Homework #7 Chapter 8 Applications of Aqueous Equilibrium

Homework #7 Chapter 8 Applications of Aqueous Equilibrium 15. solution: A solution that resists change in ph when a small amount of acid or base is added. solutions contain a weak acid and its conjugate base or a weak base and its conjugate acid. For weak acid/conjugate base systems, when a base is added to the solution it reacts with the weak acid to form more conjugate base. When an acid is added to the same solution it reacts with conjugate base to form more weak acid. In both cases the ph remains approximately constant since the number of OH - /H + ions in solution is approximately constant. A similar system is set up for weak base/ conjugate acid buffers. In buffer systems the larger the amount of weak acid/conjugate base or weak base/conjugate acid the better the buffer. In addition, the best buffers have equal amounts of weak acid and conjugate base or weak base and conjugate acid. If the buffer solution is made of NaHCO 3 and Na 2CO 3 then the following equation will happen in solution HCO 3- (aq) + OH - (aq) CO 3 2- (aq) + H 2O(l) CO 3 2- (aq) + H + (aq) HCO 3- (aq) 16. Capacity: An indication of the amount of acid or base that can be added before a buffer loses its ability to resist the change in ph. The buffer capacity is greatest when there are equal amount of weak acid and conjugate base or weak base and conjugate acid. In addition, as the overall amount of weak acid and conjugate base goes up the buffering capacity of the solutions go up (same as for weak base and conjugate acid). Therefore, solution C which has 1.0 M solutions of a weak acid and its conjugate base will have the greatest buffering capacity. The Henderson-Hasselbalch equation allows us to calculate the ph of a buffered system. log A ph pk a HA Systems with the greatest buffer capacity will have [A - ]=[HA] causing the ph=pka. Therefore, when choosing a buffer system you should choose a system that has a pka close to what the ph of the overall solution should be. Since all of the 3 systems have equal amounts of [A - ] and [HA] the ph for all the systems is the same. 1

21. a) Major Species: HC 3H 5O 2 (no conjugate base present) Use an ICE table to determine ph. HC 3H 5O 2(aq) H + (aq) + C 3H 5O 2- (aq) HC 3H 5O 2 H + C 3H 5O K a Initial M 0 0 Change -x +x +x Equilibrium -x x x H C H O HC H O x 3 3 5 5 2 2 xx Since K a is small assume x = 2 x 5 1.310 x 0.0011 Check assumption 0.0011 100% 1.1% Good Concentration of H + H x 0. 0011M 1.310 Calculate the ph ph logh log0.0011 2. 96 b) Major Species: Na + and C 3H 5O (no conjugate acid present) Use an ICE table to determine ph. Need to determine K b of: C 3H 5O 2- (aq) + H 2O(l) HC 3H 5O 2(aq) + OH - (aq) K b =? 14 K 1.010 K b w 7.710 5 K 1.310 a 10 5 C 3H 5O HC 3H 5O 2 OH - Initial M 0 0 Change -x +x +x Equilibrium -x x X HC H O OH xx Kb Since K b is small assume x = 2 x 10 7.7 10 6 x 8.810 3 5 2 7.7 10 10 C x 3H5O 2 2

Check assumption 6 8.810 0.0088% Good Concentration of OH - 6 OH x 8.810 M Calculate poh 6 poh log OH log 8.810 5.06 Calculate ph ph 14 poh 14 5.06 8.94 c) For pure water [H + ] = [OH - ] = 1.0 10-7 7 H log1.0 10 7. 00 ph log d) Major Species: HC 3H 5O 2, Na +, and C 3H 5O Use Henderson-Hasselbalch equation to solve for ph. A ph pka log HA ph log K a log ph 5 log 1.3 10 4.89 You could have also used and ice table to solve for ph. HC 3H 5O 2(aq) H + (aq) + C 3H 5O 2- (aq) HC 3H 5O 2 H + C 3H 5O Initial M 0 Change -x +x +x Equilibrium -x x +x H C H O 3 5 2 x K a 1.310 HC3H 5O2 x Since K a is small assume x = x 5 1.310 x 5 5 x 1.310 Check assumption 5 1.310 100% 0.013% Good Concentration H + 5 x 1.310 5 ph log H log 1.310 4. 89 3

