1. Buffers and titrations
Chapter 17

2. Common ion effect 𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
What happens if NaC2H3O2 is added?
What happens to [H+]?
What happens to pH?

3. example
What is the pH of a 0.1M solution of Acetic acid (Ka = 1.8 x 10-5)?
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
Initial
0.1
Change -x +x
Equilibrium x

4. example
What is the pH of a 0.1M solution of Acetic acid (Ka = 1.8 x 10-5)?
1.8𝑥 10 −5 = (𝑥)(𝑥) (0.1)
𝑥=1.34𝑥 10 −3
[ 𝐻 + ]=1.34𝑥 10 −3
𝑝𝐻=2.87

5. example
What is the pH of a 0.1M solution of sodium acetate (Kb = 5.6 x 10-10)?
𝐶 2 𝐻 3 𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)⇄ 𝑂𝐻 − 𝑎𝑞 +𝐻 𝐶 2 𝐻 3 𝑂 2 (𝑎𝑞)
Initial
0.1
Change -x +x
Equilibrium x

6. example
What is the pH of a 0.1M solution of sodium acetate (Kb = 5.6 x 10-10)?
5.6𝑥 10 −10 = (𝑥)(𝑥) (0.1)
𝑥=7.48𝑥 10 −6
[ 𝑂𝐻 − ]=7.48𝑥 10 −6
𝑝𝑂𝐻=5.13
𝑝𝐻=8.87

7. example
What is the pH of a solution that is 0.1M Acetic acid (Ka = 1.8 x 10-5) and 0.1M sodium acetate?
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
Initial
0.1
Change -x +x
Equilibrium
0.1-x x
0.1+x

8. example
What is the pH of a solution that is 0.1M Acetic acid (Ka = 1.8 x 10-5) and 0.1M sodium acetate?
1.8𝑥 10 −5 = (𝑥)(0.1+𝑥) (0.1−𝑥)
𝑥=1.8𝑥 10 −5
[ 𝐻 + ]=1.8𝑥 10 −5
1.8𝑥 10 −5 = (𝑥)(0.1) (0.1)
𝑝𝐻=4.74

9. example
What is the pH of a solution that is 0.1M Acetic acid (Ka = 1.8 x 10-5) and 0.1M HCl?
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
Initial
0.1
Change -x +x
Equilibrium
0.1-x
0.1+x x

10. example
What is the pH of a solution that is 0.1M Acetic acid (Ka = 1.8 x 10-5) and 0.1M HCl?
1.8𝑥 10 −5 = (0.1+𝑥)(𝑥) (0.1−𝑥)
𝑥=1.8𝑥 10 −5
[ 𝐶 2 𝐻 3 𝑂 2 − ]=1.8𝑥 10 −5
1.8𝑥 10 −5 = (0.1)(𝑥) (0.1)
[ 𝐻 + ]=0.1
𝑝𝐻=1

12. Buffers Solution that resists changes in pH from added acid or base.
Weak Acid and conjugate base
Weak Base and conjugate acid
𝐻𝐴 / 𝐴 −
𝐻𝐴 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐻 2 𝑂 𝑙 + 𝐴 − (𝑎𝑞)
𝐴 − (𝑎𝑞) + 𝐻 + 𝑎𝑞 →𝐻𝐴 𝑎𝑞

13. Buffer examples 𝐻 𝐶 2 𝐻 3 𝑂 2 / 𝐶 2 𝐻 3 𝑂 2 −
𝐻 𝐶 2 𝐻 3 𝑂 2 / 𝐶 2 𝐻 3 𝑂 2 −
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐻 2 𝑂 𝑙 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞) + 𝐻 + 𝑎𝑞 →𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞
𝑁𝐻 3 / 𝑁𝐻 4 +
𝑁𝐻 3 𝑎𝑞 + 𝐻 + (𝑎𝑞)→ 𝑁𝐻 4(𝑎𝑞) +
𝑁𝐻 4 + 𝑎𝑞 + 𝑂𝐻 (𝑎𝑞) − → 𝑁𝐻 3 𝑎𝑞 + 𝐻 2 𝑂 𝑙

