1. Buffer Effectiveness, Titrations, and pH Curves
Chapter

2. 16.3 - Buffer Effectiveness
Effective buffers neutralize small to moderate amounts of added acid or base
Two factors influence the effectiveness:
Relative Amounts of the Acid and Conjugate Base
Absolute Concentrations of the Acid and Conjugate Base
*Remember adding too much of an acid or base can result in a buffer being destroyed*

3. Relative Amounts of Acid and Base
Buffers are most effective when the concentrations of the acid and the base are equal
Example:
A generic buffer containing HA and A- for which pKa= Both solutions contain 0.20 mol of total acid and conjugate base, but Solution 1 contains equal amounts while Solution 2 contains differing amounts. We will calculate the pH after the addition of mol of NaOH and the percent change in pH.
Meaning that the closer the concentrations are to each other, the more effective it is.
0.100 M
0.180 M
0.100 M
0.020 M

4. pH = pKa + log([base]/[acid]) = 5.00 + log(0.110/0.090) = 5.09
Solution 1: 0.10 mol HA and 0.10 mol A- ; initial pH = 5.00
Solution 2: 0.18 mol HA and mol A- ; initial pH = 4.05
OH- (aq) + HA (aq) → H2O (l) + A- (aq)
OH- (aq) + HA (aq) → H2O (l) + A- (aq)
Before Addition
~0.00 mol
0.100 mol
Addition
0.010 mol
-----
After Addition
0.090 mol
0.110 mol
Before Addition
~0.00 mol
0.18 mol
0.020 mol
Addition
0.010 mol
-----
After Addition
0.17 mol
0.030 mol
pH = pKa + log([base]/[acid])
= log(0.110/0.090)
= 5.09
% change = x 100%
= 1.8%
pH = pKa + log([base]/[acid])
= log(0.030/0.17)
= 4.25
% change = x 100%
= 5.0%
5.00
4.05
Larger Difference

5. As we just saw, with more equal concentrations,
the more resistant the buffer becomes.
As a guideline:
“In order for a buffer to be reasonably effective, the relative concentrations of acid and conjugate base should not differ by more than a factor of 10.”

6. Absolute Concentrations of Acid and Base
From the textbook: “A buffer is most effective (most resistant to pH changes) when the concentrations of acid and conjugate base are high.”
Example:
A generic buffer containing HA and A- for which pKa= Solution 1 contains both acid and conjugate base that are 10 times stronger than the concentrations in Solution 2. We will calculate the pH after the addition of mol of NaOH and the percent change in pH. (Both have equal relative amounts so they have the same initial pH of 5)
0.500 M
0.050 M
0.500 M
0.050 M

7. pH = pKa + log([base]/[acid]) = 5.00 + log(0.51/0.49) = 5.02
Solution 1: 0.50 mol HA and 0.50 mol A- ; initial pH = 5.00
Solution 2: mol HA and mol A- ; initial pH = 5.00
OH- (aq) + HA (aq) → H2O (l) + A- (aq)
OH- (aq) + HA (aq) → H2O (l) + A- (aq)
Before Addition
~0.00 mol
0.500 mol
Addition
0.010 mol
-----
After Addition
0.490 mol
0.510 mol
Before Addition
~0.00 mol
0.050 mol
Addition
0.010 mol
-----
After Addition
0.040mol
0.060 mol
pH = pKa + log([base]/[acid])
= log(0.51/0.49)
= 5.02
% change = x 100%
= 0.4%
pH = pKa + log([base]/[acid])
= log(0.060/ )
= 5.18
% change = x 100%
= 3.6%
5.00
5.00
Larger Difference

8. Buffer Range
Taking into account that the relative concentrations shouldn’t differ by more than a factor of 10, we can calculate the pH range over which a buffer is considered effective.
Lowest pH (when the base is ⅒ as concentrated as the acid):
Highest pH (when the base is ten times as concentrated as the acid):
pH = pKa + log([base]/[acid])
= pKa + log(0.10)
= pKa - 1
pH = pKa + log([base]/[acid])
= pKa + log(10)
= pKa + 1
“The effective range for a buffering system is one pH unit on either side of pKa.”
If pKa = 5.0 for a weak acid, we can use it to prepare a buffer in a range of , it would be most effective at pH 5.0.

9. Buffer Capacity
Buffer Capacity is the amount of acid of base that you can add to a buffer without causing a large change in pH.
“Buffer capacity increases with increasing absolute concentrations of the buffer components.”
“Overall buffer capacity increases as the relative concentrations of the buffer components become more similar to each other.”

10. 16.4 - Titrations and pH Curves
Acid-Base Titration: a basic (or acidic) solution of unknown concentration reacts with an acidic (or basic) solution of known concentration
Indicator: a substance whose color depends on the pH
Equivalence Point: the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid

11. Titration of a Strong Acid with a Strong Base
For this lesson, we will use a titration of 25.0mL of 0.100M HCl with 0.100M NaOH.
We will first calculate the amount of base we need to neutralize the acid (reach the equivalence point), and then calculate the pH at several points of the titration.

