1. Entry Task: Jan 22 nd Thursday Turn in Determine Ka Lab

2. Agenda Notes on Buffers In class practice HW: Buffer ws

3. pH of a salt (review) What is the pH of a 0.100 M Solution of a NaCN. The ka for HCN is 5.8 x 10 -10. CN -  HCN + OH - Kb = [OH-] [HCN] [CN - ] Kb = [1.0 x 10 -14 ] [5.8 x 10 -10 ] Kb = 1.7 x 10 -5 [x 2 ] [0.100] = 1.7 x 10 -5 pOH = 2.88 or pH = 11.12 x 2 = 1.7 x 10 -6 = 1.3 x 10 -3 [OH - ]

4. Buffers Resist the change in pH Contains acid/conjugate base Or Contains base/conjugate acid

5. MAYHAN

6. Buffers If you add OH-, the HF (acid) will neutralize the OH- (base), thus the HF level will decrease. MAYHAN

7. Buffers If an acid (H+) is added, the F − neutralize it to form HF and water. MAYHAN

8. Calculating the pH of Buffer What is the pH of a buffer that is 0.12M lactic acid- HC 3 H 5 O 3 and 0.10M in sodium lactate, NaC 3 H 5 O 3. The ka: 1.4x 10 -4. HC 3 H 5 O 3  H 3 O+ + C 3 H 5 O 3 - Ka = [H 3 O+] [C 3 H 5 O 3 - ] [HC 3 H 5 O 3 ] [x][0.10] [0.12] = 1.4 x 10 -4 1.7 X 10 -4 or pH = 3.77 x= (1.4x 10 -4 )(0.12) 0.10 = 1.4 x 10 -4 Common Ion way-

9. Calculating the pH of Buffer What is the pH of a buffer that is 0.12M lactic acid- HC 3 H 5 O 3 and 0.10M in sodium lactate, NaC 3 H 5 O 3. The ka for HCN is 1.4 x 10 -4. HC 3 H 5 O 3  H 3 O+ + C 3 H 5 O 3 - pH = pKa + log 3.85 [base] [acid] pH = 1.4 x 10 -4 + log pH = 3.77 [0.10] [0.12] + (-0.0792) Alternative way- Henderson-Hasselbalch equation-

10. Prepare a Buffer- How many grams of NH 4 Cl must be added to a 2.0 liter of 0.10 M NH 3 to form a buffer whose pH is 9.00. The kb is 1.8 x 10 -5. NH 3  NH 4 + + OH - pH 14-9 = pOH of 5 = 1.8 x 10 -5 = x = 0.18 M [x][1.0x10 -5 ] [0.10] OH- = 1.0 x 10 -5 = 19.24 grams of NH 4 Cl (0.10)(1.8 x 10 - 5 ) 1.0 x 10 -5 X 2.0 L = 0.18 M 0.36 mol 53.45g 1.0 mol

11. Buffer Capacity the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. It depends on the amount of acid or base from which the buffer is made. 0.10 M HC 2 H 3 O 2 and 0.1 M NaC 2 H 3 O 2 1.0 M HC 2 H 3 O 2 and 1.0 M NaC 2 H 3 O 2 Which of the solutions have a greater buffer capacity? The larger molarity because it would resist MORE

12. pH Range For a buffer system, pick a pKa close to the desired pH. If you want a pH buffer system with a pH of 4, which acid/c. base par would you choose? Benzoic Acid- HC 7 H 5 O 6 / NaC 7 H 5 O 6 Ka= 6.3 x 10 -5 Phosphoric acid- H 3 PO 4 / NaH 3 PO 4 ka= 7.5 x 10 -3 Benzoic Acid Its pKa is 4.2 Hydrofluoric Acid- HF / NaF Ka= 6.8 x 10 -4

13. Buffer Problems Hurdles- 1 st - Which species in problem is “acid” or “base”? 2 nd Kb value for OH - or Ka for H + 3 rd Setting up Ka or Kb expression correctly (1 st ) 4 th Is there any changes in concentrations from given: *Grams  Moles  Molarity *Mixing two different volumes (M 1 V 1 = M 2 V 2 ) MAYHAN

14. 1a. Calculate the pH of a buffer that is 0.100 M in NaHCO 3 and 0.125M in Na 2 CO 3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO 3 with 65mls of 0.15 M Na 2 CO 3. Ka = [H+][CO 3 -2 ] [HCO 3 - ] 5.6 x 10 -11 = [x][0.125] [0.100] Who is the conjugate in this reaction? x= (5.6 x 10 -11 )(0.100) 0.125 x = [H+]= 4.48 x 10 -11 pH = -log(4.48 x 10 -11 ) = 10.35 HCO 3 - so we use H 2 CO 3 Ka value= 5.6x10 -11 Provide the Ka expression Provide the Ka express with # Rearrange to get X by itself MAYHAN

15. 1a. Calculate the pH of a buffer that is 0.100 M in NaHCO 3 and 0.125M in Na 2 CO 3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO 3 with 65mls of 0.15 M Na 2 CO 3. NOTICE we have different volumes!! What will be the TOTAL VOLUME? 120 mls We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the same!! (0.20M)(0.055L) = (0.15M)(0.065L) = (x)(0.120 L) MAYHAN (x) (0.120L) 9.17x10 -2 M of NaHCO 3 (new M) 8.125x10 -2 Mof Na 2 CO 3 (new M)

16. 1a. Calculate the pH of a buffer that is 0.100 M in NaHCO 3 and 0.125M in Na 2 CO 3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO 3 with 65mls of 0.15 M Na 2 CO 3. MAYHAN 5.6 x 10 -11 = [x][8.125x10 -2 ] [9.17x10 -2 ] x= (5.6 x 10 -11 )(9.17x10 - 2 ) 8.125x10 -2 x = [H+]= 6.32 x 10 -11 pH = -log(6.32 x 10 -11 ) = 10.20 Plug Back into Ka expression Rearrange to get X by itself

17. 2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4 10 -4 1.4 x 10 -4 = [x][0.11] [0.12] x= (1.4 x 10 -4 )(0.12) 0.11 x = [H+]= 1.52 x 10 -4 pH = -log(4.48 x 10 -11 ) = 3.82 Provide the Ka express with # Rearrange to get X by itself MAYHAN

18. 2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4 10 -4 What will be the TOTAL VOLUME? 180mls We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the same!! (0.13M)(0.085L) = (0.15M)(0.095L) = (x)(0.180 L) MAYHAN (x) (0.180L) 6.14x10 -2 M of lactic acid (new M) 7.92x10 -2 M of sodium lactate (new M)

19. 2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4 10 -4 MAYHAN 1.4 x 10 -4 = [x][7.92x10 -2 ] [6.14x10 -2 ] x= (1.4 x 10 -4 )(6.14x10 - 2 ) 7.92x10 -2 x = [H+]= 1.09 x 10 -4 pH = -log(1.09 x 10 -4 ) = 3.96 Plug Back into Ka expression Rearrange to get X by itself