22. a) Major Species: H +, Cl -, and HC 3H 5O 2 (no conjugate acid present) The majority of the H + ions are coming from the strong acid therefor calculate ph as you would for a strong acid. ph log H log 0.020 1.70 We can check this by using the long way HC 3H 5O 2(aq) H + (aq) + C 3H 5O 2- (aq) HC 3H 5O 2 H + C 3H 5O Initial M 0.020 M 0 Change -x +x +x Equilibrium -x 0.020+x X K a H C H O x x 1.310 HC H O x 3 5 2 0.020 5 3 5 2 Since K a is small assume 0.023+x = 0.023 and -x = x0.020 5 1.310 5 x 6.510 Check assumptions 5 6.510 100% 0.33% Good 0.020 5 6.510 100% 0.065% Good Concentration of H + Calculate the ph 5 H 0.020 x 0.020 6.5 10 0.020M ph log H log 0.020 1.70 b) Major Species: H +, Cl -, and C 3H 5O H + + C 3H 5O HC 3H 4O 2(aq) C 3H 5O H + HC 3H 5O 2 Initial mol 0.020 mol 0 Final 0.080 mol 0 0.020 mol Major Species after reaction: C 3H 5O and HC 3H 5O 2 A ph pk a log HA 0.080 mol 5 0.020 mol ph log 1.310 log 5.49 c) Major Species: H + and Cl - 4

ph log H log 0.020 1.70 d) Major Species: H +, Cl -, Na +, and C 3H 5O C 3H 5O 2- (aq) + H + (aq) HC 3H 5O 2(aq) C 3H 5O H + HC 3H 5O 2 Initial mol 0.020 mol mol Final 0.080 mol 0 0.120 mol Major Species: Cl -, Na +, C 3H 5O and HC 3H 5O 2 A ph pk a log HA 0.080 mol 5 0.120 mol ph log 1.310 log 4.71 23. a) Major Species: HC 3H 5O 2, Na +, and OH - HC 3H 5O 2(aq) + OH - (aq) C 3H 5O 2- (aq) + H 2O(l) HC 3H 5O 2 OH - C 3H 5O Initial mol 0.020 mol 0 Final 0.080 mol 0 0.020 mol Major Species: HC 3H 5O 2, Na +, and C 3H 5O A ph pk a log HA 0.020 mol 5 0.080 mol ph log 1.310 log 4.28 b) Major Species: C 3H 5O 2-, Na +, and OH -.. The majority of the OH- ions are from the strong base. Therefore, use the strong base concentration to calculate ph poh log OH log 0.020 1.70 ph poh 14 ph 14 poh 14 1.70 12.30 We can check this C 3H 5O 2- (aq) + H 2O(l) OH - (aq) + HC 3H 5O 2(aq) K b 7.7 10 C 3H 5O OH - HC 3H 5O 2 Initial M 0.020 M 0 Change -x +x +x Equilibrium -x 0.020+x X 10 5

0.020 x OH HC3H 5O2 x x 10 Kb 7.710 C3H5O2 Since K is small assume 0.020+x = 0.020 and -x = x0.020 10 7.710 10 x 3.910 Check assumptions 10 7.710 6 100% 3.9 10 % Good 0.020 10 7.710 7 100% 7.7 10 % Good Concentration of OH - [OH ] = 0.20 + x = 0.020 + 3.9 10 10 = 0.020M Calculate the ph poh log OH log 0.020 1.70 ph poh 14 ph 14 poh 14 1.70 12.30 c) Major Species: Na +, and OH - Not a poh log OH log 0.020 1.70 ph poh 14 ph 14 poh 14 1.70 12.30 d) Major Species: HC 3H 5O 2, C 3H 5O 2-, Na +, and OH - HC 3H 5O 2(aq) + OH - (aq) C 3H 5O 2- (aq) + H 2O(l) HC 3H 5O 2 OH - C 3H 5O Initial mol 0.020 mol mol Final 0.080 mol 0 0.120 mol Major Species: HC 3H 5O 2, C 3H 5O 2-, and Na + A ph pk a log HA 25. Major Species: HF, K +, and F - K a = 7.2 10-4 (Table 7.2) 0.120 mol 5 0.080 mol ph log 1.310 log 5.06 6

ph ph A pka log HA 4 1.00M 0.60M log 7.210 log 3.36 27. Major Species: HF, K +, F -, Na +, and OH - HF(aq) + OH - (aq) F - (aq) + H 2O(l) HF OH - F - Initial 0.60 mol 0.10 mol 1.00 mol Final 0.50 mol 0 1.10 mol Major Species: HF, K +, F -, and Na + A ph pk a log HA 1.10 mol 4 0.50 mol ph log 7.210 log 3.49 Major Species: HF, K +, F -, H +, and Cl - F - (aq) + H + (aq) HF(aq) F - H + HF Initial 1.00 mol 0.20 mol 0.60 mol Final 0.80 mol 0 0.80 mol Major Species: HF, K +, F -, and Na + A ph pk a log HA 0.80 mol 4 0.80 mol ph log 7.210 log 3.14 33. All of the solutions are buffer solutions because there is a weak base (C 5H 5N) and its conjugate acid (C 5H 5NH + ) present, therefore, you can use Henderson-Hasselbalch equation. log A ph pk a HA A ph pk a 10 HA Need to determine pk a of C 5H 5NH + K a K b Kw 5.9 10 14 1.010 9 1.710 Calculate pk a 6 pk log log5.9 10 5. 23 a) a K a 6 A ph pk 4.50 5.23 10 a 10 0.19 HA 7

