# Homework #7 Chapter 8 Applications of Aqueous Equilibrium

Save this PDF as:

Size: px
Start display at page:

Download "Homework #7 Chapter 8 Applications of Aqueous Equilibrium"

## Transcription

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

### Problem 1 C 6 H 5 [ COOH C 6 H[H 5 COO + ] - + H [ I C - x + x + x E x x x

Problem 1 What is the ph of a 291mL sample of 2.993M benzoic acid (C 6 H 5 COOH) (K a =6.4x10 5 )? Write out acid dissociation reaction: C 6 H 5 COOH C 6 H 5 COO H Make an ICE chart since this is a weak

### CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, (i) What is the conjugate base of each of the following species?

CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, 2001 M.A. Brook B.E. McCarry A. Perrott 1. (i) What is the conjugate base of each of the following species? (a) H 3 O + (b) NH 4 + (c) HCl

### Chapter 17 Homework Problem Solutions

Chapter 17 Homework Problem Solutions 17.40 D 2 O D + + OD, K w = [D + ] [OD ] = 8.9 10 16 Since [D + ] = [OD ], we can rewrite the above expression to give: 8.9 10 16 = ([D + ]) 2, [D + ] = 3.0 10 8 M

### CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold.

CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold. 1. Consider the equilibrium: PO -3 4 (aq) + H 2 O (l) HPO 2-4 (aq)

### ph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin)

JASPERSE CHEM 210 PRACTICE TEST 3 VERSION 2 Ch. 17: Additional Aqueous Equilibria Ch. 18: Thermodynamics: Directionality of Chemical Reactions Key Equations: For weak acids alone in water: [H + ] = K a

### Chapter 16 Aqueous Ionic Equilibrium

Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall The Danger of Antifreeze

### AP Chapter 15 & 16: Acid-Base Equilibria Name

AP Chapter 15 & 16: Acid-Base Equilibria Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 15 & 16: Acid-Base Equilibria 2 Warm-Ups (Show

### CHAPTER 13: ACIDS & BASES. Section Arrhenius Acid & Bases Svante Arrhenius, Swedish chemist ( ).

CHAPTER 13: ACIDS & BASES Section 13.1 Arrhenius Acid & Bases Svante Arrhenius, Swedish chemist (1839-1927). He understood that aqueous solutions of acids and bases conduct electricity (they are electrolytes).

### AP Chemistry: Acid-Base Chemistry Practice Problems

Name AP Chemistry: Acid-Base Chemistry Practice Problems Date Due Directions: Write your answers to the following questions in the space provided. For problem solving, show all of your work. Make sure

### AP Chemistry. CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect Buffered Solutions. Composition and Action of Buffered Solutions

AP Chemistry CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak electrolyte.

### ACIDS AND BASES. HCl(g) = hydrogen chloride HCl(aq) = hydrochloric acid HCl(g) H + (aq) + Cl (aq) ARRHENIUS THEORY

ACIDS AND BASES A. CHARACTERISTICS OF ACIDS AND BASES 1. Acids and bases are both ionic compounds that are dissolved in water. Since acids and bases both form ionic solutions, their solutions conduct electricity

### 16.3 Weak Acids Weak Bases Titration

16.3 Weak Acids Weak Bases Titration Titration of Weak Acid with Strong Base Titration of Base Acid with Strong Acid Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Weak Acids Weak Bases Titration

### Strong and Weak. Acids and Bases

Strong and Weak Acids and Bases Strength of Acids H2SO4 HSO4 - + H + HNO3 NO3 - + H + Strong Acids HCl Cl - + H + H3PO4 H2PO4 - + H + Phosphoric acid Moderate Acid CH3COOH CH3COO - + H + Acetic acid HF

### Department of Chemistry University of Texas at Austin

Polyprotic and Special Cases Calculations Supplemental Worksheet KEY For the following polyprotic acid questions: Citric acid (H3C6H5O6) Ka1 = 8.4 x 10 4 Ka2 = 1.8 x 10 5 Ka3 = 4.0 x 10 6 Oxalic acid (H2C2O4)

### Acid Base Equilibria

Acid Base Equilibria Acid Ionization, also known as acid dissociation, is the process in where an acid reacts with water to produce a hydrogen ion and the conjugate base ion. HC 2 H 3 O 2(aq) H + (aq)

