Chapter 4 Acid-Base Equilibrium DRAFT Answer Key. p. 6. BC Science Chemistry 12 Edvantage Interactive

Transcription

1 p. 2 p. 3 p. 6 BC Science Chemistry 12 Edvantage Interactive

2 p. 7 p. 8 p. 9 BC Science Chemistry 12 Edvantage Interactive

3 BC Science Chemistry 12 Edvantage Interactive

4 p. 12 p. 14 p. 16 p. 18 BC Science Chemistry 12 Edvantage Interactive

5 p. 20 p. 22 p. 23 BC Science Chemistry 12 Edvantage Interactive

6 BC Science Chemistry 12 Edvantage Interactive

7 p. 26 p. 27 p. 28 p. 29 BC Science Chemistry 12 Edvantage Interactive

8 p. 30 p. 31 BC Science Chemistry 12 Edvantage Interactive

9 BC Science Chemistry 12 Edvantage Interactive

10 Section 4.4 Answer Key Warm Up (page 33) 1. Solution [H 3 O ] [OH ] A. 1.0 M NaOH 1.0 x M 1.0 M B. 1.0 M HCl 1.0 M 1.0 x M 2. The moles of hydronium ions present in solution A are equal to the moles of hydroxide ions present in solution B. Therefore, upon mixing, neither will be in excess as a result of the following net ionic equation for the neutralization reaction: H 3 O (aq) OH (aq) 2 H 2 O(l) The solution will therefore be neutral with [H 3 O ] = [OH ] = 1.0 x 10 7 M mol H 3 O x 1.0 L solution = 1.0 x L solution 1.0 x mol H 3 O Quick Check (page 35) 1. Solution [H 3 O ] ph orange juice 3.2 x 10 4 M 3.49 milk of magnesia 2.52 x M stomach acid M The [H 3 O ] increased because the ph decreased. A 5 unit ph decrease corresponds to a 10 5 times increase in hydronium concentration. 3. Initial ph (prior to dilution) = log (0.10) = 1.00 After the dilution, the [H 3 O ] = M. So final ph (after dilution) = log (0.010) = 2.00 (Note 2 decimal places in the answer corresponding to 2 significant figures in each concentration) Practice Problems Converting ph to [H 3 O ] (page 36) 1. [H 3 O ] No. of sig. figures ph 5.00 x 10 5 M x 10 4 M x M x 10 7 M M BC Science Chemistry 12 Edvantage Interactive

11 2. Just before the equivalence point was reached: [H 3 O ] = = 4.5 x 10 5 M Just after the equivalence point was reached: [H 3 O ] = = 2.2 x M 3. Because ph is defined as a negative logarithm, converting a negative ph value to a concentration involves the following: [H 3 O ] = 10 (1.20) = = 16 M (2 sig. figures) The ph scale is not necessary to conveniently express concentrations this large. Practice Problems ph and poh (page 40) 1. Solution [H 3 O ] [OH ] ph poh Acidic/Basic/Neutral? Orange Juice 3 x 10 4 M 3 x M Acidic Tears 3.98 x 10 8 M 2.51 x 10 7 M Basic Blood 4.0 x 10 8 M 2.5 x 10 7 M Basic Milk 3.16 x 10 7 M 3.16 x 10 8 M Acidic 2. a. As a solution becomes more acidic, both [H 3 O ] and poh increase and both [OH ] and ph decrease. b. A basic solution has a poh value which is less than 7, and [H 3 O ] which is greater than 10 7 M. c. If the ph of a solution equals 14.0, the [OH ] equals 1 M. (1 sig. figure) d. If the poh of a solution decreases by 5, then the [H 3 O ] has decreased by a factor of We would expect the K w for water at 10 o C to be less than because the autoionization of water is endothermic. Regardless of the temperature, pure water is always neutral and so: For pure water at any temperature: [H 3 O ] = [OH ] and ph = poh Therefore, because ph poh = pk w, when we calculate the value of pk w, the ph (and also the poh) must be half the value of pk w. pk w = log (2.55 x ) = Therefore, ph (and poh) = = Practice Problems Mixing Strong Acids and Bases (page 43) 1. [H 3 O ] ST = 0.40 M x 25.0 ml = M (in excess) 40.0 ml [OH ] ST = 0.30 M x 15.0 ml = M 40.0 ml [H 3 O ] XS = M M = M so ph = log (0.138) = 0.86 (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

