Buffer Solutions
• Buffer solutions are really a common ion effect
problem.
• Buffers are made when you add a salt that is a
weak base or acid to an acid-base solution to
control the pH.
• For example, if you have an aquarium and fish
need a certain pH to thrive, you would add salt to
change the pH-not an acid!
• Buffers are made by adding a conjugate acid to a
weak base and visa versa.
• Most of the time buffers are added to control
changes in pH.
Buffer Solutions
A buffer solution is a special case of the
common ion effect.
The function of a buffer is to resist or assist
changes in the pH of a solution.
solution.
Buffer CompositionComposition-examples
+
Conj. Base
Weak Acid
CH3CO2H
+
CH3CO2H2PO4
+
HPO42Weak Base
+
Conj. Acid
NH3
+
NH4+
Buffer Solutions
Consider CH3CO2H / CH3CO2- to see how buffers
work
ACID USES UP ADDED OHWe know that
CH3CO2- + H2O
CH3CO2H + OH10
has Kb = 5.6 x 10
Therefore, the reverse reaction of the WEAK
ACID with added OHhas Kreverse = 1/ Kb = 1.8 x 109
Kreverse is VERY LARGE, so CH3CO2H
completely snarfs up OH- !!!!
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Buffer Solutions
Consider CH3CO2H/CH3CO2- to see how buffers
work
CONJ. BASE USES UP ADDED H+
CH3CO2H + H2O
CH3CO2- + H3O+
5
has Ka = 1.8 x 10
Therefore, the reverse reaction of the WEAK
BASE with added H+
has Kreverse = 1/ Ka = 5.6 x 104
Kreverse is VERY LARGE, so CH3CO2- completely
snarfs up H+ !
Buffer Solutions
Problem: What is the pH of a buffer that has
[CH3CO2H] = 0.700 M and [CH
[CH3CO2-] = 0.600 M?
CH3CO2H + H2O
CH3CO2- + H3O+
Ka = 1.8 x 10-5
[CH3CO2H] [CH3CO2- ] [H3O+]
initial
change
equilib
0.700
-x
0.700 - x
0.600
0
+x
+x
x
0.600 + x
Buffer Solutions, con’t.
Assuming that x << 0.700 and 0.600,
we have
[H3O+ ](0.600)
-5
K a = 1.8 x 10
=
0.700
[H3O+] = 2.1 x 10-5 M and pH = 4.68
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Henderson--Hasselbalch Equation
Henderson
[H3O + ] =
[Acid]
•Ka
[Conj. base]
Take the negative log of both sides of this
equation
[Acid]
pH = pK a - log
[Conj. base]
pH = pK a + log
[Conj. base]
[Acid]
The pH is determined largely by the pKa of the
acid and then adjusted by the ratio of acid and
conjugate base.
Adding an Acid to a Buffer
Problem: What is the pH when 1.00 mL of 1.00 M
HCl is added to
a) 1.00 L of pure water (before HCl, pH = 7.00)
b) 1.00 L of buffer that has [CH3CO2H] = 0.700
M and [CH3CO2-] = 0.600 M (pH = 4.68)
Solution to Part (a)
Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L
of water
M1•V1 = M2 • V2
M2 = 1.00 x 10-3 M
pH = 3.00
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [CH3CO2H] = 0.700 M
and [CH3CO2-] = 0.600 M (pH before = 4.68)
Solution to Part (b)
Step 1 — do the stoichiometry
H3O+ (from HCl) + CH3CO2- (from buffer) ----->
>
CH3CO2H (from buffer) + H2O
The reaction occurs completely because K is
very large.
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Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (pH = 3.00)
b)
1.00 L of buffer that has [CH3CO2H] = 0.700 M
and [CH3CO2 -] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 1—
1—Stoichiometry
[H3O+] + [CH3CO2-] ----->
> [CH3CO2H]
I
C
After rxn
0.00100
0.600
0.700
-0.00100
-0.00100
+0.00100
0
0.599
0.701
Because [H3O+] = 2.1 x 10-5 M BEFORE
adding HCl, we again neglect x relative to
0.701 and 0.599.
pH = 4.68 before adding HCl and
pKa = - log (1.8 x 10-5) = 4.74
pH = 4.74 + Log (.599/.701) = 4.68
The pH has not changed on adding HCl to the
buffer!
KaHNO2 = 4.0 x 10-4
• If the total volume of a buffer made of NO2- and HNO2
is 1.35 L, what is the volume of the acid and the
volume of the base that will make a buffer with a final
pH of 3.55? Assume both acid and CB are the same
molarity.
pKa = 3.40 pH = 3.55 pH = pKa + Log [A-]
[HA]
pH – pKa = Log [A-]/ [HA]
10.15
1.41 =
= 1.41 =
[A-]/[HA]
.15 = Log [A-] / [HA]
[A-]/[HA]
and A- + HA = 1.35 L
A- = 0.79 L and HA = 0.56 L
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