### Chapter 17 Additional Aspects of

Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17 Additional Aspects of AP Chemistry 2014-15 North Nova Education Centre Mr. Gauthier

### ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor

ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate acid Conjugate

### Acid-Base Titration Solution Key

Key CH3NH2(aq) H2O(l) CH3NH3 (aq) OH - (aq) Kb = 4.38 x 10-4 In aqueous solution of methylamine at 25 C, the hydroxide ion concentration is 1.50 x 10-3 M. In answering the following, assume that temperature

### IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water.

IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water. The Brønsted-Lowry definition of an acid is a species that can donate an H + ion to any

### Chemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B

Chemical Equilibrium Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product formation,

### Acids, Bases and Buffers

1 Acids, Bases and Buffers Strong vs weak acids and bases Equilibrium as it relates to acids and bases ph scale: [H+(aq)] to ph, poh, etc ph of weak acids ph of strong acids Conceptual about oxides (for

### Worksheet 4.1 Conjugate Acid-Base Pairs

Worksheet 4.1 Conjugate AcidBase Pairs 1. List five properties of acids that are in your textbook. Acids conduct electricity, taste sour, neutralize bases, change the color of indicators, and react with

### Chapter 16 Homework Solutions

//05 Chapter 16 Homework Solutions 6. a) H AsO b) CH 3 NH 3 + c) HSO d) H 3 PO 8. acid base conj. base conj. acid a) H O CHO OH CH O a) HSO HCO 3 SO H CO 3 b) H 3 O + HSO 3 H O H SO 3 10. a) H C 6 H 7

### CHAPTER 7.0: IONIC EQUILIBRIA

Acids and Bases 1 CHAPTER 7.0: IONIC EQUILIBRIA 7.1: Acids and bases Learning outcomes: At the end of this lesson, students should be able to: Define acid and base according to Arrhenius, Bronsted- Lowry

### A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base.

1 A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base. after addition of H 3 O + equal concentrations of weak

### molality: m = = 1.70 m

C h e m i s t r y 1 2 U n i t 3 R e v i e w P a g e 1 Chem 12: Chapters 10, 11, 12, 13, 14 Unit 3 Worksheet 1. What is miscible? Immiscible? Miscible: two or more substances blend together for form a solution

### Chemistry 102 Chapter 15 ACID-BASE CONCEPTS

General Properties: ACID-BASE CONCEPTS ACIDS BASES Taste sour Bitter Change color of indicators Blue Litmus turns red no change Red Litmus no change turns blue Phenolphtalein Colorless turns pink Neutralization

### 5.111 Lecture Summary #22 Wednesday, October 31, 2014

5.111 Lecture Summary #22 Wednesday, October 31, 2014 Reading for Today: Sections 11.13, 11.18-11.19, 12.1-12.3 in 5 th ed. (10.13, 10.18-10.19, 11.1-11.3 in 4 th ed.) Reading for Lecture #23: Sections

### Acid-Base Equilibria. 1.NH 4 Cl 2.NaCl 3.KC 2 H 3 O 2 4.NaNO 2. Acid-Ionization Equilibria. Acid-Ionization Equilibria

Acid-Ionization Equilibria Acid-Base Equilibria Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion. (See Animation:

### CH 15 Summary. Equilibrium is a balance between products and reactants

CH 15 Summary Equilibrium is a balance between products and reactants Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios. Capital K is used to represent the equilibrium

### Acids and Bases Review Worksheet II Date / / Period. Molarity. moles L. Normality [H 3 O +1 ] [OH -1 ] ph poh

Honors Chemistry Name Acids and Bases Review Worksheet II Date / / Period Solute Name of Solute Molar Mass grams mole Molarity moles L Normality [H 3 O +1 ] [OH ] ph poh Acidic or Basic 1. HCl Hydrochloric

### School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban. CHEM191 Tutorial 1: Buffers

School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban CHEM191 Tutorial 1: Buffers Preparing a Buffer 1. How many moles of NH 4 Cl must be added to 1.0 L of 0.05 M NH 3 to form

### Chapter 8: Applications of Aqueous Equilibria

Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations

### Chem12 Acids : Exam Questions M.C.-100

Chem12 Acids : Exam Questions M.C.-100 1) Given : HPO 4 2- (aq) + NH 4 + (aq) H 2 PO 4 - (aq) + NH 3 (aq), the strongest acid in the above equation is : a) NH 4 + b) HPO 4 2- c) NH 3 d) H 2 PO 4-2)