12 2. [H 3 O ] ST = [HCl] = = M (in excess) [OH ] ST = 1.0 g NaOH x 1mol NaOH = M L 40.0 g [H 3 O ] XS = M M = M ph = log (0.151) = 0.82 (2 sig. figures) 3. The first solution has a ph = 2.00 and is therefore acidic with hydronium ions being the major species in the solution. The second solution has a poh = 3.00 and is therefore basic with hydroxide ions as the major species. In the first solution: [H 3 O ] = = M In the second solution: [OH ] = = M Therefore: [H 3 O ] ST = M x 25.0 ml = 3.57 x 10 3 M (in excess) 70.0 ml [OH ] ST = M x 45.0 ml = 6.43 x 10 4 M 70.0 ml [H 3 O ] XS = 3.57 x 10 3 M 6.43 x 10 4 M = 2.93 x 10 3 M ph = log (2.93 x 10 3 ) = 2.53 (2 sig. figures) 4. The final solution is basic so the equation which applies is: [OH ] XS = [OH ] ST [H 3 O ] ST As both [OH ] XS and [OH ] ST are available to us, we must solve for [H 3 O ] ST. Therefore: [H 3 O ] ST = [OH ] ST [OH ] XS Initial poh = = So [OH ] ST = Final poh = = So [OH ] XS = [H 3 O ] ST = (1.500 x 10 3 M) (5.623 x 10 4 M) = x 10 4 M = x 10 3 M = x 10 4 M x 10 4 mol H 3 O = x 10 4 mol HCl x 36.5 g HCl x 1.50 L = g HCl L L mol (3 sig. figures) BC Science Chemistry 12 Edvantage Interactive

13 4.4 Activity: Finding Acidic and Basic Common Solutions (page 44) Solution ph poh [H 3 O ] M [OH ] M Acidic/Basic/Neutral Unpolluted Rainwater x x 10 9 Acidic Saliva x x 10 8 Basic Stomach Acid x Acidic Tears x x 10 7 Basic Vinegar x x Acidic Milk x x 10 8 Acidic Milk of Magnesia x x 10 4 Basic Lemon Juice x x Acidic Tomato Juice x x Acidic Orange Juice x x Acidic Grapefruit Juice x Acidic Liquid Drain Cleaner x Basic Black Coffee x x 10 9 Acidic Urine x x 10 8 Acidic Blood x x 10 7 Basic Laundry Bleach x Basic Windex x x 10 4 Basic Pepto Bismol x x 10 9 Acidic Household Ammonia x x 10 3 Basic Red Wine x x Acidic Results and Discussion 1. None of the solutions are neutral, although some are only slightly acidic or basic. 2. The majority of household solutions used for cleaning purposes are quite basic. In the table above, the solutions having hazardous warning labels are italicized. 3. The consumable solutions are usually acidic. 4.4 Review Questions (page 46) 1. The ph of a solution is defined as the negative logarithm of the concentration of hydronium ions. Equation: ph = log [H 3 O ] The poh of a solution is defined as the negative logarithm of the concentration of hydroxide ions. Equation: poh = log [OH ] 2. Sorenson s logarithmic ph and poh scales are a convenient and compact way of expressing the typically very small concentrations of hydronium and hydroxide ions respectively and the extent by which they can change in aqueous solutions. BC Science Chemistry 12 Edvantage Interactive

14 3. 4. [H 3 O ] ph Acidic/Basic/Neutral 3.50 x 10 6 M Acidic 3.1 x M Basic M Acidic 1.0 M 0.00 Acidic 6. 8 x 10 9 M 8.17 Basic [OH ] poh Acidic/Basic/Neutral 7.2 x 10 9 M 8.14 Acidic 2.8 x M 9.55 Acidic X 10 4 M Basic 1.0 x M Acidic M Basic 5. a) As a solution s poh value and [H 3 O ] both decrease, the solution becomes more basic. b) As a solution s ph value and [OH ] both decrease, the solution becomes more acidic. c) The product of the [H 3 O ] and [OH ] equals K w. d) The sum of ph and poh equals pk w. 6. [H 3 O ] poh Acidic/Basic/Neutral M Acidic 2.51 x 10 6 M Acidic 7.2 x M 2.86 Basic 1.64 x M Basic 7. Pure water is always neutral at any temperature and so ph = poh. Therefore ph = pk w = 6.51 (2 sig. figures) 2 8. The dissociation equation is: Sr(OH) 2 (s) Sr 2 (aq) 2 OH (aq) Therefore: [OH ] = 2 x 0.30 M = 0.60 M poh = log (0.60) = ph = = (2 sig. figures) 9. [OH ] = 2.00 g NaOH x 1 mol NaOH = M L 40.0 g poh = log (0.100) = ph = = (3 sig. figures) BC Science Chemistry 12 Edvantage Interactive