14. Calculating pH of a buffer
𝐻𝐴 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐴 − (𝑎𝑞)
𝐾 𝑎 = 𝐻 + [ 𝐴 − ] [𝐻𝐴]
𝐻 + = 𝐾 𝑎 [𝐻𝐴] [ 𝐴 − ]

15. Calculating pH of a buffer
𝐻𝐴 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐴 − (𝑎𝑞)
𝐻 + = 𝐾 𝑎 [𝐻𝐴] [ 𝐴 − ]
− log 𝐻 + =−log⁡( 𝐾 𝑎 𝐻𝐴 𝐴 − )
𝑝𝐻= 𝑝𝐾 𝑎 −log⁡( 𝐻𝐴 𝐴 − )

16. Calculating pH of a buffer
𝐻𝐴 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐴 − (𝑎𝑞)
𝑝𝐻= 𝑝𝐾 𝑎 −log⁡( 𝐻𝐴 𝐴 − )
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡( 𝐴 − 𝐻𝐴 )
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡( 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑 )

17. Henderson-Hasselbalch EqN (HH)
𝐻𝐴 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐴 − (𝑎𝑞)
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡( 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑 )

18. example
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 ⇄ 𝐻 + 𝑎𝑞 + 𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞)
Using the Henderson-Hasselbalch equation, determine the pH of a solution that is 0.1M Acetic acid (Ka = 1.8 x 10-5) and 0.1M sodium acetate?
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡( 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑 )
𝑝𝐻=−log⁡(1.8𝑥 10 −5 )+log⁡( 0.1𝑀 𝑀 )
𝑝𝐻=4.74

19. example
Using the Henderson-Hasselbalch equation, determine the pH of a solution that is 0.50M Ammonia (Kb = 1.8 x 10-5) and 0.20M ammonium chloride?
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡( 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑 )
𝑝𝐻=9.26+log⁡( 0.50𝑀 𝑀 )
𝑝𝐻=
𝑝𝐻=9.66

21. Calculating ph change in buffer
Determine the pH of a 0.1M acetic acid and 0.1M sodium acetate buffer solution after 0.025mol of HCl has been added to 1L of the solution.
𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞) + 𝐻 + 𝑎𝑞 →𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞
Before
0.1
Added
0.025
After
0.075
0.125
This is not an ICE box Equilibrium calculation!!

22. Calculating ph change in buffer
Determine the pH of a 0.1M acetic acid and 0.1M sodium acetate buffer solution after 0.025mol of HCl has been added to 1L of the solution.
𝐶 2 𝐻 3 𝑂 2 − (𝑎𝑞) + 𝐻 + 𝑎𝑞 →𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡ 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑
𝑝𝐻=−log⁡(1.8𝑥 10 −5 )+log⁡
𝑝𝐻=4.52

23. Calculating ph change in buffer
Determine the pH of a 0.1M acetic acid and 0.1M sodium acetate buffer solution after 0.01mol of NaOH has been added to 0.50L of the solution.
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐶 2 𝐻 3 𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)
Before
0.05
Added
0.01
After
0.04
0.06
This is not an ICE box Equilibrium calculation!!