b) c) d) A ph pk 5.00 5.23 10 a 10 0.59 HA A ph pk 5.23 5.23 10 a 10 1.0 HA A ph pk 5.50 5.23 10 a 10 1.9 HA 35. Major Species: Na +, C 2H 3O 2-, H +, and Cl - Yes a reaction will go to completion H + + C 2H 3O HC 2H 3O 2 H + C 2H 3O HC 3H 5O 2 Initial x 1.0 mol 0 Final 0 1.0 mol-x X Major Species: C 2H 3O, HC 2H 3O 2, Na +, and Cl - a) When the ph = pk a [HC 2H 3O 2] = [C 2H 3O 2- ] A 1.0x 1 HA x x 0.5M Since it is a 1.0 L solution 0.5 moles of HCl is needed b) K a = 1.8 10-5 c) A ph pka log HA A 1.0x 0.29 HA x 5 A HA 4.20 log 1.810 log A log 0.54 HA x 0.78M Since it is a 1.0 L solution 0.78 moles of HCl is needed. log A ph pk a HA 5 A 5.00 log 1.810 log HA A log 0.25 HA A 1.0x 1.8 HA x x 0.36M Since it is a 1.0 L solution 0.36 moles of HCl is needed. 37. Major Species: Na=, C 2H 3O 2-, and HC 2H 3O 2 8

K a = 1.8 10-5 A ph pka log HA 5 A 5.00 log 1.8 10 log 0.200M A 0.36M They asked for the mass not the molarity M NaC2H3O2 82.04 g mol 0.36 mol C2H3O2 1 mol NaC2H3O2 82.04 g NaC2H3O2 1 L C H O 1 mol C H O 1 mol NaC 2H3O2 0.500 L C H O 15 g NaC H O 2 3 2 2 3 2 2 3 2 2 3 2 38. a) Major Species: H +, Cl -, NH 3, + and NH 4 A reaction will go to completion NH 3(aq) + H + (aq) NH 4+ (aq) NH 3 H + + NH 4 Initial 0.0125 mol 0.010 mol 0.0375 mol Final 0.0025 mol 0 0.0475 mol Major Species: Cl -, NH 3, + and NH 4 A ph pk a log HA 0.0025 mol 10 0.0475 mol ph log 5.610 log 7.97 b) Major Species: H +, Cl -, NH 3, + and NH 4 A reaction will go to completion NH 3(aq) + H + (aq) NH 4+ (aq) NH 3 H + + NH 4 Initial 0.125 mol 0.010 mol 0.375 mol Final 0.115 mol 0 0.385 mol Major Species: Cl -, NH 3, + and NH 4 A ph pk a log HA 0.115 mol 10 0.385 mol ph log 5.610 log 8.73 40. Major Species: HNO 3, Na +, and NO 2 log A ph pk a HA a 4 Ka pk log log 4.010 3.40 (Table 7.2) 9

Total volume x + y = 1.00 x = volume of HNO 2 - y = volume of NO 2 Molarity of HNO 2 in final solution M n x0.50 x0.50 x y 1.0 L - Molarity of NO 2 in final solution M n y0.50 y0.50 x y 1.0 L Use the Henderson-Hasselbalch equation to find another relationship between x and y A ph pk a log HA 3.55 3.40 log 0.50 y 1.0 L 0.50 x 1.0 L y 0.15 log x y 1.4x The total volume of the solution is 1 L (x+y=1). Use this equation to solve for x and y. y 1.4x x y 1.00 x 1.4x 2.4x 1.00 x 0.42L HNO2 y.58l NO2 41. a) Major Species: H 2PO - 4 and HPO 4. pk log log 6.2 10 2 HPO 4 H2PO 4 a K a A ph pka log HA 8 7. 21 2 HPO 4 7.15 7.21log H2PO 4 0.87 H2PO 4 1 2 1.1 HPO 4 0.87 b) The ph of intracellular fluid is ~7.15 (part b). The pk a of H 3PO 4 is 2.12. The best buffers have equal amounts of A - and HA, therefore, the ph = pk a. In order to 10