15 10. [H 3 O ] = = 3.16 x 10 3 M = [HI] because HI is a strong acid x 10 3 mol HI x g HI x 2.0 L = 0.81 g (2 sig. figures) L mol 11. [H 3 O ] [OH ] poh ph Acidic/Basic/Neutral x 10 5 M x M Acidic 2.22 x M M Basic 3.2 x 10 2 M 3.2 x M Acidic 3 x M 3 x 10 4 M Basic 12. [H 3 O ] ST = 0.50 M x 75.0 ml = M (in excess) ml [OH ] ST = 0.20 g NaOH x 1 mol x ml = M L 40.0 g ml [H 3 O ] XS = M M = M ph = log (0.163) = 0.79 (2 sig. figures) 13. Initial [H 3 O ] in acidic solution = = M [H 3 O ] ST = M x ml = M ml Initial [OH ] in basic solution = = M [OH ] ST = M x ml = M (in excess) ml [OH ] XS = M M = M (note 1 sig. figure after subtraction) poh = log (0.0064) = 2.19 so ph = = 11.8 (1 sig. figure) 14. [HI] = [H 3 O ] ST because HI is a strong acid. [H 3 O ] ST = 3.2 g HI x 1 mol HI = M L g Initial poh = = 1.00 so [OH ] ST = = 0.10 M (in excess) [OH ] XS = 0.10 M M = 0.05 M (note 1 sig. figure after subtraction) So poh = log (0.050) = 1.30 so ph = = 12.7 (1 sig. figure) BC Science Chemistry 12 Edvantage Interactive

16 15. One solution method is to combine both acidic solutions (call them A and B ) to determine initial [H 3 O ] and then use the initial [H 3 O ] to determine [H 3 O ] ST. Combining the acidic solutions gives: [H 3 O ] A = 0.20 M x 25.0 ml = M 60.0 ml [H 3 O ] B = 0.15 M x 35.0 ml = M 60.0 ml So initial [H 3 O ] = M M = M [H 3 O ] ST = M x 60.0 ml = M ml [OH ] ST = 0.30 M x 40.0 ml = M (in excess) ml [OH ] XS = M M = M So poh = log (0.0175) = so ph = = (2 sig. figures) An alternative approach would be to calculate the total number of moles of hydronium ions from the acid solutions and then subtract that total from the moles of hydroxide ions from the basic solution. The excess moles of hydroxide ions present in the final ml are then used to calculate final ph. 16. The final solution is basic so the equation which applies is: [OH ] XS = [OH ] ST [H 3 O ] ST As both [OH ] XS and [OH ] ST are available to us, we must solve for [H 3 O ] ST. Therefore: [H 3 O ] ST = [OH ] ST [OH ] XS [OH ] ST = [KOH] = M Final poh = = So [OH ] XS = = M [H 3 O ] ST = ( M) ( M) = M [H 3 O ] ST = [HCl] = mol HCl x 36.5 g HCl x L = g HCl L mol (3 sig. figures) BC Science Chemistry 12 Edvantage Interactive

17 17. The final solution is acidic so the equation which applies is: [H 3 O ] XS = [H 3 O ] ST [OH ] ST As both [H 3 O ] XS and [H 3 O ] ST are available to us, we must solve for [OH ] ST. Therefore: [OH ] ST = [H 3 O ] ST [H 3 O ] XS [H 3 O ] ST = [HCl] = M Final ph = 2.75 so [H 3 O ] XS = = M [OH ] ST = M M = [OH ] ST = [LiOH] = mol LiOH x 23.9 g Li x L = 0.13 g LiOH L mol (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

