Buffer Systems
Fig. 19.5
BUFFERS
Mixture of an acid and its conjugate base.
Buffer solution  resists change in pH when acids
or bases are added or when dilution occurs.
Mix:
A moles of weak acid + B moles of conjugate base
Find:
• moles of acid remains close to A, and
• moles of base remains close to B
 Very little reaction
HA  H+ + A-
Le Chatelier’s principle
?
Why does a buffer resist change in pH when small
amounts of strong acid or bases is added?
The acid or base is consumed
by A- or HA respectively
A buffer has a maximum capacity to resist change to pH.
How a Buffer Works
Consider adding H3O+ or OH- to water and also to a buffer
For 0.01 mol H3O+ to 1 L water:
[H3O+] = 0.01 mol/1.0 L = 0.01 M
pH = -log([H3O+]) = 2.0
So, change in pH from pure water: DpH = 7.00 – 2.00 =
5.0
For the H2CO3- / HCO3- system:
pH of buffer = 7.38
Addition of 0.01 mol H3O+ changes pH to 7.46
So change in pH from buffer:
DpH = 7.46 – 7.38 = 0.08 !!!
How a Buffer Works
Consider a buffer made from acetic acid and sodium acetate:
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Ka =
[H3
O+ ]
[CH3COO-] [H3O+]
[CH3COOH]
= Ka x
[CH3COOH]
[CH3COO-]
or
How a Buffer Works
Let’s consider a buffer made by placing 0.25 mol of acetic acid and
0.25 mol of sodium acetate per liter of solution.
What is the pH of the buffer?
And what will be the pH of 100.00 mL of the buffer before and
after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer?
What will be the pH of 300.00 mL of pure water if the same acid
is added?
1.8 x
10-5
=
(0.25x)
(0.25)
x = 1.8 x 10-5 = [H3O+]
pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74
Before acid added!
How a Buffer Works
What is pH if added to pure water?
1.00 mL conc. HCl
.001 L x 12.0 mol/L = 0.012 mol H3O+
Added to 300.00 mL of water :
0.012 mol H3O+ = 0.0399 M H O+
3
.301 L soln.
pH = -log(0.0399 M)
pH = 1.40 Without
buffer!
How a Buffer Works
After acid is added to buffer:
Conc. (M)
CH3COOH(aq) + H2O(aq)
Initial
0.250
---Change
+0.012
---Equilibrium
0.262
---Solving for the quantity ionized:
Conc. (M)
CH3COOH(aq) + H2O(aq)
Initial
Change
Equilibrium
0.262
-x
0.262 - x
----------
CH3COO- + H3O+
0.250
0
-0.012
0.012
0.238
0.00
CH3COO- + H3O+
0.238
+x
0.238 + x
Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238
(0.238 x)
-5
1.8 x 10 =
-5 = [H O+]
X
=
1.982
x
10
3
(0.262)
pH = -log(1.982 x 10-5) = 4.70 After the acid is added!
0
+x
x
How a Buffer Works
Suppose we add 1.0 mL of a concentrated base instead of an acid.
Add 1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see
what the impact is:
1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OHThis will reduce the quantity of acid present and force the equilibrium
to produce more hydronium ion to replace that neutralized by the
addition of the base!
Conc. (M)
Initial
Change
Equilibrium
CH3COOH(aq) + H2O(aq)
0.250
- 0.012 - x
0.238 - x
----------
CH3COO- + H3O+
0.250
+0.012 +x
0.262 +x
0
+x
x
Assuming: Again, using x as the quantity of acid dissociated we get:
our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238
1.8 x
10-5
(0.262) x
=
0.238
pH = -log(1.635 x 10-5) = 4.79
x = 1.635 x 10-5 = [H3O+]
After base is added!
How a Buffer Works
By adding the 1.00mL base to 300.00 mL of pure water we would get a
hydroxide ion concentration of:
[OH-]
0.012
mol
OH
=
301.00 mL
= 3.99 x 10-5 M OH-
The hydrogen ion concentration is:
-14
Kw
1
x
10
+
[H3O ] =
=
[OH ]
3.99 x 10-5 M
= 2.506 x 10-10
This calculates out to give a pH of:
pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the
base in pure water!
In summary:
Buffer alone pH = 4.74
Buffer plus 1.0 mL base pH = 4.79
Buffer plus 1.0 mL acid pH = 4.70
Base alone, pH = 9.59
Acid alone, pH = 1.40
Problem:
Calculate the pH of a solution containing 0.200 M NH3
and 0.300 M NH4Cl given that the acid dissociation
constant for NH4+ is 5.7x10-10.
NH3 + H2O  NH4+ + OHbase
Ka
acid
pH  pK a  log
[B]
[BH ]
pH = 9.244 + log
pH = 9.07
pKa = 9.244
(0.200)
(0.300)
pKa applies
to this acid