24. Calculating ph change in buffer
Determine the pH of a 0.1M acetic acid and 0.1M sodium acetate buffer solution after 0.01mol of NaOH has been added to 0.50L of the solution.
𝐻 𝐶 2 𝐻 3 𝑂 2 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐶 2 𝐻 3 𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡ 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑
𝑝𝐻=−log⁡(1.8𝑥 10 −5 )+log⁡
𝑝𝐻=4.92

26. titrations
Process of adding an acid to neutralize a base, or adding a base to neutralize an acid
Often used to determine the concentration of an unknown base or acid

27. Titration curve
Graph representing how the pH changes as acid or base is added during a titration

28. Titration vocab
Titrant – solution added in small increments, usually known concentration
End Point – point where indicator changes color
Equivalence Point – point where moles of H+ equals moles of OH-
Half Equivalence Point – point when half of the titrant needed for equivalence has been added

29. Types of titration Strong Acid & Strong Base Weak Acid & Strong Base
Weak Base & Strong Acid

30. Titration #𝑚𝑜𝑙 𝐻 + =#𝑚𝑜𝑙 𝑂𝐻 − 𝑀 𝐴 𝑉 𝐴 = 𝑀 𝐵 𝑉 𝐵 𝑀 𝐴 (20)=(0.1)(20)
20mL of HCl is titrated with 20mL of 0.1M NaOH. What is the concentration of the HCl?
#𝑚𝑜𝑙 𝐻 + =#𝑚𝑜𝑙 𝑂𝐻 −
𝑀 𝐴 𝑉 𝐴 = 𝑀 𝐵 𝑉 𝐵
𝑀 𝐴 (20)=(0.1)(20)
𝑀 𝐴 =0.1𝑀 𝐻𝐶𝑙

31. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
Initial pH
pH = -log(0.1) = 1

32. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
After 5mL of NaOH has been added
𝐻 + 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐻 2 𝑂(𝑙)
Before
0.002
Added
0.0005
After
0.0015

33. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
After 5mL of NaOH has been added
pH = -log(0.0015/.025)
pH = -log(0.06)
pH = 1.22

34. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
At equivalence point
Since all HCl has been neutralized by NaOH, all H+ ions are due to autoionization of water
pH = 7

35. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
After 30mL of NaOH has been added
𝐻 + 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐻 2 𝑂(𝑙)
Before
0.002
Added
0.003
After
0.001

36. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.
After 30mL of NaOH has been added
pOH = -log(0.001/.050)
pOH = -log(0.02)
pOH = 1.70
pH = 14-pOH = = 12.30

37. Calculating pH during titration
Strong Acid / Strong Base
20mL of 0.1M HCl titrated with 0.1M NaOH.

38. Calculating pH during titration
Strong Acid / Strong Base
Before Equivalence point – H+ dominates
At Equivalence point – neutral
After Equivalence point – OH- dominates

39. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.
Initial pH
𝐾 𝑎 = 𝐻 + [ 𝐴 − ] [𝐻𝐴]
𝑥=4.24𝑥 10 −3
[ 𝐻 + ]=4.24𝑥 10 −3
1.8𝑥 10 −4 = (𝑥)(𝑥) (0.1)
𝑝𝐻=2.37

40. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.
After 5mL of NaOH has been added
𝐻𝐶𝐻𝑂 2 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐶𝐻𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)
Before
0.002
Added
0.0005
After
0.0015

41. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.
After 5mL of NaOH has been added
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡ 𝐵𝑎𝑠𝑒 𝐴𝑐𝑖𝑑
𝑝𝐻=−log⁡(1.8𝑥 10 −4 )+log⁡
𝑝𝐻=3.26

42. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.
After 20mL of NaOH has been added (equivalence point)
𝐻𝐶𝐻𝑂 2 𝑎𝑞 + 𝑂𝐻 − 𝑎𝑞 → 𝐶𝐻𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)
Before
0.002
Added
After

43. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.
After 20mL of NaOH has been added (equivalence point)
𝐶𝐻𝑂 2 − 𝑎𝑞 + 𝐻 2 𝑂(𝑙)⇄ 𝑂𝐻 − 𝑎𝑞 + 𝐻𝐶𝐻𝑂 2 (𝑎𝑞)
𝑥=1.7𝑥 10 −6
5.6𝑥 10 −11 = (𝑥)(𝑥) (0.05)
[ 𝑂𝐻 − ]=1.7𝑥 10 −6
𝑝𝑂𝐻=5.77
𝑝𝐻=8.23

