Homework #7 Chapter 8 Applications of Aqueous Equilibrium

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1 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. [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

2 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 K a = HC 3H 5O 2 H + C 3H 5O Initial M 0 0 Change -x +x +x Equilibrium x x x K a = [H+ ][C 3 H 5 O + 2 ] xx = = [HC 3 H 5 O 2 ] (0.100 x Since K a is small assume x = x 2 = x = Check assumption % = 1.1% Good Concentration of H + [H + ] = x = M Calculate the ph ph = log[h + ] = log( = 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 =? K b = K w = = K a C 3H 5O HC 3H 5O 2 OH - Initial M 0 0 Change -x +x +x Equilibrium x x X K b = [HC 3H 5 O 2 ][OH ] xx [C 3 H 5 O = = ] (0.100 x Since K b is small assume x = x 2 = x = Check assumption % = % Good Concentration of OH - [OH ] = x = M Calculate poh poh = log[oh ] = log( =

3 Calculate ph ph = poh = = 8.94 c For pure water [H + ] = [OH - ] = ph = log[h + ] = log( = 7.00 d Major Species: HC 3H 5O 2, Na +, and C 3H 5O Use Henderson-Hasselbalch equation to solve for ph. [HA] = log( log ( = 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 K a = HC 3H 5O 2 H + C 3H 5O Initial M Change -x +x +x Equilibrium x X x K a = [H+ ][C 3 H 5 O + 2 ] x( x = = [HC 3 H 5 O 2 ] (0.100 x Since K a is small assume x = and x=0.100 x(0.100 = (0.100 x = Check assumption % = 0.013% Good Concentration H + [H + ] = x = M Calculate ph ph = log[h + ] = log( = 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. Therefore, calculate ph as you would for a strong acid. ph = log[h + ] = log( = 1.70 We can check this by using the long way HC 3H 5O 2(aq H + (aq + C 3H 5O 2- (aq K a = HC 3H 5O 2 H + C 3H 5O Initial M M 0 Change -x +x +x Equilibrium x x X K a = [H+ ][C 3 H 5 O + 2 ] x( x = = [HC 3 H 5 O 2 ] (0.100 x 3

4 Since K a is small assume x = and x = x(0.020 = (0.100 x = Check assumptions % = 0.33% Good 100% = 0.065% Good Concentration of H + [H + ] = x = = M Calculate the ph 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 mol 0 Final mol mol Major Species after reaction: C 3H 5O and HC 3H 5O 2 [HA] = log( log ( = 5.49 c Major Species: H + and Cl - 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 mol mol Final mol mol Major Species: Cl -, Na +, C 3H 5O and HC 3H 5O 2 [HA] = log( log ( = mol mol mol mol 4

5 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 mol 0 Final mol mol Major Species: HC 3H 5O 2, Na +, and C 3H 5O [HA] = log( log ( mol mol = 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 = ph = poh = = We can check this C 3H 5O 2- (aq + H 2O(l OH - (aq + HC 3H 5O 2(aq C 3H 5O OH - HC 3H 5O 2 Initial M M 0 Change -x +x +x Equilibrium x x X K b = [OH ][HC 3 H 5 O 2 ] x( x [C 3 H 5 O = = ] (0.100 x Since K is small assume x = and x = x(0.020 = (0.100 x = Check assumptions % = % Good 100% = % Good Concentration of OH - [OH ] = x = = M Calculate the ph poh = log[oh ] = log(0.020 = 1.70 ph + poh = ph = poh = = c Major Species: Na +, and OH - Not a poh = log[oh ] = log(0.020 = 1.70 K b =

6 ph + poh = ph = poh = = 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 mol mol Final mol mol Major Species: HC 3H 5O 2, C 3H 5O 2-, and Na + [HA] = log( log ( = mol mol 27. Major Species: HF, K +, and F - K a = (Table 7.2 [HA] = log( log ( = 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 mol Major Species: HF, K +, F -, and Na + [HA] = log( log ( 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 mol Major Species: HF, K +, F -, and Na + [HA] = log( log ( 1.10 mol 0.50 mol 0.80 mol 0.80 mol = 3.49 =

7 35. 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. [HA] [A ] [HA] = 10pH pk a Need to determine pk a of C 5H 5NH + K a = K w = K b = Calculate pk a pk a = log(k a = log( = 5.23 a b c d [A ] [HA] = 10pH pk a = = 0.19 [A ] [HA] = 10pH pk a = = 0.59 [A ] = [HA] 10pH pk a = = 1.0 [A ] = [HA] 10pH pk a = = 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 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 ] [HA] = 1 = 1.0 mol x x x = 0.5 mol 0.5 moles of HCl is needed b K a = Still have a buffer [HA] pk a = log( = mol x 4.20 = log ( x 1.0 mol x 0.29 = x x = 0.78 M 0.85 moles of HCl is needed. 7