### Problem Solving. ] Substitute this value into the equation for poh.

Skills Worksheet Problem Solving In 1909, Danish biochemist S. P. L Sørensen introduced a system in which acidity was expressed as the negative logarithm of the H concentration. In this way, the acidity

### Acid and Base Titrations - Equation Guide

Acid and Base Titrations - Equation Guide Strong Acid + Strong Base: Initial Region: ph = - log (n sa / V sa ) or ph = - log (C sa ) Pre-Equivalence Region: ph = - log sa # or ph = - log Equivalence: ph

### Acids and Bases Written Response

Acids and Bases Written Response January 1999 4. Consider the salt sodium oxalate, Na2C2O4. a) Write the dissociation equation for sodium oxalate. (1 mark) b) A 1.0M solution of sodium oxalate turns pink

### Chpt 16: Acids and Bases

Chpt 16 Acids and Bases Defining Acids Arrhenius: Acid: Substances when dissolved in water increase the concentration of H+. Base: Substances when dissolved in water increase the concentration of OH- Brønsted-Lowry:

### minocha (am56888) Topic 08 - ph Calculations brakke (2012SL) 1 1. an acid. correct 2. a solvent. 3. a base. 4. a salt. 1. hydrogen.

minocha (am56888) Topic 08 - ph Calculations brakke (2012SL) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.

### Mixtures of Acids and Bases

Mixtures of Acids and Bases CH202, lab 6 Goals : To calculate and measure the ph of pure acid and base solutions. To calculate and measure the ph of mixtures of acid and base solutions. Safety : Hydrochloric

### K w. Acids and bases 8/24/2009. Acids and Bases 9 / 03 / Ionization of water. Proton Jumping Large proton and hydroxide mobility

Chapter 2 Water Acids and Bases 9 / 03 / 2009 1. How is the molecular structure of water related to physical and chemical behavior? 2. What is a Hydrogen Bond? 3Wh 3. What are Acids Aid and db Bases? 4.

### Brønsted-Lowry Acid-Base Model. Chapter 13 Acids and Bases. The Nature of H + Outline. Review from Chapter 4. Conjugate Pairs

Brønsted-Lowry Acid-Base Model William L Masterton Cecile N. Hurley Edward J. Neth cengage.com/chemistry/masterton Chapter 13 Acids and Bases Brønsted-Lowry Johannes Brønsted (1879-1947) Thomas Lowry (1874-1936)

### Acids, Bases and Salts

(Hebden Unit 4 page 109 182) 182) We will cover the following topics: 1. Definition of Acids and Bases 2. Bronsted-Lowry Acids and Bases 2 1 Arrhenius Definition of Acids and Bases An acid is a substance

### First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, 11-13

Announcements First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, 11-13 Please keep up with the work (lots of problems in this Chapter) and see me if you have

### Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA

APPLICATIONS OF AQUEOUS EQUILIBRIA Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution

### Water. Water participates in H-bonding with biomolecules.

Water Most biochemical reactions occur in an aqueous environment. Water is highly polar because of its bent geometry. Water is highly cohesive because of intermolecular hydrogen bonding. Water participates

### 7. The coffee cup allows for pv work because it allows for a change in volume.

1. A black body radiator is a theoretically perfect body that absorbs all energy incident upon it (or produced within it) and then emits 100% of this energy as electromagnetic radiation. 2. First, it is

### Secondary Topics in Equilibrium

Secondary Topics in Equilibrium Outline 1. Common Ions 2. Buffers 3. Titrations Review 1. Common Ions Include the common ion into the equilibrium expression Calculate the molar solubility in mol L -1 when

### Titration Of A Weak Acid With Strong Base. BCH 312 [Practical]

Titration Of A Weak Acid With Strong Base BCH 312 [Practical] Weak Acid : Weak acids or bases do not dissociate completely, therefore an equilibrium expression with Ka must be used. The Ka is a quantitative

### Acids and Bases. Chapter 11

Acids and Bases Chapter 11 Acids and Bases in our Lives Acids and bases are important substance in health, industry, and the environment. One of the most common characteristics of acids is their sour taste.

