Advanced Chemistry Ch What is the concentration of a monoprotic weak acid if its ph is 5.50 and its K a = 5.7 x 10-10?

Transcription

1 Advanced Chemistry Ch. 15 Name 1. What is the concentration of a monoprotic weak acid if its ph is 5.50 and its K a = 5.7 x 10-10? 2. What is the concentration of a weak base if its K b = 1.4 x and its ph = 8.75? 3. What is the concentration of a weak acid if its K a = 4.5 x and its ph = 6.24? 4. A weak base with a concentration of 1.3 mol/l has a percent ionization of 0.72 %. What is the Kb of this weak base? 5. What is the percent ionization of a 1.38 mol/l weak acid if its K a = 2.7 x 10-6?

2 6. What is the percent ionization of a 0.48 mol/l weak acid if its K a = 1.4 x 10-9? 7. The ph of a solution of HCl(aq) is found to be 0.67 L. What is the [H+V] in this solution? 8. If the Kb of a 0.34 mol/l solution of weak base is 1.3 x what is its percent ionization? 9. What is the ph of a 1.24 mol/l solution of HCN (aq) if its K a = 6.2 x 10-10? 10. If the Kb of a 0.58 mol/l solution of weak base is 1.8 x 10-10, what is its percent ionization? 11. A weak base with a concentration of 0.66 mol/l has a percent ionization of 1.4 %. What is the Kb of this weak base?

3 12. A weak acid with a concentration of 0.64 mol/l has a percent ionization of 4.5 x 10-4%. What is the ph of this weak acid? 13. What is the ph of a 1.47 mol/l solution of HCN (aq) if its K a = 3.5 x 10-11? 14. A weak acid with a concentration of 0.53 mol/l has a percent ionization of %. What is the ph of this weak acid? 15. What is the ph of a 0.45 mol/l solution of acetic acid if its K a = 1.8 x 10-5? 16. What is the concentration of a weak acid if its K a = 3.7 x 10-7 and its ph = 3.38? 17. What is the ph of a 3.4 mol/l solution of KOH?

4 18. If the Kb of a mol/l solution of weak base is 7.7 x 10-7, what is its percent ionization? 19. If the Ka of a mol/l solution of a weak acid is 7.7 x 10-7, what is its percent ionization? 20. A weak acid with a concentration of 0.47 mol/l has a percent ionization of 9.5 x What is the ph of this weak acid? ml of standardized 0.45 mol/l NaOH is titrated with 21 ml of 0.35 mol/l acetic acid. Calculate the ph of the solution ml of standardized 0.43 mol/l NaOH is titrated with 23 ml of 0.36 mol/l acetic acid. Calculate the ph of the solution.

5 ml of 0.36 mol/l acetic acid is titrated with a standardized 0.43 mol/l KOH solution. Calculate the ph of the solution after 21 ml of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x ml of 0.39 mol/l acetic acid is titrated with a standardized 0.33 mol/l KOH solution. Calculate the ph of the solution after 17 ml of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x 10-5.

6 Answer Section PROBLEM 1. ANS: HX(aq) <=====> H 1+ (aq) + X 1- (aq) x 10 -ph 10 -ph [H 1+ ] = 10 -ph = [H 1+ ] = 3.16 x 10-6 mol/l (3.16 x 10-6 ) 2 / x = 5.7 x x = 1.8 x 10-2 mol/l 2. ANS: XOH(aq) <=====> X 1+ (aq) + OH 1- (aq) x 10 -poh 10 -poh poh = 14 ph = = 5.25 [OH 1- ] = 10 -poh = [OH 1- ] = 5.6 x 10-6 mol/l (5.6 x 10-6 ) 2 / x = 1.4 x mol/l x = (5.6 x 10-6 ) 2 / 1.4 x = 2.3 mol/l 3. ANS: HX(aq) <=====> H 1+ (aq) + X 1- (aq) x 10 -ph 10 -ph [H 1+ ] = 10 -ph = [H 1+ ] = 5.75 x 10-7 mol/l (5.75 x 10-7 ) 2 / x = 4.5 x x = 0.74 mol/l 4. ANS: XOH <=======> X 1+ + OH 1 initial (0.72)(1.3)/100 Ka = ((0.72)(1.3)/100) 2 / 1.3 = 6.7 x 10-5

7 5. ANS: HY(aq) <====> H 1+ (aq) + Y 1- (aq) initial 1.38 x x x x 2 / 1.38 = 2.7 x 10-6 x = (2.7 x 10-6 x 1.38) 0.5 ph = log x ph = 2.71, % Ionization = 100(x / 1.38) = 0.14% 6. ANS: HY(aq) <====> H 1+ (aq) + Y 1- (aq) initial 0.48 x x x x 2 / 0.48 = 1.4 x 10-9 x = (1.4 x 10-9 x 0.48) 0.5 ph = log x ph = 4.59, % Ionization = 100(x / 0.48) = 5.4 x 10-3 % 7. ANS: HCl is a strong acid so it is completely ionized [H 1+ ] = = 0.21 mol/l 8. ANS: XOH <=====> X 1+ + OH 1- initial 0.34 x x x 2 / 0.34 = 1.3 x x = 6.7 x 10-7 = [OH 1- ] % ionized = 100 x (6.7 x 10-7 / 0.34) 2.0 x 10-4 % ionized

