Buffers
Corn Hill, Edward Hopper, 1930
Buffers
A buffer is a chemical system that includes:
or
Why Buffers?
Why Buffers?
Buffer Example
Blood must maintain a pH of about 7.4 and so contains a buffer composed of H2CO3 (a weak acid) and HCO3- (its conjugate base).
H2CO3(aq) + H2O(l) ⇌ HCO3-(aq) + H3O+(aq)
HA(aq) + HCO3-(aq) → A-(aq) + H2CO3(aq)
(and then a new equilibrium in the buffer is established)
Buffer Example
Blood must maintain a pH of about 7.4 and so contains a buffer composed of H2CO3 (a weak acid) and HCO3- (its conjugate base).
H2CO3(aq) + H2O(l) ⇌ HCO3-(aq) + H3O+(aq)
b) If base enters the blood stream, what component of the buffer system will neutralize it?
A-(aq) + H2CO3(aq) → HA(aq) + HCO3-(aq)
(and then a new equilibrium in the buffer is established)
Identifying Possible Buffer Solutions
Which of the following combinations form a buffer?
Buffer Capacity
Buffer Capacity is the amount of acid or base that a buffer can absorb without significantly changing the pH.
Buffer Capacity
Buffer Capacity
Example:
A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which of the following actions will destroy the buffer?
HF (aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
a) Adding 0.050 mol of HCl?
Yes, the moles of added acid are equal to the moles of the buffer’s base and will eliminate all of the base present in the buffer.
Buffer Capacity
Example:
A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which of the following actions will destroy the buffer?
HF (aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
b) Adding 0.050 mol of NaOH?
No, this is fewer moles of NaOH than moles of acid present in the buffer.
Buffer Capacity
Example:
A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which of the following actions will destroy the buffer?
HF (aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
c) Adding 0.050 mol of NaF?
No, this will simply add more of the conjugate base to the buffer.
Buffer Capacity
Example:
A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which of the following actions will destroy the buffer?
HF (aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
d) Does this buffer system have a higher buffering capacity for added acids or added bases?
Bases because the buffer contains more of the acid than its conjugate base.
Calculating the pH of a Buffer Solution
The pH of a buffer solution is affected by the common ion effect. Both the acid and the base have the same ion and therefore equilibrium is affected by its presence on both sides of the equilibrium.
Calculating the pH of a Buffer Solution
i) Write the ionization reaction for the acid:
HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
Example: What is the pH of a buffer solution that is 0.150 M in HF and 0.130 M in KF? (Ka of HF is 7.2x10-4)
Calculating the pH of a Buffer Solution
ii) Fill in the RICE table with the given concentrations of each species.
(Can eliminate due to large concentrations and small Ka value)
R | HF(aq) | + | H2O(l) | ⇌ | F-(aq) | + | H3O+(aq) |
I | 0.150M | | ------- | | 0.130M | | 0M |
C | - x | | ------- | | + x | | + x |
E | 0.150M-x | | ------- | | 0.130M+x | | x |
Example: What is the pH of a buffer solution that is 0.150 M in HF and 0.130 M in KF? (Ka of HF is 7.2x10-4)
Calculating the pH of a Buffer Solution
iii) Substitute equilibrium concentrations in Ka equation and solve for x.
R | HF(aq) | + | H2O(l) | ⇌ | F-(aq) | + | H3O+(aq) |
I | 0.150M | | ------- | | 0.150M | | 0M |
C | - x | | ------- | | + x | | + x |
E | 0.150M-x | | | | 0.150M+x | | x |
R | HF(aq) | + | H2O(l) | ⇌ | F-(aq) | + | H3O+(aq) |
I | 0.150M | | ------- | | 0.130M | | 0M |
C | - x | | ------- | | + x | | + x |
E | 0.150M-x | | ------- | | 0.130M+x | | x |
x = [H3O+] = 8.3x10-4 M
Example: What is the pH of a buffer solution that is 0.150 M in HF and 0.130 M in KF? (Ka of HF is 7.2x10-4)
Calculating the pH of a Buffer Solution
iv) Calculate the pH using [H3O+].
R | HF(aq) | + | H2O(l) | ⇌ | F-(aq) | + | H3O+(aq) |
I | 0.150M | | ------- | | 0.150M | | 0M |
C | - x | | ------- | | + x | | + x |
E | 0.150M-x | | | | 0.150M+x | | x |
R | HF(aq) | + | H2O(l) | ⇌ | F-(aq) | + | H3O+(aq) |
I | 0.150M | | ------- | | 0.130M | | 0M |
C | - x | | ------- | | + x | | + x |
E | 0.150M-x | | ------- | | 0.130M+x | | x |
pH = -log[8.3x10-4] = 3.08
x = [H3O+] = 8.3x10-4 M
Example: What is the pH of a buffer solution that is 0.150 M in HF and 0.130 M in KF? (Ka of HF is 7.2x10-4)
An easier way…
The Henderson-Hasselbalch Equation:
HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(aq)
Let’s do the previous problem again, this time using the H-H equation:
Notes about using the H-H Equation:
Notes about pKa
Example
Acid name | Acid formula | Ka | pKa |
iodic | HIO3 | 1.6x10-1 | 0.80 |
chlorous | HClO2 | 1.1x10-2 | 2.0 |
hydrofluoric | HF | 7.2x10-4 | 3.1 |
benzoic | C6H5COOH | 6.5x10-5 | 4.2 |
carbonic (1) | H2CO3 | 4.3x10-7 | 6.4 |
hydrosulfuric | H2S | 1.0x10-7 | 7.0 |
hydrocyanic | HCN | 6.2x10-10 | 9.2 |
One last practice
What if two solutions are combined to make a buffer solution?
Try this:
What is the pH of a buffer solution that is 0.50 M in HC2H3O2 and 0.40 M in NaC2H3O2?