18 Section 4.5 Answer Key Warm Up (page 1) 1. a) HCl(aq) H 2 O(l) Cl (aq) H 3 O (aq) (note 1way arrow) Acid 1 Base 2 Base 1 Acid 2 b) As HCl is a strong acid and thus 100 % ionized, the [H 3 O ] = [HCl] = 0.50 M and ph = log (0.50) = a) HF(aq) H 2 O(l) F (aq) H 3 O (aq) (note 2way arrow) Acid 1 Base 2 Base 1 Acid 2 The reactants are favoured in this equilibrium because the HF is a weaker acid than H 3 O (and H 2 O is a weaker base than F ) and acidbase equilibria always favour the weaker species. b) As HF is a weak acid and thus ionizes only to a slight extent, the [H 3 O ] eq << [HF] initial. Therefore, we would expect the ph to be higher than a. c. K a for HNO 2 = [NO 2 ][H 3 O ] [HNO 2 ] K a for HC 2 O 4 = [C 2 O 4 2 ][H 3 O ] [HC 2 O 4 ] b. d. K b for C 2 O 4 2 = [HC 2 O 4 ][OH ] [C 2 O 4 2 ] K b for NH 3 = [NH 4 ][OH ] [NH 3 ] Quick Check (page 53) 1. The amount of water that a weak acid will convert to hydronium ions is insignificant compared to the magnitude of water s concentration (55.6 M). The concentration of water is thus assumed to be constant and so is not included in an ICE table. BC Science Chemistry 12 Edvantage Interactive

19 2. Because the K a of a weak acid is normally so much greater than K w, we can assume that the initial [H 3 O ] resulting from the autoionization of water is insignificant compared to the equilibrium [H 3 O ] resulting from the ionization of this weak acid. 3. The value of K a is small enough compared to the initial concentration of the acid that the percent of the acid which actually ionizes will not significantly change that original concentration. Remember that this assumption is only valid if the percent ionization of the weak acid is 5 %. Practice Problems Calculating ph given K a and [HA] initial (page 54) 1. Let x = [H 3 O ] eq HCOOH H 2 O HCOO H 3 O I C x x x E 0.50 x x x Assume 0.50 x 0.50 K a = [HCOO ] [H 3 O ] = x 2 _ = 1.8 x 10 4 [HCOOH] 0.50 x = (1.8 x 10 4 ) (0.50) = 9.49 x 10 3 M ph = log (9.49 x 10 3 ) = 2.02 (2 sig. figures) 2. Let x = [H 3 O ] eq CH 3 COOH H 2 O CH 3 COO H 3 O I C x x x E x x x Assume x K a = [CH 3 COO ] [H 3 O ] = x 2 _ = 1.8 x 10 5 [CH 3 COOH] x = (1.8 x 10 5 ) (0.850) = 3.91 x 10 3 M ph = log (3.91 x 10 3 ) = 2.41 (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

20 3. [H 3 O ] = = 5.37 x 10 3 M Percent ionization = [H 3 O ] eq x 100 % = 5.37 x 10 3 M x 100 % = 5.4 % [HA] initial 0.10 M The simplifying assumption would not be valid in this case because the percent ionization is greater than 5 % Practice Problems Calculating [HA] initial, given K a and ph (page 56) 1. Let x = [H 3 C 6 H 5 O 7 ] initial [H 3 O ] eq = = M H 3 C 6 H 5 O 7 H 2 O H 2 C 6 H 5 O 7 H 3 O I x 0 0 C E x K a = [H 3 C 6 H 5 O 7 ] [H 3 O ] = ( ) 2 = 7.1 x 10 4 [H 3 C 6 H 5 O 7 ] (x ) x = ( ) x 10 4 x = M (2 sig. figures) 2. Let x = [H 2 C 2 O 4 ] initial [H 3 O ] eq = = M H 2 C 2 O 4 H 2 O HC 2 O 4 H 3 O I x 0 0 C E x K a = [HC 2 O 4 ] [H 3 O ] = (0.282) 2 = [H 2 C 2 O 4 ] (x 0.282) x = (0.282) x = 1.6 M (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

