1. Buffer This

2. There are two common kinds of buffer solutions: 1Solutions made from a weak acid plus a soluble ionic salt of the weak acid. 2Solutions made from a weak base plus a soluble ionic salt of the weak base 1.Solutions made of weak acids plus a soluble ionic salt of the weak acid One example of this type of buffer system is: – The weak acid - acetic acid CH 3 COOH – The soluble ionic salt - sodium acetate NaCH 3 COO The Common Ion Effect and Buffers

3. This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of H + and the pH of a solution that is 0.15 M in both acetic acid sodium acetate yields: RCH 3 COOH H 3 O + + CH 3 COO - I C E One example of the type I of buffer system is: The weak acid - acetic acid CH 3 COOH The soluble ionic salt - sodium acetate NaCH 3 COO CH 3 COOH H 3 O + + CH 3 COO - NaCH 3 COO →Na + + CH 3 COO - ~100% Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

4. Solution[H + ]pH 0.15 M CH 3 COOH1.6 x 10 -3 2.80 0.15 M CH 3 COOH & 0.15 M NaCH 3 COO buffer 1.8 x 10 -5 4.74 [H + ] is ~90 times greater in pure acetic acid than in buffer solution. Note that the pH of the buffer equals the pK a of the buffering acid. Alternatively you might have noticed: [base]/[acid] = 1 = 10 0 log 10 0 = 0 pH = pK a + 0 = pK a = 4.74 [H + ] = K a = 1.8E-5 Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

5. The general expression for the ionization of a weak monoprotic acid is: The generalized ionization constant expression for a weak acid is: The Common Ion Effect and Buffers

6. If we solve the expression for [H + ], this relationship results: By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to: The Common Ion Effect and Buffers

7. The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt. If the salt’s cation is not univalent the relationship changes to: The Common Ion Effect and Buffers

8. Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation. The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. The Common Ion Effect and Buffers

9. The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. A special case exists for the Henderson-Hasselbalch equation when [base]/[acid] = some power of 10, regardless of the actual concentrations of the acid and base, where the Henderson- Hasselbalch equation can be interpreted without the need for calculations : [base]/[acid] = 10 x log 10 x = x in general pH = pK a + x Examples: 1. when [base] = [acid], [base]/[acid] = 1 or 10 0, log 1 = 0, pH = pK a, (corresponds to the midpoint in the titration of a weak acid or base) 2. when [base]/[acid] = 10 or 10 1, log 10 = 1 then pH = pK a + 1 3. when [base]/[acid] =.001 or 10 -2, log 10 = -2 then pH = pK a -2 Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their K a ’s Henderson-Hasselbalch - Caveats and Advantages

10. Buffer Solutions There are two common kinds of buffer solutions: I.Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid. II.Less common, solutions made from a weak base plus a soluble ionic salt of the conjugate acid of the weak base. Both of the above may also be prepared by starting with a weak acid (or weak base) and add half as many moles of strong base (acid)

11. This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of OH- and the pOH of the solution that is 0.15 M in aqueous ammonia, NH 3, and 0.30 M in ammonium nitrate, NH 4 NO 3 yeilds: R NH 3 NH 4 + + OH - I C E NH 4 NO 3 → NH 4 + + NO 3 - ~100% NH 3 NH 4 + + OH - Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids One example of the type II of buffer system is: The weak base – ammonia NH 3 The soluble ionic salt – ammonium nitrate NH 4 NO 3 Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

12. Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation. The Common Ion Effect and Buffers We can derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship. –The general ionization equation for weak bases is:

13. A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. Solution[OH - ]pH 0.15 M NH 3 1.6 x 10 -3 M11.20 0.15 M NH 3 & 0.15 M NH 4 NO 3 buffer 9.0 x 10 -6 M8.95 The [OH-] in aqueous ammonia is 180 times greater than in the buffer. The Common Ion Effect and Buffers

14. Buffering Action If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl. 1Calculate the pH of the original buffer solution. Substitute the quantities determined in the previous relationship into the ionization expression for ammonia. R NH 3 NH 4 + + OH - I C E NH 4 NO 3 → NH 4 + + NO 3 - ~100% NH 3 NH 4 + + OH -

15. NH 3 + H + NH 4 + Buffering Action 2Next, calculate the concentration of all species after the addition of the gaseous HCl. – The HCl will react with some of the ammonia and change the concentrations of the species. – This is another limiting reactant problem. R NH 3 NH 4 + + OH - I C E HCl → H + + Cl - ~100%

16. Buffering Action 3Using the concentrations of the salt and base and the Henderson- Hassselbach equation, the pH can be calculated. 4Finally, calculate the change in pH.

17. Buffering Action 1.If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH. NH 4 + + OH - NH 3 R NH 3 NH 4 + + OH - I C E NaOH → Na + + OH - ~100%

18. Buffering Action 3.Finally, calculate the change in pH. 2.Using the concentrations of the salt and base and the Henderson- Hassselbach equation, the pH can be calculated.

19. Buffering Action Notice that the pH changes only slightly in each case. Original Solution Original pH Acid or base added New pH  pH 1.00 L of solution containing 0.100 M NH 3 and 0.200 M NH 4 Cl 8.95 0.020 mol NaOH9.08+0.13 0.020 mol HCl8.81-0.14

20. Preparation of Buffer Solutions Calculate the concentration of H + and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. NaOH and CH 3 COOH react in a 1:1 mole ratio. After the two solutions are mixed, Calculate total volume. The concentrations of the acid and base are: Substitution of these values into the ionization constant expression (or the Henderson- Hasselbach equation) permits calculation of the pH.

21. Preparation of Buffer Solutions For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Starting with a solution that is 0.100M in aqueous ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using ammonium chloride as the source of the soluble ionic salt of the conjugate weak acid. The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair pOH can be determined from the pH: pKb can be looked up in a table: [base] concentration is provided: Solve for [acid]: Does this result make sense? NH 4 Cl NH 4 + + Cl - ~100% NH 3 NH 4 + + OH - H2OH2O