have a ph of 7.15 with a solution containing H 3PO 4/H 2PO - 4 there would have to be much more H 2PO - 4 in solution that H 3PO 4 making it an ineffective buffer. 42. a) Major Species: H 2CO - 3 and HCO 3 pk log K 7 log 4.310 6.37 a A ph pka log HA 3 a HCO 3 HCO 3 7.40 6.37 log 2 3 6.37 log H CO 0.0012M HCO 0.013M 44. a) Major Species: K +, OH -, CH 3NH 3+, and Cl - OH - (aq) + CH 3NH 3+ (aq) H 2O(l) + CH 3NH 2(aq) OH - CH 3NH + 3 CH 3NH 2 Initial 0.1 mol 0.1 mol 0 Final 0 0 0.1 mol Major Species: K +, CH 3NH 2, and Cl - b) Major Species: K +, OH -, and CH 3NH 2 c) Major Species: K +, OH -, CH 3NH 3+, and Cl - OH - (aq) + CH 3NH 3+ (aq) H 2O(l) + CH 3NH 2(aq) OH - CH 3NH + 3 CH 3NH 2 Initial 0.2 mol 0.1 mol 0 Final 0.1 mol 0 0.1 mol Major Species: K +, OH -, CH 3NH 2, and Cl - d) Major Species: K +, OH -, CH 3NH 3+, and Cl - OH - (aq) + CH 3NH 3+ (aq) H 2O(l) + CH 3NH 2(aq) OH - CH 3NH + 3 CH 3NH 2 Initial 0.1 mol 0.2 mol 0 Final 0 0.1 mol 0.1 mol Major Species: K +, CH 3NH 2, Cl -, and CH 3NH 2 45. a) Major Species: H +, NO 3-, Na + -, and NO 3 b) Major Species: H +, NO 3-, HF 11

c) Major Species: H +, NO 3-, Na +, and F - H + (aq) + F - (aq) HF(aq) H + F - HF Initial 0.2 mol 0.4 mol 0 Final 0 0.2 mol 0.2 mol Major Species: NO 3-, Na +, and F - d) Major Species: H +, NO 3-, Na +, and OH - H + (aq) + OH - (aq) H 2O(aq) H + OH - Initial 0.2 mol 0.4 mol Final 0 0.2 mol Major Species: NO 3-, Na +, and OH - Not a 46. Major Species: Na +, F -, H +, and Cl - H + + F - HF H + F - HF Initial 0.0025 mol 0.0100 mol 0 Final 0 0.0075 mol 0.0025 mol Major Species: Na +, F -, Cl -, and HF K a = 7.2 10-4 (Table 7.2) A ph pk a log HA 0.0075 mol 4 0.0025 mol ph log 7.210 log 3.62 47. A buffer has the greatest buffer capacity when [HA] = [A - ]. When this happens the ph = pk a. Therefore, the best acid would have a pk a of 7.00. The acid that has the closest pk a value is HOCl which has a pk a of 7.46. In order to make a 1 L solution you would mix equal amounts of HOCl with NaOCl (or another salt containing OCl - ). 52. a) The blue line is the weak acid and the red line is the weak acid. You can tell the difference between the two plots because of the following 3 reasons. 1) The equivalence point of a weak acids ph curve is at a ph great than 7 and the equivalence point of a strong acid ph curve is at 7. 2) At the start of the ph curve of a weak acid titration the ph changes quickly and then levels off. This does not happen in a strong acid titration curve. 3) Since both acids have the same initial concentration the starting point of the strong acid should be at a lower ph than the weak acid. 12