21 3. Let x = [HNO 2 ] initial [H 3 O ] eq = = M HNO 2 H 2 O NO 2 H 3 O I x 0 0 C E x K a = [NO 2 ] [H 3 O ] = (0.0141) 2 = 4.6 x 10 4 [HNO 2 ] (x ) x = (0.0141) x 10 4 x = 0.45 M (2 sig. figures) Practice Problems Calculating K a, given [HA] initial and ph (page 57) 1. [H 3 O ] eq = = M H 2 C 6 H 6 O 6 H 2 O HC 6 H 6 O 6 H 3 O I C E (2 s.d.) K a = [HC 6 H 6 O 6 ] [H 3 O ] = ( ) 2 = 1.0 x 10 5 [H 2 C 6 H 6 O 6 ] (2 sig. figures) 2. [H 3 O ] eq = = M C 2 H 5 OCOOH H 2 O C 2 H 5 OCOO H 3 O I C E K a = [C 2 H 5 OCOO ] [H 3 O ] = ( ) 2 = 1.4 x 10 4 [C 2 H 5 OCOOH] (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

22 3. [H 3 O ] eq = = M HA H 2 O A H 3 O I C E K a = [A ] [H 3 O ] = ( ) 2 = 6.56 x 10 5 (The acid is most likely benzoic acid) [HA] 1.49 (3 sig. figures) Quick Check (page 59) 1. Weak Base Conjugate Acid Appropriate K a Value pk a Value Calculated K b pk b Value HC 2 O 4 H 2 C 2 O x x H 2 PO 4 H 3 PO x x HPO 4 H 2 PO x x NO 2 HNO x x HC 6 H 5 O 7 H 2 C 6 H 5 O x x HCO 3 H 2 CO x x CN HCN 4.9 x x Practice Problems Calculating [OH ] and ph using K b and [B] initial (page 61) g N 2 H 4 x 1 mol = M L 32.0 g N 2 H 4 Let x = [OH ] eq N 2 H 4 H 2 O N 2 H 5 OH I C x x x E x x x Assume x K b = [N 2 H 5 ] [OH ] = _x 2 _ = 1.7 x 10 6 [N 2 H 4 ] x = 1.13 x 10 3 M OH so poh = log (1.13 x 10 3 ) = ph = = (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

23 2. K b for HCOO = K w = 1.00 x = 5.56 x K a for HCOOH 1.8 x 10 4 Let x = [OH ] eq HCOO H 2 O HCOOH OH I C x x x E 0.60 x x x Assume 0.60 x 0.60 K b = [HCOOH] [OH ] = _x 2 _ = 5.56 x [HCOO ] 0.60 x = [OH ] = 5.8 x 10 6 M (2 sig. figures) so poh = log (5.8 x 10 6 ) = 5.24 ph = = 8.76 and [H 3 O ] = = 1.7 x 10 9 M 3. Calculate the K a and pk a values for (a) Methylammonium, CH 3 NH 3 K a = K w = 1.0 x = 2.3 x K b for CH 3 NH x 10 4 pk a = log (2.3 x ) = (b) Hydrazinium, N 2 H 5 K a = K w = 1.0 x = 5.9 x 10 9 K b for N 2 H x 10 6 pk a = log (5.9 x 10 9 ) = 8.23 Practice Problems Calculating [B] initial given K b and ph (or poh) (page 62) 1. poh = = 2.20 [OH ] = = M Let x = [C 2 H 5 NH 2 ] initial C 2 H 5 NH 2 H 2 O C 2 H 5 NH 3 OH I x 0 0 C E x K b = [C 2 H 5 NH 3 ] [OH ] = ( ) 2 = 5.6 x 10 4 [C 2 H 5 NH 2 ] x BC Science Chemistry 12 Edvantage Interactive

24 x = ( ) x 10 4 so x = M mol C 2 H 5 NH 2 x 45.0 g C 2 H 5 NH 2 x L = 0.87 g C 2 H 5 NH 2 (2 sig. figures) L mol 2. K b for CN = K w = 1.00 x = 2.04 x 10 5 K a for HCN 4.9 x poh = = 2.50 [OH ] = = M Let x = [CN ] initial CN H 2 O HCN OH I x 0 0 C E x K b = [HCN] [OH ] = ( ) 2 = 2.04 x 10 5 [CN ] x x = ( ) 2 so x = 0.49 M (2 sig. figures) 2.04 x [OH ] = 1.0 x x Let x = [N 2 H 4 ] initial = 1.0 x 10 4 M N 2 H 4 H 2 O N 2 H 5 OH I x 0 0 C 1.0 x x x 10 4 E x 1.0 x x x 10 4 K b = [N 2 H 5 ] [OH ] = (1.0 x 10 4 ) 2 = 1.7 x 10 6 [N 2 H 4 ] x 1.0 x 10 4 x 1.0 x 10 4 = (1.0 x 10 4 ) 2 so x = 6.0 x 10 3 M (2 sig. figures) 1.7 x 10 6 Practice Problems Calculating K b, given [B] initial and ph (or poh) (page 64) 1. poh = = 1.10 [OH ] eq = = M CH 3 NH 2 H 2 O CH 3 NH 3 OH BC Science Chemistry 12 Edvantage Interactive