12. Volume of NaOH Required
Titration Reaction: HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
Initial Number of Moles of Acid:
Initial mol HCl = L x = mol HCl
We need mol of NaOH, so we calculate the volume of NaOH that we need:
Volume NaOH solution = mol x = L
We will need 25 mL of NaOH to reach the equivalence point. In this case, both the acid and the base have equal concentrations so the volume will be equal.
25 mL NaOH
0.100 mol
1.00 L
1.00 L
0.100 mol
25 mL HCl
As said before, The equivalence point is reached when the number of moles of base added equals the number of moles of acid initially in the solution.

13. Initial pH (before Adding Any Base)
The initial pH of the solution is the pH of a 0.100M HCl solution (HCl is a strong acid so [H3O+] = [HCl])
pH = -log[H3O+]
= -log(0.100)
= 1.00
pH = 1.00
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

14. pH after Adding 5.00 mL NaOH
In the titration, H3O+ is neutralized by NaOH: OH- (aq) + H3O+ (aq) → 2 H2O (l)
Calculate the amount of H3O+ at any given point by using the reaction stoichiometry ー 1 mol of NaOH neutralizes 1 mol of H3O+
Calculate the number of moles of base added at 5.00mL by multiplying the added volume (in L) by the concentration of the basic solution
mol NaOH added = L x = mol NaOH
0.100 mol
1.00 L
mol NaOH
As said before, The equivalence point is reached when the number of moles of base added equals the number of moles of acid initially in the solution.

15. pH after Adding 5.00 mL NaOH
OH- (aq) H3O+ (aq) → 2 H2O (l)
Calculate H3O+ concentration by dividing the number of moles of H3O+ remaining by the total volume (initial volume plus added volume), then you can find pH.
Before Addition
~ 0.00 mol
mol
Addition
mol
-----
After Addition
0.002 mol
pH = 1.18
0.002 mol H3O+
L L
[H3O+] = = M
As said before, The equivalence point is reached when the number of moles of base added equals the number of moles of acid initially in the solution.
pH = -log(0.0667)
= 1.18

16. pH’s after Adding 10.0, 15.0, and 20.0 mL NaOH
To calculate the pH after these additions, you do the same calculations as before
These are the results:
Volume (mL) pH
10.0
1.37
15.0
1.60
20.0
1.95
pH change
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

17. Reaching the Equivalence Point
After adding 25.0 mL, we will reach the equivalence point where the pH will always equal 7.00 (at 25°).
Since the acid is now fully neutralized, the only source of hydronium ions is from water.
The [H3O+] is 1.00 x10-7, so the pH is 7.00
pH = 7.00
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

18. pH after Adding 30.0 mL NaOH
As NaOH is added beyond the equivalence point, it becomes an excess reagent.
Calculate the amount of OH- at any given point (past equivalence point) by subtracting the initial amount of H3O+ from the amount of OH- added
OH- (aq) H3O+ (aq) → 2 H2O (l)
Before Addition
~ 0.00 mol
mol
Addition
0.003 mol
-----
After Addition
~ mol
mol OH-
0.100 mol
1.00 L
mol OH- added = L x = mol OH-
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

19. pH after Adding 30.0 mL NaOH
Then calculate the OH- concentration by dividing the number of moles of OH- remaining by the total volume, then we are able to calculate [H3O+] and pH.
[OH-] = = M
[OH-][H3O+] = 10-14
[H3O+] = =
= 1.10 x M
pH = -log(1.10 x 10-12)
= 11.96
mol OH-
L L
10-14
[OH-]
10-14
pH = 11.96
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

20. pH after Adding 35.0, 40.0, and 50.0 mL NaOH
Again, this is the same as the previous slide.
These are the results:
Volume (mL) pH
35.0
12.22
40.0
12.36
45.0
12.52
Change in pH
Because HCl is a strong Acid, the concentration of H3O+ is also 0.100M

21. Volume of NaOH added (mL)
The Overall pH Curve
The overall pH curve for the titration of a strong acid with a strong base has the characteristic S-shaped.
When a strong base is titrated with a strong acid, the graph looks the same except it is reflected over the equivalence point. pH
Volume of NaOH added (mL)

22. Summarization
The initial pH is simply the pH of the strong acid solution to be titrated
Before the equivalence point, H3O+ is in excess. Calculate the [H3O+] by subtracting the number of moles of added OH- from the initial number of moles of H3O+ and dividing by the total volume.
At the equivalence point, neither reactant is in excess and the pH = 7.00
Beyond the equivalence point, OH-is in excess. Calculate the [OH-] by subtracting the initial number of moles of H3O+ from the number of moles of added OH- and dividing by the total volume

23. Practice Problem
A 50.0 mL sample of a M sodium hydroxide is titrated with M nitric acid
Calculate the pH of:
(a) after adding 30.00mL of HNO3
(b) at the equivalence point

24. Practice Problem [OH-] = = 0.05M pOH = -log(0.05) = 1.30 pH = 14 - pOH
Initial amount of NaOH:
Moles NaOH = L x = mol = mol OH-
Moles HNO3 = L x = mol HNO3
mol
0.05 L L
0.200 mol
1.00 L
[OH-] =
= 0.05M
pOH = -log(0.05)
= 1.30
pH = 14 - pOH =
= 12.70
0.200 mol
1.00 L
OH- (aq) H3O+ (aq) → 2 H2O (l)
Before Addition
mol
~0.00 mol
Addition
-----
0.006 mol
After Addition
0.004 mol
(b) pH = 7.00 because at the equivalence point, the pH will always be 7.00 for strong acid/base titrations