b) A buffer solution is a solution that resist change in ph. In the weak acid/strong base titration, as the strong base is added to the weak acid resulting in the formation of the conjugate base of the weak acid which will be a weak base. When there is equal amount of the weak acid and weak base this is called the half equivalence point. At this point if an acid is added the weak base can react with the acid to absorb the added H + in the solution resulting in approximately the same ph. If a base is added the weak acid can react with the base to absorb the added OH - in the solution resulting in approximately the same ph. Therefore, the middle of the buffer region sits at the half equivalence point. Although the ph appears to be stable at the half equivalence point for the strong acid/strong base titration this system is not considered a buffer. When strong base is added to the acid system the H + reacts with the OH - to form water. However, water is neutral therefore, if an acid was added to the system there would be nothing to react with it causing the ph to go down dramatically. c) The statement is true. Since both acid have the same initial amount and initial concentrations the amount of base added to get to the equivalence point will be the same for both systems. The definition of the equivalence point is the point at which enough titrant has been added to fully react the analyte. Since the moles of acid is the same for both the strong and weak acid the number of moles of base needed to fully react each system is the same. d) The statement is false. When a strong acid reacts with a strong base a neutral salt and water is formed. This results in a ph of 7.00 at the equivalence point. When a weak acid reacts with a strong base the conjugate base of the weak acid is formed. At the equivalence point all of the weak acid will be converted into the conjugate base of the acid resulting in a ph that is greater than 7.00 at the equivalence point. 53. a) The acid is a weak acid. You can tell this because at the beginning of the titration (beaker 1) all of the hydrogen atoms (blue atoms) are attached to their counter ions (green atoms). As strong base is added, hydrogen atoms are removed from their counter ions forming water and leaving behind the conjugate base (or counter ion) of the acid. If the system was a strong acid solution add you would see only hydrogen ions (blue atoms) in the solution at the beginning of the titration. As strong base is added, the hydrogen atoms should disappear because they are reacting with OH - to form water which is not shown in the picture. b) c(start of titration) ae (half equivalence point)b (equivalence point)d(end of titration) c) The ph=pk a for beaker e. This is the half equivalence point or the point where half of the weak acid (green atom bonded to blue atom) is converted to the conjugate base (green atom) of the weak acid. d) Beaker b represents the equivalence point. The equivalence point occurs when all of the weak acid (green atom bonded to blue atom) is converted to the conjugate base (green atom) of the weak acid. e) You would not need to know the K a value of the weak acid to determine the ph of the system for beaker d. At this point all of the weak acid has been converted into the conjugate base of the weak acid. In addition, more strong base has been added. In a system with a strong base and a weak base the strong base 13

determines the ph of the system due to the dramatically larger number of OH - ions from the strong base. 54. a) The equivalence point occurs when equal mole amounts of weak acid and strong base have been added to the system. For this plot it occurs after ~22 ml of base are added. The ph at the equivalence point will be greater than 7. b) & c) The maximum buffering occurs when the [HA] = [A - ] or ph = pk a. This will be in the middle of the flat region (~12). d) The ph only depends on [HA] on the far left of the plot because no [A - ] is present. e) The ph only depends on [A - ] at the equivalence point, when ~22 ml of base are added f) The ph depends only on the amount of excess base on the far right of the plot. 55. For weak base titrated with a strong acid the equivalence point should occur at a ph lower than 7 due to the weak conjugate acid that is generated. 59. a) f (The ph initially increases more rapidly for weak acids than it does for strong acids. In, addition the weaker the acid the stronger the conjugate base therefore the weakest acid should have the highest ph at the equivalence point) b) a. If you did not know the initial concentration of the acids and wanted to know if the acid was strong or weak you could look at the ph at the equivalence point. If the ph at the equivalence point is 7.0 then it is a strong acid if the ph at the equivalence point is greater than 7.0 it is a weak acid. c) d (The pk a of the system is 6.0, therefore, when [HA] = [A - ] the ph of the system should be 6.0. This is referred to as the ½ equivalence point because ½ the amount of base has been added that would be needed to get to the equivalence point. For this graph the ½ equivalence point occurs after ~25 ml of NaOH has been added. The line that has a ph closest to 6 at the ½ equivalence point is d. 14

61. This is a strong acid strong base titration Calculate the initial moles of HClO 4 0.200 mol HClO 4 1 mol H L HClO mol HClO 0.0400 L HClO 0.00800 mol H 4 1 4 1 4 a) Major Species: H + -, and ClO 4 0.00800mol ph log H log 0.699 0.0400L b) Major Species: H +, ClO 4-, K +, and OH - H + (aq) + OH - (aq) H 2O(l) H + OH - Initial 0.00800 mol 0.00100 mol Final 0.00700 mol 0 Major Species: H +, ClO 4-, and K + 0.00700mol ph log H log 0.854 0.0400L 0.0100L c) Major Species: H +, ClO 4-, K +, and OH - H + (aq) + OH - (aq) H 2O(l) H + OH - Initial (mol) 0.00800 mol 0.00400 mol Final (mol) 0.00400 mol 0 Major Species: H +, ClO 4-, and K + 0.00400mol ph log H log 1.301 0.0400L 0.0400L d) Major Species: H +, ClO 4-, K +, and OH - H + (aq) + OH - (aq) H 2O(l) H + OH - Initial (mol) 0.00800 mol 0.00800 mol Final (mol) 0 0 Major Species: ClO 4-, and K + Since the moles of OH - equal the moles of H + the solution will be neutral 7.00. Another way to state this is that we are at the equivalence point. e) Major Species: H +, ClO 4-, K +, and OH - H + (aq) + OH - (aq) H 2O(l) H + OH - Initial (mol) 0.00800 mol 0.01000 mol 15