8 c Still have a buffer [HA] 5.00 = log ( 1.0 mol x 1.8 = x x = 0.36 M 0.36 moles of HCl is needed. 1.0 mol x x 39. Major Species: Na=, C 2H 3O 2-, and HC 2H 3O 2 K a = [HA] 5.00 = log( log ( [A ] M [A ] = 0.36 M They asked for the mass not the molarity L C 2 H 3 O 2 ( 0.36 mol C 2H 3 O 2 1 L C 2 H 3 O ( 1 mol NaC 2H 3 O mol C 2 H 3 O ( g NaC 2H 3 O mol NaC 2 H 3 O 2 = 15 g NaC 2 H 3 O 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 mol mol mol Final mol mol Major Species: Cl -, NH 3, + and NH 4 [HA] = log( 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 mol mol mol Final mol mol Major Species: Cl -, NH 3, + and NH mol mol 8

9 [HA] = log( log ( = 8.73 The ph of the original buffer solutions are: a = log(5.6 [HA] log ( b mol mol M 0.15 M = M = log(5.6 [HA] log ( = M Therefore, they start out with the same ph. Solution b is able to keep the ph closer to the original ph because it has a greater buffer capacity due to the larger amounts of NH 3 and NH 4Cl. 42. Major Species: HNO 3, Na +, and NO 2 [HA] pk a = log(k a = log( = 3.40 (Table 7.2 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 (0.50 Mx (0.50 Mx = = x + y 1.0 L - Molarity of NO 2 in final solution M = n (0.50 My (0.50 My = = x + y 1.0 L Use the Henderson-Hasselbalch equation to find another relationship between x and y [HA] (0.50 My 3.55 = log ( 1.0 L (0.50 Mx 1.0 L 3.55 = log ( y 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 = 0.42 L = HNO2 y = 0.58 L = NO2 9

10 43. a Major Species: H 2PO - 4 and HPO 4. [HA] pk a = log(k a = log( = = log ( [HPO 4 2 ] [H 2 PO 4 ] [HPO 2 4 ] [H 2 PO 4 ] = 0.87 [H 2 PO 4 ] [HPO 2 4 ] = = 1.1 b The ph of intracellular fluid is ~7.15 (part b. The pk a of H 3PO 4 is The best buffers have equal amounts of A - and HA, therefore, the ph = pk a. In order to 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. 44. a Major Species: H 2CO - 3 and HCO 3 [HA] pk a = log(k a = log( = = log ( [HCO 3 ] [H 2 CO 3 ] = log ( [HCO 3 ] M [HCO 3 ] = M 46. 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 mol Major Species: K +, CH 3NH 2, and Cl - (weak base b Major Species: K +, OH -, and CH 3NH 2 (weak base and strong base 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 mol Major Species: K +, OH -, CH 3NH 2, and Cl - (weak base and strong base 10

11 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 mol 0.1 mol Major Species: K +, CH 3NH 2, Cl -, and CH 3NH a Major Species: H +, NO 3-, Na + -, and NO 3 (strong acid b Major Species: H +, NO 3-, HF (strong acid and weak acid 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 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 mol Major Species: NO 3-, Na +, and OH - (strong base 48. Major Species: Na +, F -, H +, and Cl - H + + F - HF H + F - HF Initial mol mol 0 Final mol mol Major Species: Na +, F -, Cl -, and HF K a = (Table 7.2 [HA] = log( log ( mol mol =

12 49. 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 The acid that has the closest pk a value is HOCl which has a pk a of In order to make a 1 L solution you would mix equal amounts of HOCl with NaOCl (or another salt containing OCl a The blue line is the weak acid and the red line is the strong 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. b While the definition of a buffer solution is a solution that resist change in ph a better definition would be a solution in which the H + and OH - concentrations are constant. In the weak acid/strong base titration, as the strong base is added to the weak acid the weak acid is converted into its conjugate 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 amount of H +. 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 OH -. Therefore, the middle of the buffer region sits at the half equivalence point. Although the ph is stable at the half equivalence point for the strong acid/strong base titration this system is not considered a buffer because at very low ph s a very small change in ph results in a large change in H + concentration causing the concentration of the H + and OH - to change. 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 H + concentration 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. 55. a The acid is a weak acid. You can tell this because at the beginning of the titration (beaker c 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 you 12

13 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 a e (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 determines the ph of the system due to the dramatically larger number of OH - ions from the strong base. 56. a The equivalence point occurs when there are equal mole of weak acid and strong base. For this plot it occurs after ~22 ml of base is 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. 13