25 I C E K b = [CH 3 NH 3 ] [OH ] = (0.0794) 2 [CH 3 NH 2 ] = = (2 sig. figures) Percent ionization = _[OH ] eq x 100 % = x 100 % = 19.8 % = 20. % [CH 3 NH 2 ] initial (2 sig. figures) This calculated K b value is greater than the K b given in the sample problem (4.4 x 10 4 ). This indicates that the temperature of the solution is higher than 25 o C. 2. [OH ] eq = = 1.26 x 10 4 M Mor H 2 O HMor OH I C 1.26 x x x 10 4 E 9.87 x x x 10 4 K b = [HMor ] [OH ] = (1.26 x 10 4 ) 2 = 1.6 x 10 6 (2 sig. figures) [Mor] 9.87 x 10 3 pk b = log (1.6 x 10 6 ) = poh = = 4.16 [OH ] eq = = 6.92 x 10 5 M Qui H 2 O HQui OH I C 6.92 x x x 10 5 E 1.43 x x x 10 5 K b = [HQui ] [OH ] = (6.92 x 10 5 ) 2 = 3.3 x 10 6 (2 sig. figures) [Qui] 1.43 x 10 3 BC Science Chemistry 12 Edvantage Interactive

26 4.5 Activity: An Organically Grown Table of Relative Acid Strengths (page 65) Compound Name Formula for Conjugate Acid Formula for Conjugate Base Calculate Ka for Acid or Conjugate Acid given: Acids chloroacetic acid ClCH 2 COOH ClCH 2 COO 1.4 x 10 3 phenylacetic acid C 7 H 7 COOH C 7 H 7 COO 4.9 x 10 5 propanoic acid C 2 H 5 COOH C 2 H 5 COO 1.3 x 10 5 pyruvic acid C 2 H 3 OCOOH C 2 H 3 OCOO 2.8 x 10 3 lactic acid C 2 H 5 OCOOH C 2 H 5 OCOO 1.4 x 10 4 acetylsalicylic acid C 8 H 7 O 2 COOH C 8 H 7 O 2 COO 3.6 x 10 4 glycolic acid CH 3 OCOOH CH 3 OCOO 1.5 x 10 4 glyoxylic acid CHOCOOH CHOCOO 3.5 x 10 4 glyceric acid C 2 H 5 O 2 COOH C 2 H 5 O 2 COO 2.9 x 10 4 Bases pyridine C 5 H 5 NH C 5 H 5 N 5.9 x 10 6 trimethylamine (CH 3 ) 3 NH (CH 3 ) 3 N 1.5 x piperidine C 5 H 10 NH 2 tertbutylamine (CH 3 ) 3 CNH 3 ethanolamine C 2 H 5 ONH 3 npropylamine C 3 H 7 NH 3 C 5 H 10 NH 7.8 x (CH 3 ) 3 CNH x C 2 H 5 ONH x C 3 H 7 NH x BC Science Chemistry 12 Edvantage Interactive

