1. Chapter 16 Aqueous Ionic Equilibrium

2. 2 Buffers  buffers are solutions that resist changes in pH when an acid or base is added  they act by neutralizing the added acid or base  but just like everything else, there is a limit to what they can do, eventually the pH changes  many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion  buffers are solutions that resist changes in pH when an acid or base is added  they act by neutralizing the added acid or base  but just like everything else, there is a limit to what they can do, eventually the pH changes  many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion

3. Demo of a pH Buffer Assume 1 drop = 0.05mL When this 1 drop is dissolved in 30 mL of H 2 O What should the pH be when we add 1 drop of 0.1M HNO 3 to 30mL of deionized water ?

4. Why are pH buffers important? Life on Earth is water-based  Human 48-75% water  Plants as high as 95% water  H + and OH - are chemically and structurally reactive in cells The functioning of the cell is very pH dependent  Blood plasma pH = 7.4  pH < 6.9 fatal (acidosis)  pH > 7.9 fatal (alkalosis)  Fish die if the pH of the water goes below 5 or above 9 Life on Earth is water-based  Human 48-75% water  Plants as high as 95% water  H + and OH - are chemically and structurally reactive in cells The functioning of the cell is very pH dependent  Blood plasma pH = 7.4  pH < 6.9 fatal (acidosis)  pH > 7.9 fatal (alkalosis)  Fish die if the pH of the water goes below 5 or above 9

5. 5 Making an Acid Buffer

6. 6 How Acid Buffers Work HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq)  buffers work by applying Le Châtelier’s Principle to weak acid equilibrium  buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it  the buffer solutions also contain significant amounts of the conjugate base anion, A − - these ions combine with added acid to make more HA and keep the H 3 O + constant  buffers work by applying Le Châtelier’s Principle to weak acid equilibrium  buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it  the buffer solutions also contain significant amounts of the conjugate base anion, A − - these ions combine with added acid to make more HA and keep the H 3 O + constant

7. 7 H2OH2O How Buffers Work HA + H3O+H3O+ A−A− A−A− Added H 3 O + new HA

8. 8 H2OH2O How Buffers Work HA + H3O+H3O+ A−A− Added HO − new A − A−A−

9. 9 Common Ion Effect HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq)  adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left  this causes the pH to be higher than the pH of the acid solution  lowering the H 3 O + ion concentration  adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left  this causes the pH to be higher than the pH of the acid solution  lowering the H 3 O + ion concentration

10. 10 Common Ion Effect

11. 11 Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change equilibrium What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

12. 12 [HA][A - ][H 3 O + ] initial 0.100 0 change equilibrium represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x-x-x 0.100 -x0.100 + x x What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5 HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O +

13. 13 determine the value of K a since K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 x 0.100 -x 0.100 +x What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

14. 14 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init the approximation is valid x = 1.8 x 10 -5 [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 x What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

15. Tro, Chemistry: A Molecular Approach 15 x = 1.8 x 10 -5 substitute x into the equilibrium concentration definitions and solve [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 1.8E-5 0.100 + x x0.100 -x What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

16. 16 substitute [H 3 O + ] into the formula for pH and solve [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 1.8E-5 What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

17. 17 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 1.8E-5 What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

18. Tro, Chemistry: A Molecular Approach 18 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF?

19. 19 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HF + H 2 O F - + H 3 O + [HA][A - ][H 3 O + ] initial 0.140.071≈ 0 change equilibrium

20. 20 [HA][A - ][H 3 O + ] initial 0.140.071 0 change equilibrium What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x -x-x 0.14 -x 0.071 + x x HF + H 2 O F - + H 3 O +

21. 21 determine the value of K a since K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.012 0.100 x 0.14  x What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 0.071 +x K a for HF = 7.0 x 10 -4

22. 22 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? K a for HF = 7.0 x 10 -4 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init the approximation is valid x = 1.4 x 10 -3 [HA][A 2 - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.140.071 x

23. Tro, Chemistry: A Molecular Approach 23 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? x = 1.4 x 10 -3 substitute x into the equilibrium concentration definitions and solve [HA][A 2 - ][H 3 O + ] initial 0.140.071≈ 0 change -x-x+x+x+x+x equilibrium 0.140.0721.4E-3 0.071 + x x 0.14  x

24. 24 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? substitute [H 3 O + ] into the formula for pH and solve [HA][A - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.140.0721.4E-3

25. 25 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values are close enough [HA][A - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.140.0721.4E-3 K a for HF = 7.0 x 10 -4

26. 26 Henderson-Hasselbalch Equation  calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson-Hasselbalch Equation  the equation calculates the pH of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base  as long as the “x is small” approximation is valid  calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson-Hasselbalch Equation  the equation calculates the pH of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base  as long as the “x is small” approximation is valid

27. 27 Deriving the Henderson-Hasselbalch Equation

28. 28 What is the pH of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 where K a =6.5x10 -5 for HC 7 H 5 O 2 ?

29. 29 What is the pH of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 where K a =6.5x10 -5 for HC 7 H 5 O 2 ? Assume the [HA] and [A - ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the “x is small” approximation HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 - + H 3 O +

30. What is the pH of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 where K a =6.5x10 -5 for HC 7 H 5 O 2 ? HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 - + H 3 O + [HA][A - ][H 3 O + ] initial 0.0500.150 ≈ 0 change -x-x+x+x+x+x equilibrium 0.050-x0.150-xx

31. 31 Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?  It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA] o and [A - ] o into the equation assuming they are equilibrium values), makes it an approximation  For this reason the Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable  generally, the “x is small” approximation will work when both of the following are true: a)the initial concentrations of acid and salt are not very dilute b)the K a is fairly small  for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of K a  It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA] o and [A - ] o into the equation assuming they are equilibrium values), makes it an approximation  For this reason the Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable  generally, the “x is small” approximation will work when both of the following are true: a)the initial concentrations of acid and salt are not very dilute b)the K a is fairly small  for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of K a