Final (mol) 0 0.00200 mol Major Species: ClO 4-, K +, and OH - 0.0020mol poh log OH log 0.014 1.85 0.0400L 0.0800L ph 14 poh 14 1.85 12.15 62. Ba(OH) 2 is a strong base and HCl is a strong acid Calculate the moles of Ba(OH) 2 that you start with 2 mol BaOH 2 2 mol OH 1 L BaOH 1 mol BaOH 0.0800 L Ba OH 0.0160 mol OH 2 2 a) Major Species: Ba 2+, and OH -.0160mol poh log OH log 0.699 0.0800L ph 14 poh 14 0.699 13.301 b) Major Species: Ba 2+, OH -, H +, and Cl - H + (aq) + OH - (aq) H 2O(l) OH - H + Initial (mol) 0.0160 mol 0.00800 mol Final (mol) 0.0080 mol 0 Major Species: Ba 2+, OH -, and Cl - 0.0080mol poh log OH log 1.10 0.0800L 0.0200L ph 14 poh 14 1.10 12.90 c) Major Species: Ba 2+, OH -, H +, and Cl - H + (aq) + OH - (aq) H 2O(l) OH - H + Initial (mol) 0.0160 mol 0.0120 mol Final (mol) 0.0040 mol 0 Major Species: Ba 2+, OH -, and Cl - 0.0040mol poh log OH log 1.44 0.0800L 0.0300L ph 14 poh 14 1.44 12.56 d) Major Species: Ba 2+, OH -, H +, and Cl - H + (aq) + OH - (aq) H 2O(l) 16

OH - H + Initial (mol) 0.0160 mol 0.0160 mol Final (mol) 0 0 Major Species: Ba 2+, and Cl - Since the moles of OH - equal the moles of H + the solution will be neutral 7.00. Another way to state this is that we are at the equivalence point. e) Major Species: Ba 2+, OH -, H +, and Cl - H + (aq) + OH - (aq) H 2O(l) OH - H + Initial (mol) 0.0160 mol 0.0320 mol Final (mol) 0 0.0160 mol Major Species: Ba 2+, H +, and Cl - 0.0160mol ph log H log 1.000 0.0800L 0.0800L 63. This is a weak acid/strong base titration a) Major Species: HC 2H 3O 2 HC 2H 3O 2(aq) H+(aq) + C 2H 3O 2- (aq) HC 2H 3O 2 H + C 2H 3O 2 - Initial 0.200 M 0 0 Change -x +x +x Equilibrium 0.200-x x x xx 5 K a 1.810 (0.200 x) Since K a is small assume 0.200-x = 0.200 xx 5 1.810 (0.200) x 0.0019 Check assumption 0.0019 100% 0.95% Good 0.200 Calculate ph H x 0.0019 H log0.0019 2. 72 ph log b) Major Species: HC 2H 3O 2, K +, and OH - HC 2H 3O 2(aq) + OH - (aq) H 2O(l) + C 2H 3O 2- (aq) 17

HC 2H 3O 2 OH - C 2H 3O Initial (mol) 0.0200 mol 0.00500 mol 0 Final (mol) 0.0150 mol 0 0.00500 mol Major Species: HC 2H 3O 2, K +, and C 2H 3O A 0.00500 mol 5 ph pk a log HA log 1.810 log 0.0150 mol 4.27 c) Major Species: HC 2H 3O 2, K +, and OH - HC 2H 3O 2(aq) + OH - (aq) H 2O(l) + C 2H 3O 2- (aq) HC 2H 3O 2 OH - C 2H 3O Initial (mol) 0.0200 mol 0.01000 mol 0 Final (mol) 0.0100 mol 0 0.01000 mol Major Species: HC 2H 3O 2, K +, and C 2H 3O A 0.01000 mol 5 ph pk a log HA log 1.810 log 0.01000 mol 4.74 d) Major Species: HC 2H 3O 2, K +, and OH - HC 2H 3O 2(aq) + OH - (aq) H 2O(l) + C 2H 3O 2- (aq) HC 2H 3O 2 OH - C 2H 3O Initial (mol) 0.0200 mol 0.01500 mol 0 Final (mol) 0.0050 mol 0 0.01500 mol Major Species: HC 2H 3O 2, K +, and C 2H 3O A 0.01500 mol 5 ph pk a log HA log 1.810 log 0.0050 mol 5.22 e) Major Species: HC 2H 3O 2, K +, and OH - HC 2H 3O 2 OH - C 2H 3O Initial (mol) 0.0200 mol 0.02000 mol 0 Final (mol) 0 0 0.02000 mol Major Species: C 2H 3O 2-, and K + C 2H 3O 2- (aq) + H 2O(l) HC 2H 3O 2(aq) + OH - (aq) K b C 2H 3O HC 2H 3O 2 OH - Initial (mol) 0.02000 mol 0 0 Initial (M) 0.0667 M 0 0 Change -x +x +x Equilibrium 0.0667-x x x 14 Kw 1.010 10 Kb 5.610 5 K 1.810 Calculate K b a 18