14 57. 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. 61. a f. The ph initially increases more rapidly for weaker acids than it does for stronger 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. 63. This is a strong acid strong base titration Calculate the initial moles of HClO 4 1 mol H L HClO 4 ( mol HClO 4 ( = mol H + 1 L HClO 4 1 mol HClO 4 a Major Species: H + -, and ClO 4 ph = log[h mol ] = log ( L = b Major Species: H +, ClO 4-, K +, and OH - H + (aq + OH - (aq H 2O(l H + OH - Initial mol mol Final mol 0 Major Species: H +, ClO 4-, and K + ph = log[h mol ] = log ( L L =

15 c Major Species: H +, ClO 4-, K +, and OH - H + (aq + OH - (aq H 2O(l H + OH - Initial (mol mol mol Final (mol mol 0 Major Species: H +, ClO 4-, and K + ph = log[h + ] = log ( mol L L = d Major Species: H +, ClO 4-, K +, and OH - H + (aq + OH - (aq H 2O(l H + OH - Initial (mol mol 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 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 mol mol Final (mol mol Major Species: ClO 4-, K +, and OH - poh = log[oh mol ] = log ( L L = 1.85 ph = poh = = 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 OH L Ba(OH 2 ( mol Ba(OH 2 ( = mol OH 1 L Ba(OH 2 1 mol Ba(OH 2 a Major Species: Ba 2+, and OH - poh = log[oh mol ] = log ( L = ph = poh = =

16 b Major Species: Ba 2+, OH -, H +, and Cl - H + (aq + OH - (aq H 2O(l OH - H + Initial (mol mol mol Final (mol mol 0 Major Species: Ba 2+, OH -, and Cl - poh = log[oh ] = log ( mol L L = 1.10 ph = poh = = c Major Species: Ba 2+, OH -, H +, and Cl - H + (aq + OH - (aq H 2O(l OH - H + Initial (mol mol mol Final (mol mol 0 Major Species: Ba 2+, OH -, and Cl - poh = log[oh ] = log ( mol L L = 1.44 ph = poh = = d Major Species: Ba 2+, OH -, H +, and Cl - H + (aq + OH - (aq H 2O(l OH - H + Initial (mol mol 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 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 mol mol Final (mol mol Major Species: Ba 2+, H +, and Cl - ph = log[h mol ] = log ( L L = This is a weak acid/strong base titration a Major Species: HC 2H 3O 2 16

17 HC 2H 3O 2(aq H + (aq + C 2H 3O 2- (aq K a = HC 2H 3O 2 H + C 2H 3O Initial M 0 0 Change -x +x +x Equilibrium x x x K a = [HC 2H 3 O 2 ] [H + ][C 2 H 3 O 2 ] = xx = (0.200 x Since K a is small assume x = xx = (0.200 x = Check assumption % = 0.95% Good Calculate ph [H + ] = x = ph = log[h + ] = log( = 2.72 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 Calculate Initial moles of HC 2H 3O 2 and KOH L HC 2 H 3 O 2 ( mol HC 2H3O2 = mol HC 1 L HC2H3O2 2 H 3 O L KOH ( mol KOH 1 L KOH ( 1 mol OH 1 mol KOH = mol OH HC 2H 3O 2 OH - C 2H 3O Initial (mol mol mol 0 Final (mol mol mol Major Species: HC 2H 3O 2, K +, and C 2H 3O [HA] = log( log ( = 4.26 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 2 - Initial (mol mol mol 0 Final (mol mol mol Major Species: HC 2H 3O 2, K +, and C 2H 3O [HA] = log( log ( = mol L mol L mol L mol L 17

18 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 mol mol 0 Final (mol mol mol Major Species: HC 2H 3O 2, K +, and C 2H 3O [HA] = log( log ( = 5.22 e 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 2 - Initial (mol mol mol 0 Final (mol mol Major Species: C 2H 3O 2-, and K + C 2H 3O 2- (aq + H 2O(l HC 2H 3O 2(aq + OH - (aq Calculate K b K b = K w = K a = C 2H 3O HC 2H 3O 2 OH - Initial (mol mol 0 0 Initial (M M 0 0 Change -x +x +x Equilibrium x x x [C 2 H 3 O 2 ] K b = [HC 2 H 3 O 2 ][OH ] = = Since K b is very small assume x = xx = ( x x = Check Assumption % = % Good Calculate ph [OH ] = x = M poh = log[oh ] = log( = 5.21 ph + poh = ph = poh = = 8.79 xx ( x mol L mol L 18

19 f 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 mol mol 0 Final (mol mol Major Species: C 2H 3O 2-, K +, and OH - The majority of the OH - will come from OH - Determine [OH - ] M = n mol = = M L Calculate ph poh = log[oh ] = log( = 1.84 ph + poh = ph = poh = = 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 K b = H 2NNH 2 H 2NNH + 3 OH - Initial M 0 0 Change -x +x +x Equilibrium x x x K b = [H 2NNH + 3 ][OH ] xx = = [H 2 NNH 2 ] (0.100 x Since K a is small assume x = xx = (0.100 x x = Check assumption % = 0.55% Good Calculate ph [OH ] = x = M poh = log[oh ] = log( = 3.26 ph + poh = ph = poh = = 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 mol mol 0 Final (mol mol mol Major Species: H 2NNH 2, NO 3-, and H 2NNH