8 9. ANS: HCN(aq) <====> H 1+ (aq) + CN 1- (aq) initial 1.24 x x x x 2 / 1.24 = 6.2 x x = (6.2 x x 1.24) 0.5 ph = log x ph = ANS: XOH <=====> X 1+ + OH 1 initial 0.58 x x x 2 / 0.58 = 1.8 x x = 1.0 x 10-5 = [OH 1- ] % ionized = 100 x (1.0 x 10-5 / 0.58) 1.8 x 10-3 % ionized 11. ANS: XOH <=======> X 1+ + OH 1- initial (0.66)(1.4)/100 (0.66)(1.4)/100 Ka = ((0.66)(1.4)/100) 2 / 0.66 = 1.3 x ANS: HX <=======> H 1+ + X 1- initial (4.5 x 10-4 )(0.64)/100 (4.5 x 10-4 )(0.64)/100 [H 1+ ] = 2.9 x 10-6

9 ph = log 2.9 x 10-6 = ANS: HCN(aq) <====> H 1+ (aq) + CN 1- (aq) initial 1.47 x x x x 2 / 1.47 = 3.5 x x = (3.5 x x 1.47) 0.5 ph = log x ph = 4.56 ph = ANS: HX <=======> H 1+ + X 1- initial (0.062)(0.53)/100 (0.062)(0.53)/100 [H 1+ ] = 3.3 x 10-4 ph = log 3.3 x 10-4 = ANS: HC 2 H 3 O 2 (aq) <====> H 1+ (aq) + C 2 H 3 O 1-2 (aq) initial 0.45 x x x x 2 / 0.45 = 1.8 x 10-5 x = (1.8 x 10-5 x 0.45) 0.5 ph = log x ph = ANS: HX(aq) <=====> H 1+ (aq) + X 1- (aq) x 10 -ph 10 -ph

10 [H 1+ ] = 10 -ph = [H 1+ ] = 4.17 x 10-4 mol/l (4.17 x 10-7 ) 2 / x = 3.7 x 10-7 x = 0.47 mol/l 17. ANS: KOH is completely ionized since it is a strong acid. [OH 1- ] = 3.4 mol/l poh = log [OH 1- ] = 0.53 ph = 14 poh ph = ANS: XOH <=====> X 1+ + OH 1- initial x x x 2 / = 7.7 x 10-7 x = 1.5 x 10-4 = [OH 1- ] % ionized = 100 x (1.5 x 10-4 / 0.029) 0.52 % ionized 19. ANS: HX <=====> H 1+ + X 1 initial x x x 2 / = 7.7 x 10-7 x = 1.6 x 10-4 = [OH 1- ] % ionized = 100 x (1.6 x 10-4 / 0.034) 0.48 % ionized 20. ANS: HX <=======> H 1+ + X 1- initial 0.47

11 HX <=======> H 1+ + X 1- initial (9.5 x 10-5 )(0.47)/100 (9.5 x 10-5 )(0.47)/100 [H 1+ ] = 4.5 x 10-7 ph = log 4.5 x 10-7 = ANS: NaOH + HC 2 H 3 O > NaC 2 H 3 O 2 + H 2 O n=cv (0.45 mol/l x L) (0.35 mol/l x L) (mol) initial reacted left over COH 1- = n/v = mol / L = mol/l poh = log[0.0848] = 1.07 ph = = PTS: 1 REF: I OBJ: 8.4 STA: CS ANS: NaOH + HC 2 H 3 O > NaC 2 H 3 O 2 + H 2 O n=cv (0.43 mol/l x L) (0.36 mol/l x L) (mol) initial reacted left over COH 1- = n/v = mol / L = mol/l poh = log[0.0737] = ph = = PTS: 1 REF: I OBJ: 8.4 STA: CS ANS: KOH + HC 2 H 3 O > KC 2 H 3 O 2 + H 2 O n=cv (0.43 mol/l x L) (0.36 mol/l x L)

12 (mol) initial reacted left over CHC 2 H 3 O 2 = n/v = mol / L = mol/l (mol/l) HC 2 H 3 O 2 <=====> H C 2 H 3 O 2 initial change x x x assume x is negligible when subtracted from x 2 / = 1.8 x 10-5 x = 6.2 x 10-4 = CH 1+ ph = log[6.2 x 10-4 ] = 3.21 PTS: 1 REF: I OBJ: 8.4 STA: CS ANS: KOH + HC 2 H 3 O > KC 2 H 3 O 2 + H 2 O n=cv (0.33 mol/l x L) (0.39 mol/l x L) (mol) initial reacted left over CHC 2 H 3 O 2 1- = n/v = mol / L = mol/l (mol/l) HC 2 H 3 O 2 <=====> H C 2 H 3 O 2 initial change x x x assume x is negligible when subtracted from x 2 / = 1.8 x 10-5 x = 1.28 x 10-3 = CH 1+ ph = log[1.28 x 10-3 ] = 2.89 PTS: 1 REF: I OBJ: 8.4 STA: CS2.08