### Chem 106 Thursday, March 10, Chapter 17 Acids and Bases

Chem 106 Thursday, March 10, 2011 Chapter 17 Acids and Bases K a and acid strength Acid + base reactions: Four types (s +s, s + w, w + s, and w + w) Determining K from concentrations and ph ph of aqueous

### Acids and Bases. Reviewing Vocabulary CHAPTER ASSESSMENT CHAPTER 19. Compare and contrast each of the following terms.

Acids and Bases Reviewing Vocabulary Compare and contrast each of the following terms. 1. Arrhenius model, Brønsted-Lowry model 2. acid ionization constant, base ionization constant 3. conjugate acid,

### Unit 4-1 Provincial Practice Questions Page 1

Page 1 Page 2 Page 3 Page 4 Page 5 1.00 Page 6 Page 7 Page 8 55. The conjugate base of HAsO 4 2 is A. H 3 O + B. 3 AsO 4 C. H 3 AsO 4 D. H 2 AsO 4 56. Consider the following acidbase equilibrium: HC 6

### Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations

Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations Sec 1 The Common-Ion Effect: The dissociation of a weak electrolyte decreases when a strong electrolyte that has an ion in common with

### Buffers. How can a solution neutralize both acids and bases? Beaker B: 100 ml of 1.00 M HCl. HCl (aq) + H 2 O H 3 O 1+ (aq) + Cl 1 (aq)

Buffers How can a solution neutralize both acids and bases? Why? Buffer solutions are a mixture of substances that have a fairly constant ph regardless of addition of acid or base. They are used in medicine,

### Titration of a weak acid with strong base

Titration of a weak acid with strong base - Objectives: - To study titration curves. - Determine the pka value of a weak acid. - Reinforce the understanding of buffers. - Titration Curves: - Titration

### More About Chemical Equilibria

1 More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 15 & 16 1 Objectives Chapter 15 Define the Common Ion Effect (15.1) Define buffer and show how a buffer controls ph of a solution

### Department of Chemistry University of Texas at Austin

Titrations and Buffers Supplemental Worksheet KEY HINT: When calculating the ph of a solution use the following 3 steps Titrations The next six problems represent many points along a titration curve of

### Review: Acid-Base Chemistry. Title

Review: Acid-Base Chemistry Title Basics General properties of acids & bases Balance neutralization equations SA + SB water + salt Arrhenius vs. Bronsted-Lowry BL plays doubles tennis match with H+) Identify

### AP Chemistry - Packet #1 - Equilibrium. 1) a) 5 points K p = (P NH3 ) (P H2S )

1) a) 5 points K p = (P NH3 ) (P H2S ) P NH3 = P H2S = (0.659 atm / 2) = 0.330 atm K p = (0.330 atm) (0.330 atm) = 0.109 atm 2 b) 5 points P NH3 = 2 P H2S (2x) (x) = 0.109 x = 0.233 atm = P H2S 2x = 0.466

### CHAPTER 14 ACIDS AND BASES. Questions

CHAPTER 1 ACIDS AND BASES Questions 19. Acids are proton (H donors, and bases are proton acceptors. HCO as an acid: HCO (aq H O(l CO (aq H O (aq HCO as a base: HCO (aq H O(l H CO (aq OH (aq H PO as an

### Chemistry 12 Provincial Exam Workbook Unit 04: Acid Base Equilibria. Multiple Choice Questions

R. Janssen, MSEC Chemistry 1 Provincial Workbook (Unit 0), P. 1 / 69 Chemistry 1 Provincial Exam Workbook Unit 0: Acid Base Equilibria Multiple Choice Questions 1. Calculate the volume of 0.00 M HNO needed

### Review of Chemistry 11

Review of Chemistry 11 HCl C 3 H 8 SO 2 NH 4 Cl KOH H 2 SO 4 H 2 O AgNO 3 PbSO 4 H 3 PO 4 Ca(OH) 2 Al(OH) 3 P 2 O 5 Ba(OH) 2 CH 3 COOH 1. Classify the above as ionic or covalent by making two lists. Describe

### K A K B = K W pk A + pk B = 14

Relationship between the ionization constants of an acid and its conjugate base HCN (aq) H 2 O(l) CN (aq) H O (aq) Conjugate couple The product between of an acid and of its conjugate base is : p p 14

### Problems -- Chapter Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of the following.

Problems -- Chapter 1 1. Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of the following. (a) NaNO and HNO answers: see end of problem set (b)

### [H ] [OH ] 5.6 " 10

Howemork set solutions 10: 11.1 Table 11.5 of the tet contains a list of important Brønsted acids and bases. (a) both, base, (c) acid, (d) base, (e) acid, (f) base, (g) base, (h) base, (i) acid, (j) acid.