27 Relative Strengths of Some Organic Acids and Bases Strength of Acid Stronger Equilibrium Reaction With Water Acid H 2 O H 3 O K a Value Base C 2 H 3 OCOOH H 2 O H 3 O C 2 H 3 OCOO 2.8 x 10 3 ClCH 2 COOH H 2 O H 3 O ClCH 2 COO 1.4 x 10 3 C 8 H 7 O 2 COOH H 2 O H 3 O C 8 H 7 O 2 COO 3.6 x 10 4 CHOCOOH H 2 O H 3 O CHOCOO 3.5 x 10 4 C 2 H 5 O 2 COOH H 2 O H 3 O C 2 H 5 O 2 COO 2.9 x 10 4 CH 3 OCOOH H 2 O H 3 O CH 3 OCOO 1.5 x 10 4 C 2 H 5 OCOOH H 2 O H 3 O C 2 H 5 OCOO 1.4 x 10 4 C 7 H 7 COOH H 2 O H 3 O C 7 H 7 COO 4.9 x 10 5 C 2 H 5 COOH H 2 O H 3 O C 2 H 5 COO 1.3 x 10 5 C 5 H 5 NH H 2 O H 3 O C 5 H 5 N 5.9 x 10 6 Strength of Base Weaker C 2 H 5 ONH 3 H 2 O H 3 O C 2 H 5 ONH x (CH 3 ) 3 NH H 2 O H 3 O (CH 3 ) 3 N 1.5 x C 3 H 7 NH 3 H 2 O H 3 O C 3 H 7 NH x (CH 3 ) 3 CNH 3 H 2 O H 3 O (CH 3 ) 3 CNH x Weaker C 5 H 10 NH 2 H 2 O H 3 O C 5 H 10 NH 7.8 x Stronger Results and Discussion 1. The K a values of the weak monoprotic carboxylic acids are within approximately 10 2 times of each other. 2. The strongest acid is pyruvic acid and the strongest base piperidine. 3. The K a values allow all of the above weak acids to be incorporated into the Table of Relative Acid Strengths from weakest to strongest acids. 4.5 Review Questions (page 68) g HCOOH x 1 mol = 1.00 M HCOOH L 46.0 g HCOOH BC Science Chemistry 12 Edvantage Interactive

28 Let x = [H 3 O ] eq HCOOH H 2 O HCOO H 3 O I C x x x E 1.00 x x x Assume 1.00 x 1.00 K a = [HCOO ] [H 3 O ] = x 2 _ = 1.8 x 10 4 [HCOOH] 1.00 x = (1.8 x 10 4 ) (1.00) = M ph = log ( ) = 1.87 (2 sig. figures) [H 3 O ] = = M poh = = [OH ] = = 7.41 x M L H 2 S x 1 mol = 0.10 M H 2 S 2.50 L 22.4 L H 2 S Let x = [H 3 O ] eq H 2 S H 2 O HS H 3 O I C x x x E 0.10 x x x Assume 0.10 x 0.10 K a = [HS ] [H 3 O ] = x 2 _ = 9.1 x 10 8 [H 2 S] 0.10 x = (9.1 x 10 8 ) (0.10) = 9.54 x 10 5 M ph = log (9.54 x 10 5 ) = 4.02 (2 sig. figures) Percent ionization = 9.54 x 10 5 x 100% = % % ionization = [H 3 O ] eq x 100 % so [H 3 O ] eq = (% ionization) [HA] initial [HA] initial 100 % [H 3 O ] eq = (1.34)(0.100) = M so ph = log ( ) = (3 sig. figures) 100 BC Science Chemistry 12 Edvantage Interactive

29 As 1.34% < 5%, we can assume that [HA] eq [HA] initial and [H 3 O ] eq = [A ] eq This allows us to calculate K a and thereby identify the unknown acid. K a = ( ) 2 = 1.80 x 10 5 (The unknown acid is acetic acid.) Note: We could have also chosen not to make the assumption because all of the equilibrium concentration values are available to us. The resulting K a calculation would have yielded the following: K a = ( ) 2 = 1.82 x 10 5 (The unknown acid would still represent acetic acid given that the K a values on the Table are given to 2 sig. figures.) 4. Let x = [HF] initial [H 3 O ] eq = = M HF H 2 O F H 3 O I x 0 0 C E x K a = [F ] [H 3 O ] = (0.0100) 2 = 3.5 x 10 4 [HF] (x ) x = (0.0100) x 10 4 so x = M HF mol HF x 20.0 g HF x 1.5 L = 8.9 g HF (2 sig. figures) L mol 5. Let x = [H 3 PO 4 ] initial ph = = 1.50 so [H 3 O ] eq = = M H 3 PO 4 H 2 O H 2 PO 4 H 3 O I x 0 0 C E x K a = [H 2 PO 4 ] [H 3 O ] = (0.0316) 2 = 7.5 x 10 3 [H 3 PO 4 ] (x ) x = (0.0316) x 10 3 so x = 0.16 M H 3 PO 4 (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