xx 10 K b 5.6 10 0.0667 x Since K b is very small assume 0.0600-x = 0.0600 xx 10 5.610 0.0667 6 x 6.110 Check Assumption 6 6.110 100% 0.0091% Good 0.0667 Calculate ph 6 [ OH ] x 6.110 M 6 poh log OH log 6.110 5.21 ph 14 poh 14 5.21 8.79 f) Major Species: HC 2H 3O 2, K +, and OH - HC 2H 3O 2 OH - C 2H 3O Initial (mol) 0.0200 mol 0.02500 mol 0 Final (mol) 0 0.00500 0.02000 mol Major Species: C 2H 3O 2-, K +, and OH - The majority of the OH- will come from OH - 0.00500mol poh log OH log 1.84 L 0.2500L ph 14 poh 14 1.84 12.16 64. This is a weak base/strong acid titration a) Major Species: H 2NNH 2. H 2NNH 2(aq) + H 2O(l) H 2NNH 3+ (aq) + OH - (aq) H 2NNH 2 H 2NNH 3 + OH - Initial M 0 0 Change -x +x +x Equilibrium -x X X xx 6 Kb 3.010 ( x) Since K a is small assume -x = xx 6 3.010 () x 5.510 4 19

Check assumption 4 5.510 100% 0.55% Good Calculate ph 4 OH x 5.510 4 poh log OH log 5.510 3.26 ph 14 poh 14 3.26 10.74 b) Major Species: H 2NNH 2, H + -, and NO 3 H 2NNH 2(aq) + H + (aq) H 2NNH 3+ (aq) H 2NNH 2 H + H 2NNH + 3 Initial (mol) 0.0100 mol 0.00400 mol 0 Final (mol) 0.0060 mol 0 0.00400 mol Major Species: H 2NNH 2, NO 3-, and H 2NNH + 3 Determine the K a 14 K 1.010 w 9 K a K 3.310 b 6 3.0 10 A 0.0060 mol 9 ph pk a log HA log 3.310 log 0.00400 mol 8.66 c) Major Species: H 2NNH 2, H + -, and NO 3 H 2NNH 2(aq) + H + (aq) H 2NNH 3+ (aq) H 2NNH 2 H + H 2NNH + 3 Initial (mol) 0.0100 mol 0.00500 mol 0 Final (mol) 0.0050 mol 0 0.00500 mol Major Species: H 2NNH 2, NO 3-, and H 2NNH + 3 A 0.0500 mol 9 ph pk a log HA log 3.310 log 0.0500 mol 8.48 d) Major Species: H 2NNH 2, H + -, and NO 3 H 2NNH 2(aq) + H + (aq) H 2NNH 3+ (aq) H 2NNH 2 H + H 2NNH + 3 Initial (mol) 0.0100 mol 0.00800 mol 0 Final (mol) 0.0020 mol 0 0.00800 mol Major Species: H 2NNH 2, NO 3-, and H 2NNH + 3 A 0.0020 mol 9 ph pk a log HA log 3.310 log 0.00800 mol 7.88 e) Major Species: H 2NNH 2, H + -, and NO 3 H 2NNH 2(aq) + H + (aq) H 2NNH 3+ (aq) 20

H 2NNH 2 H + H 2NNH + 3 Initial (mol) 0.0100 mol 0.0100 mol 0 Final (mol) 0 0 0.0100 mol Major Species: H 2NNH - 3+, and NO 3 H 2NNH 3+ (aq) H 2NNH 2(aq) + H + (aq) H 2NNH + 3 H 2NNH 2 H + Initial (mol) 0.0100 mol 0 0 Initial (M) 0.0667 0 0 Change -x +x +x Equilibrium 0.0667-x x x xx 9 K a 3.310 0.0667 x Since K a is small assume 0.0667-x = 0.0667 xx 9 3.310 (0.0667) 5 x 1.510 Check assumption 5 1.510 100% 0.022% Good 0.0667 Calculate ph 5 H x 1.510 5 H log1.5 10 4. 82 ph log f) Major Species: H 2NNH 2, H + -, and NO 3 H 2NNH 2(aq) + H + (aq) H 2NNH 3+ (aq) H 2NNH 2 H + H 2NNH + 3 Initial (mol) 0.0100 mol 0.0200 mol 0 Final (mol) 0 0.0100 mol 0.0100 mol Major Species: H 2NNH 3+, H + -, and NO 3 The H + concentration will be mainly from the excess strong acid. 0.0100mol ph log H log 1.301 L L 95. a) CaC 2O 4(s) Ca 2+(aq) + C 2O 4 2- (aq) K sp 2 2 Ca C O 2 Ca 2+ C 2O 2-4 Initial 0 0 Change +x +x Equilibrium x x * x is the solubility 4 21