### Mr. Storie 40S Chemistry Student Acid and bases Unit. Acids and Bases

Acids and Bases 1 UNIT 4: ACIDS & BASES OUTCOMES All important vocabulary is in Italics and bold. Outline the historical development of acid base theories. Include: Arrhenius, BronstedLowry, Lewis. Write

### HALFWAY to EQUIVALENCE POINT: ph = pk a of the acid being titrated.

CHEMISTRY 109 Help Sheet #33 Titrations Chapter 15 (Part II); Section 15.2 ** Cover topics appropriate for your lecture** Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets:

### EQUILIBRIUM GENERAL CONCEPTS

017-11-09 WHEN THE REACTION IS IN EQUILIBRIUM EQUILIBRIUM GENERAL CONCEPTS The concentrations of all species remain constant over time, but both the forward and reverse reaction never cease When a system

### Page 1. Spring 2002 Final Exam Review Palmer Graves, Instructor MULTIPLE CHOICE

Page 1 MULTIPLE CHOICE 1. Which one of the following exhibits dipole-dipole attraction between molecules? a) XeF b) AsH c) CO d) BCl e) Cl 2. What is the predominant intermolecular force in AsH? a) London-dispersion

### Name: Per: Date: Unit 11 - Acids, Bases and Salts Chemistry Accelerated Chemistry I Define each of the following: 1. Acidic hydrogens.

Name: Per: Date: Unit 11 - Acids, Bases and Salts Chemistry Accelerated Chemistry I Define each of the following: 1. Acidic hydrogens 2. Binary acids 3. Oxyacids 4. Carboxylic acid 5. Amines Name the following

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which one of the following is the weakest acid? 1) A) HF (Ka = 6.8 10-4) B) HNO2 (Ka

### 1) What is the Arrhenius definition of an acid? Of a base? 2) What is the Bronsted-Lowry definition of an acid? Of a base?

Problems, Chapter 16 (with solutions) NOTE: Unless otherwise stated, assume T = 25. C in all problems) 1) What is the Arrhenius definition of an acid? Of a base? An Arrhenius acid is a substance that produces

### CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.3 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00

### Acid-Base Character of Salt Solutions. Cations. Cations are potentially acidic, but some have no effect on ph.

Acid-Base Character of Salt Solutions The ph of a salt solution will depend on the acidbase nature of both the cation and anion. Cations Cations are potentially acidic, but some have no effect on ph. M(H

### 1. Know and be capable of applying the Bronsted-Lowery model of acids and bases (inculdig the concepts related to conjugate acid-base pairs.

Acid-Base Equilibria You have just completed a chapter on equilibrium. That chapter focused primarily on gas phase reactions (with a few exceptions). This section on Acid-Base equilibria (along with the

### ACIDS AND BASES CONTINUED

ACIDS AND BASES CONTINUED WHAT HAPPENS WHEN AN ACID DISSOLVED IN WATER? Water acts as a Brønsted Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a

### Do Now April 24, 2017

Do Now April 24, 2017 Obj: Observe and describe neutralization reactions. Copy: Neutralization is when an acid and base react to product a salt and water. e.g. HCl + NaOH NaCl + H 2 O acid base salt water

### NATURE OF ACIDS & BASES

General Properties: NATURE OF ACIDS & BASES ACIDS BASES Taste sour Bitter Change color of indicators Blue Litmus turns red no change Red Litmus no change turns blue Phenolphtalein Colorless turns pink

### HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x

Chpter 15 : xx, 4, 8, 0, 8, 44, 48, 50, 51, 55, 76, 8, 90, 96(90), 1(111) xx. NH 4 OH NH H O NH H O NH 4 H O 4.. HONH (q) H O(l) OH (q) HONH (q) HONH H O OH HONH initil 0.100 M ~0 0 x x x equilibrium 0.100

### Chem 5 PAL Worksheet Acids and Bases Smith text Chapter 8

D.CHO3HE.KOHB.NHC.CHC3OHHCl3F.H.CHHCH3COG.H2HCHEM 5 PAL Worksheet Acids and Bases Fall 2017 Chem 5 PAL Worksheet Acids and Bases Smith text Chapter 8 Many substances in the body are acids and bases. Many

