonable
because this reaction is producing 4 moles of gas from 2
moles of liquid and 1 mole of gas. (b) This problem involves solving for the standard entropy of phosgene, using the entropy change for the reaction and the standard entropies of the other reactants and products. 0.3329 kJ = 2(S°phosgene) + 2(+0.1868) — 2(+0.2017) — (+0.2050) S°phosgene = 0.2838 kJ/K (c) This problem involves solving for the standard enthalpy of formation of phosgene, using the standard enthalpy change for the reaction and the standard enthalpies of formation of the other reactants and products. —353.2 kJ = 2(DH°f phosgene) + 2(—92.3 kJ) — 2(—134.5 kJ) — 0.0 kJ DH°f phosgene = —218.8 kJ |