44. Calculating pH during titration
Weak Acid / Strong Base
20mL of 0.1M HCHO2 titrated with 0.1M NaOH.

45. Calculating pH during titration
Weak Acid / Strong Base
Before Equivalence point, after some titrant added – Buffer solution
At Half Equivalence point – [HA] = [A-]
pH =pKa
At Equivalence point – only conjugate base
Slightly basic
After Equivalence point – OH- dominates

47. Titration of polyprotic acids
There is an equivalence point for each proton (H+) in polyprotic acids

48. Indicators
Indicators are often weak organic acids that are a different color from their conjugate base
HIn – generic indicator
In- – conjugate base
Use a small amount to not contribute to pH of titration reaction
𝐻𝐼𝑛+ 𝐻 2 𝑂 ⇄ 𝐼𝑛 − + 𝐻 3 𝑂 +
Color 1
Color 2

49. Indicators
When doing a titration, you should choose an indicator that changes color at the same pH as the equivalence point of the titration
At equivalence point [HIn] = [In-]
pH (equiv pt) = pKa (indicator)
𝑝𝐻= 𝑝𝐾 𝑎 +log⁡ 𝐼𝑛 − 𝐻𝐼𝑛

50. indicators 𝐻𝐼𝑛+ 𝐻 2 𝑂 ⇄ 𝐼𝑛 − + 𝐻 3 𝑂 +
If pH < pKa, indicator is color 1
If pH > pKa, indicator is color 2
If pH = pKa, indicator is intermediate color
𝐻𝐼𝑛+ 𝐻 2 𝑂 ⇄ 𝐼𝑛 − + 𝐻 3 𝑂 +
Color 1
Color 2

52. Ksp Review
Equilibrium constant for the dissolution of an ionic compound
𝐶𝑎𝐹 2 𝑠 ⇄ 𝐶𝑎 2+ 𝑎𝑞 + 2 𝐹 − (𝑎𝑞)
𝐾 𝑠𝑝 = [ 𝐶𝑎 2+ ][ 𝐹 − ] 2

53. Molar Solubility
Molar concentration of the quantity of a compound that dissolves
Molar Solubility ≠ Ksp

54. Example Ksp for AgCl is 1.77 x 10-10 𝐴𝑔𝐶𝑙 𝑠 ⇄ 𝐴𝑔 + 𝑎𝑞 + 𝐶𝑙 − (𝑎𝑞)
𝐴𝑔𝐶𝑙 𝑠 ⇄ 𝐴𝑔 + 𝑎𝑞 + 𝐶𝑙 − (𝑎𝑞)
𝐾 𝑠𝑝 = 𝐴𝑔 + 𝐶𝑙 − = 𝑥 10 −10
𝐾 𝑠𝑝 = 𝑋 𝑋 = 𝑥 10 −10
𝑋 = 𝑥 10 −5 𝑀

55. ph effect on Solubility
Common Ion Effect
𝑀𝑔 𝑂𝐻 2 𝑠 ⇄ 𝑀𝑔 2+ 𝑎𝑞 +2 𝑂𝐻 − (𝑎𝑞)
What happens when acid is added?
Equilibrium shifts to the right as the OH- is neutralized with the added acid.
Decreasing pH increases the solubility of compounds containing basic ions

56. precipitation If Q = Ksp Saturated Solution
If Q < Ksp No precipitation
If Q > Ksp Precipitation will occur

57. Selective precipitation
A solution may contain a mixture of different dissolved ions
Selective precipitation is when certain reagents are chosen to separate the ions by precipitating only certain ions
Qualitative Chemical Analysis
Using selective precipitation to determine which ions are present in an unknown mixture

59. Solubility Rules
You are expected to know that all Sodium, Potassium, Ammonium, and Nitrate salts are soluble in water
Na+, K+, NH4+, NO3-