30 6. Let x = [H 2 C 2 O 4 ] initial [H 3 O ] eq = = M H 2 C 2 O 4 H 2 O HC 2 O 4 H 3 O I x 0 0 C E x K a = [HC 2 O 4 ] [H 3 O ] = ( ) 2 = 5.9 x 10 2 [H 2 C 2 O 4 ] (x ) x = ( ) 2 so x = M H 2 C 2 O x mol H 2 C 2 O 4 x 90.0 g H 2 C 2 O 4 x L = 0.11 g H 2 C 2 O 4 (2 sig. figures) L mol 7. [H 3 O ] eq = = 2.40 x 10 5 M HClO H 2 O ClO H 3 O I C 2.40 x x x 10 5 E x x 10 5 K a = [ClO ] [H 3 O ] = (2.40 x 10 5 ) 2 = 2.9 x 10 8 (2 sig. figures) [HClO] ph = = 2.66 so [H 3 O ] eq = = M C 6 H 5 CH 2 COOH H 2 O C 6 H 5 CH 2 COO H 3 O I C E K a = [C 6 H 5 CH 2 COO ] [H 3 O ] = ( ) 2 = 4.9 x 10 5 (2 sig. figures) [C 6 H 5 CH 2 COOH] BC Science Chemistry 12 Edvantage Interactive

31 9. Conjugate Conjugate K a for acid pk a K b for base pk b Acid Base HNO 2 NO x x H 2 O 2 HO x x C 6 H 5 OH C 6 H 5 O 1.3 x x HSO 4 2 SO x x Conjugate Acid Conjugate Base H 2 PO 4 2 HPO 4 H 2 C 6 H 5 O 7 2 HC 6 H 5 O 7 H 3 BO 3 H 2 BO 3 HCO 3 2 CO 3 K a for acid pk a K b for base pk b 6.2 x x x x x x x x L NH 3 x 1 mol = M NH L 22.4 L NH 3 Let x = [OH ] eq NH 3 H 2 O NH 4 OH I C x x x E x x x Assume x K b for NH 3 = K w = 1.0 x = 1.8 x 10 5 K a for NH x K b = [NH 4 ] [OH ] = _x 2 _ = 1.8 x 10 5 [NH 3 ] x = 2.45 x 10 3 M OH so poh = log (2.45 x 10 3 ) = ph = = (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive

32 12. Let x = [OH ] eq (CH 3 ) 2 CHNH 2 H 2 O (CH 3 ) 2 CHNH 3 OH I C x x x E 0.60 x x x Assume 0.60 x 0.60 K b = [(CH 3 ) 2 CHNH 3 ] [OH ] = _x 2 _ = 4.7 x 10 4 [(CH 3 ) 2 CH NH 2 ] 0.60 x = M OH so poh = log (0.0168) = 1.77 (2 sig. figures) ph = = poh = = 4.00 [OH ] = 1.00 x 10 4 M Let x = [SO 3 2 ] initial 2 SO 3 H 2 O HSO 3 OH I x 0 0 C 1.00 x x x 10 4 E x 1.00 x x x 10 4 K b for SO 3 2 = K w = 1.0 x = 1.0 x 10 7 K a for HSO x 10 7 K b = [HSO 3 ] [OH ] = (1.00 x 10 4 ) 2 = 1.0 x 10 7 [SO 3 2 ] x 1.00 x 10 4 x 1.00 x 10 4 = (1.00 x 10 4 ) 2 so x = 0.10 M (2 sig. figures) 1.0 x [OH ] = = M Let x = [(CH 3 ) 3 N] initial (CH 3 ) 3 N H 2 O (CH 3 ) 3 NH OH I x 0 0 C E x BC Science Chemistry 12 Edvantage Interactive

33 K b = [(CH 3 ) 3 NH ] [OH ] = ( ) 2 = 6.3 x 10 5 [(CH 3 ) 3 N] x x = ( ) x 10 5 so x = M (CH 3 ) 3 N 2.5 L x mol (CH 3 ) 3 N x 22.4 L (CH 3 ) 3 N = 9.1 L (2 sig. figures) L mol 15. [OH ] = = M C 2 H 5 NH 2 H 2 O C 2 H 5 NH 3 OH I C E K b = [C 2 H 5 NH 3 ] [OH ] = ( ) 2 [C 2 H 5 NH 2 ] = 4.3 x 10 4 (2 sig. figures) BC Science Chemistry 12 Edvantage Interactive