The concentrations of Ca 2+ and C 2O 2-4 must be in M. g M 128.10 CaC O mol 2 4 3 g CaC2O4 mol CaC2O4 1 L CaC 2O4 128.10 g CaC2O4 4.810 6.1 10 1 5 2 2 2 5 2 9 2 4 4.8 10 2.3 10 Ksp Ca C O xx x b) BiI 3(s) Bi 3+ (aq) + 3I - (aq) K sp Bi 3 I 3 Bi 3+ I - Initial 0 0 Change +x +3x Equilibrium x 3x * x is the solubility K sp 3 3 3 4 5 4 19 Bi I x3x 27x 271.3210 8.2010 100. a) Ag 2SO 4(s) 2Ag + (aq) + SO 4 2- (aq) Ag + SO 4 2- Initial 0 0 Change +2x +x Equilibrium 2x x K sp 2x 2 2 2 Ag SO x 1.2 10 5 x 0.014M b) Ag 2SO 4(s) 2Ag + 2- (aq) + SO 4 (aq) Ag + 2- SO 4 Initial 0.10 M 0 Change +2x +x Equilibrium 0.10+2x x Assume that x is small due to the small equilibrium constant 4 1.2 10 5 M 2 2 5 4 1.2 10 Ksp Ag SO 2 5 0.10 2x x1.210 Assume that 0.10 +2x =0.10 2 5 0.10 x 1.210 x 0.0012M Check assumption 2 0.0012 100% 2.4% Good 0.10 22

c) Ag 2SO 4(s) 2Ag + 2- (aq) + SO 4 (aq) Ag + 2- SO 4 Initial 0 0.20 M Change +2x +x Equilibrium 2x 0.20+x Assume that x is small due to the small equilibrium constant x x 2 2 5 4 1.2 10 Ksp Ag SO 2 5 2 0.20 1.210 Assume that 0.20 +x =0.20 2 5 2x 0.20 1.210 x 0.0039M Check assumption 0.0039 100% 1.95% Good 0.20 101. a) Fe(OH) 3(s) Fe 3+ (aq) + 3OH - (aq) K sp Fe 3 OH 3 This problem is slightly different because there is some OH - in water. From the K w we know that the initial [OH - ]=1.0 10-7. Fe 3+ OH - Initial 0 1 10-7 Change +x +3x Equilibrium x 1 10-7 +3x * x is the solubility Since K sp is small assume that 1 10-7 +3x = 1 10-7 K sp 3 3 7 Fe OH x1 10 17 x 410 M Check assumption 17 410 100% 110 7 110 b) Fe(OH) 3(s) Fe 3+ (aq) + 3OH - (aq) K sp % 3 7 Fe 3 OH 3 3 Good 410 38 Calculate the concentration of OH - ions when the ph is 5.0 ph log H H 10 H OH 1 OH ph 10 H 10 110 14 5.0 14 110 110 110 14 5 5 110 9 23

Fe 3+ OH - Initial 0 1 10-9 Change +x +3x Equilibrium x 1 10-9* * x is the solubility * The question told you to assume that the ph is constant K sp 3 3 9 Fe OH x1 10 11 x 410 M c) Fe(OH) 3(s) Fe 3+ (aq) + 3OH - (aq) K sp Fe 3 OH 3 3 410 38 Calculate the concentration of OH - ions when the ph is 11.0 ph log H ph H 10 H OH 1 OH H 10 10 110 14 11.0 14 110 110 110 14 11 11 0.001 Fe 3+ OH - Initial 0 0.001 Gain/Lose +x +3x Equilibrium x 0.001 * * x is the solubility * The question told you to assume that the ph is constant Since K sp is small assume that 0.001 + 3x = 0.001 K sp x 410 3 3 Fe OH x0.001 29 M 3 410 38 118. Mn 2+ (aq) + C 2O 2-4 (aq) MnC 2O 4(aq) K 1 MnC 2O 4(aq) + C 2O 2-4 (aq) Mn(C 2O 4) 2 (aq) K 2 K K 1 K Mn 2+ (aq) +2C 2O 2-4 (aq) Mn(C 2O 4) 2 (aq) 2 3 1 5 K K1K2 7.9 10 7.9 10 6.2 10 24