### (aq)], does not contain sufficient base [C 2 H 3 O 2. (aq)] to be a buffer. If acid is added, there is too little conjugate base [C 2 H 3 O 2

PURPOSE: 1. To understand the properties of buffer solutions. 2. To calculate the ph of buffer solutions and compare the calculated values with the experimentally determined ph values. PRINCIPLES: I. Definition,

### Volume NaOH Delivered (ml)

Chemistry Spring 011 Exam 3: Chapters 8-10 Name 80 Points Complete five (5) of the following problems. Each problem is worth 16 points. CLEARLY mark the problems you do not want graded. You must show your

### 2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect

Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento,

### Acids, Bases, & Neutralization Chapter 20 & 21 Assignment & Problem Set

Acids, Bases, & Neutralization Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Acids, Bases, & Neutralization 2 Study Guide: Things You Must Know

### 15 Acids, Bases, and Salts. Lemons and limes are examples of foods that contain acidic solutions.

15 Acids, Bases, and Salts Lemons and limes are examples of foods that contain acidic solutions. Chapter Outline 15.1 Acids and Bases 15.2 Reactions of Acids and Bases 15.3 Salts 15.4 Electrolytes and

### cm mol l -1 NaOH added to 50.0 cm 3 of 0.10 mol l -1 HCl

cm 3 0.10 mol l -1 NaOH added to 50.0 cm 3 of 0.10 mol l -1 HCl Acids have been described as substances that dissolve in water to form H + (aq) ions, whilst bases are substances that react with acids.

### Chapter 15 Acids and Bases

Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Stomach Acid & Heartburn the cells

### Acid Base Titrations

ChemActivity CA47b Acid Base Titrations Model 1 Titration of a strong acid with a strong base. 20.00 ml of HNO 3 is titrated with 0.10 M NaOH. The acid-base reaction is The net ionic reaction is HNO 3

### Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 167.

Honour Chemistry Unit 8: Acids and Bases Chapter 20: Acids and Bases 20.1: Describing Acids and Bases UNIT 8: ACIDS AND BASES Physical and Chemical Properties of Acid and Base Acids Bases Taste Sour (Citric

### Ch 16: Acids and Bases

Ch 16: Acids and Bases A c i d s a n d B a s e s C h 1 6 P a g e 1 Homework: Read Chapter 16 Work out sample/practice exercises in the sections, Chapter problems: 39, 41, 49, 63, 67, 83, 91, 95, 99, 107,

### Chem 210 Jasperse Final Exam- Version 1 Note: See the very last page to see the formulas that will be provided with the final exam.

Chem 210 Jasperse Final Exam- Version 1 Note: See the very last page to see the formulas that will be provided with the final exam. 1 1. Which of the following liquids would have the highest vapor pressure,

### Chapter 7 Acids and Bases

Chapter 7 Acids and Bases 7.1 The Nature of Acids and Bases 7.2 Acid Strength 7.3 The ph Scale 7.4 Calculating the ph of Strong Acid Solutions 7.5 Calculating the ph of Weak Acid Solutions 7.6 Bases 7.7

### Cu 2+ (aq) + 4NH 3(aq) = Cu(NH 3) 4 2+ (aq) I (aq) + I 2(aq) = I 3 (aq) Fe 3+ (aq) + 6H 2O(l) = Fe(H 2O) 6 3+ (aq) Strong acids

There are three definitions for acids and bases we will need to understand. Arrhenius Concept: an acid supplies H + to an aqueous solution. A base supplies OH to an aqueous solution. This is the oldest

### A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

### Experiment 32C APPLICATIONS OF ACID-BASE EQUILIBRIA

Experiment 32C APPLICATIONS OF ACID-BASE EQUILIBRIA FV 23Feb18 MATERIALS: 50 ml buret (2), 25 ml graduated cylinder (2), 50 ml beaker (2), 150 ml beaker (2), small plastic vials (6), stirring rods (2),

### 5.111 Principles of Chemical Science

MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 24.1 5.111 Lecture

### Chemistry 12 AUGUST Course Code = CH. Student Instructions

MINISTRY USE ONLY MINISTRY USE ONLY Place Personal Education Number (PEN) here. Place Personal Education Number (PEN) here. MINISTRY USE ONLY Chemistry 12 2001 Ministry of Education AUGUST 2001 Course