20 Determine the K a K a = K w = K b = [HA] = log( log ( = 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 mol mol 0 Final (mol mol mol Major Species: H 2NNH 2, NO 3-, and H 2NNH + 3 [HA] = log( log ( = 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 mol mol 0 Final (mol mol mol Major Species: H 2NNH 2, NO 3-, and H 2NNH + 3 [HA] = log( log ( = 7.88 e 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 mol mol 0 Final (mol mol Major Species: H 2NNH - 3+, and NO 3 H 2NNH 3+ (aq H 2NNH 2(aq + H + (aq K a = H 2NNH + 3 H 2NNH 2 H + Initial (mol mol 0 0 Initial (M Change -x +x +x Equilibrium x x x mol L mol L mol L mol L mol L mol L 20

21 K a = [H 2NNH 2 ][H + ] xx [H 2 NNH + = = ] ( x Since K a is small assume x = xx = ( x x = Check assumption % = 0.022% Good Calculate ph [H + ] = x = M ph = log[h + ] = log( = 4.82 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 mol mol 0 Final (mol mol mol Major Species: H 2NNH 3+, H + -, and NO 3 The H + concentration will be mainly from the excess strong acid. ph = log[h + ] = log ( 99. a CaC 2O 4(s Ca 2+ (aq + C 2O 2-4 (aq K sp = [Ca 2+ ][C 2 O 2 4 ] Ca 2+ C 2O mol L = Initial 0 0 Change +x +x Equilibrium x x * x is the solubility The concentrations of Ca 2+ and C 2O 2-4 must be in M. ( g CaC 2 O 4 ( 1 mol CaC 2O 4 = M 1 L CaC 2 O g CaC 2 O 4 K sp = [Ca 2+ ][C 2 O 2 4 ] = xx = x 2 = ( = 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 = [Bi 3+ ][I ] 3 = x(3x 3 = 27x 4 = 27( =

22 106. a Ag 2SO 4(s 2Ag + 2- (aq + SO 4 (aq K sp = [Ag + ] 2 [SO 2 4 ] Ag + 2- SO 4 Initial 0 0 Change +2x +x Equilibrium 2x x K sp = [Ag + ] 2 [SO 2 4 ] = (2x 2 x = 4x 3 = x = M b Ag 2SO 4(s 2Ag + 2- (aq + SO 4 (aq K sp = [Ag + ] 2 [SO 2 4 ] Ag + 2- SO 4 Initial 0.10 M 0 Change +2x +x Equilibrium x x Assume that x is small due to the small equilibrium constant K sp = [Ag + ] 2 [SO 2 4 ] = ( x 2 x = Assume that x =0.10 ( x = x = M Check assumption 2( % = 2.4% Good c Ag 2SO 4(s 2Ag + 2- (aq + SO 4 (aq K sp = [Ag + ] 2 [SO 2 4 ] Ag + 2- SO 4 Initial M Change +2x +x Equilibrium 2x 0.20+x Assume that x is small due to the small equilibrium constant K sp = [Ag + ] 2 [SO 2 4 ] = (2x 2 ( x = Assume that x =0.20 (2x 2 (0.20 = x = M Check assumption % = 2.0% Good 107. 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 - ]= Fe 3+ OH - Initial Change +x +3x Equilibrium x x * x is the solubility 22

23 Since K sp is small assume that x = K sp = [Fe 3+ ][OH ] 3 = x( = x = M Check assumption 3( % = % Good b Fe(OH 3(s Fe 3+ (aq + 3OH - (aq K sp = [Fe 3+ ][OH ] 3 Calculate the concentration of OH - ions when the ph is 5.0 ph = log[h + ] [H + ] = 10 ph = = [H + ][OH ] = [OH ] = [H + = = ] Fe 3+ OH - Initial Change +x +3x Equilibrium x * * x is the solubility * The question told you to assume that the ph is constant K sp = [Fe 3+ ][OH ] 3 = x( = x = M c Fe(OH 3(s Fe 3+ (aq + 3OH - (aq K sp = [Fe 3+ ][OH ] 3 Calculate the concentration of OH - ions when the ph is 11.0 ph = log[h + ] [H + ] = 10 ph = = [H + ][OH ] = [OH ] = [H + = = ] Fe 3+ OH - Initial Gain/Lose +x +3x Equilibrium x * * x is the solubility * The question told you to assume that the ph is constant Since K sp is small assume that x = K sp = [Fe 3+ ][OH ] 3 = x( = x = M 23

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