If we have a chemical equilibrium, then we add an ionic
substance, only part of the ionic substance may affect the
equilibrium.
Example – in HF  H+ + FIf we add Na F
Na+
Fno effect adds to products
We should be able to calculate the concentration of all
reactants and products after we add the common ion.
Example – The Ka of HF is 6.8 x 10-4 at 25oC. What is the H+
concentration if we add enough NaF to make the solution
0.10 M HF and 0.20 M NaF?
The starting [HF] is 0.1 M and the starting [F-] is 0.2 M
We ignore the cation of the salt!

0.0 M
+ F0.2 M
-x
+x
+x
0.1
x
HF
0.1 M
H+
Due to the NaF!
0.2
Ka = [H+][F-]
6.8x 10-4 = 0.2 x X = 3.4 x 10-4 M
[HF]
0.1
Example – What is the pH of a solution that is 0.3 M acetic
acid and 0.3 M sodium acetate if the Ka for acetic acid is 1.8
x 10-5?
HC2H3O2  H+ + C2H3O2-
0.0 M
0.3 M
-x
+x
+x
0.3
x
0.3 M
0.3
Ka = [H+][C2H3O2-] 1.8 x 10-5 = 0.3 x X = 1.8 x 10-5
[HC2H3O2-]
0.3
pH = -log 1.8 x 10-5
pH = 4.74
In rare cases, x may represent a large amount. If this is true,
we will need to use the quadratic equation!
Solution that resists the change in pH
pH will still go up and down when you add acid or base, but
not as much as in pure water
Important in many pH dependent reactions.
Blood is buffered to keep the body at a specific pH.
Composition of Buffer
Weak acid and salt of that acid
(HC2H3O2 and NaC2H3O2)
This works by having enough HC2H3O2 to neutralize extra
OH-, and enough C2H3O2- to absorb extra H+
H+
In acids
HC2H3O2  H+ + C2H3O2Acetate ions absorb extra H+
In bases
HC2H3O2  H+ + C2H3O2OH-
H2O
OH- absorb H+, so acetic acid dissociates more
If we add too much acid or base, we use up all of the acetate
or acetic acid, and the buffer no longer works.
Buffer Capacity
Ability of the buffer to absorb strong acids or bases.
Depends on the concentrations of the acid and salt used.
Higher concentrations of acid and salt, more buffering
capacity.
Higher
0.01 M acetic acid and 0.01 M sodium acetate
buffer
0.20 M acetic acid and 0.20 M sodium acetate
capacity
pH of buffer
Important to use a buffer at the proper pH
pH depends on Ka of acid and the ratio of acid to its
conjugate base
Lots of base
Lots of acid
+
More acidic HA  H + A More basic
Can determine the pH of a buffer by using
pH = pKa + log [base]
[acid]
Henderson – Hasselbach equation
Okay, that’s the Anderson – Hasselhoff equation
If you know the Ka of the acid, and the concentration of the
acid and the concentration of the base, we can determine the
pH of the buffer.
Example – What is the pH of a buffer containing 0.10 M
HF (Ka = 6.8 x 10-4) and 1.0 M NaF?
pH = pKa + log [base] pH = - log(6.8 x 10-4) + log (1.0)
0.1
[acid]
pH = 4.167
Example – How many moles of sodium acetate are
needed to be added to 0.50 moles of acetic acid (Ka = 1.8 x
10-5) to make a buffer with a pH of 5.00?
Since the final volume will be the same for both acid and base, we can
use moles instead of molarity
pH = pKa + log [base] 5 = - log(1.8 x 10-5) + log (x)
[acid]
0.5
0.255 = log x
1.799 = x
X = 0.90 moles
0.5
0.5
Adding Strong acids and bases to buffers
Adding acid and base changes the pH of the buffers,
but not too much
To determine the pH, we need to calculate the
concentration of chemicals after adding the acid and base
All the extra H+ or OH- will be neutralized by the buffer.
We can then use buffer formula to determine final pH
Adding acids
1. Write the acid reaction
2. Write in all initial MOLES (You can’t use molarity!)
3. shift the equilibrium to use up all the extra H+
4. Determine the final [ ] of acid and base
5. Use the buffer formula to find pH
Example
If a buffer is made from 0.2 moles of acetic acid and
0.10 moles of sodium acetate, what is the pH if we add 0.02
moles of HCl?
0.02
0.1
0.2
HC2H3O2  H+ + C2H3O20.22
0.0
0.08
Shift the equilib to reactants to use up all 0.02 moles of H+
You need to use up 0.02 moles of acetate!
Now use the buffer formula
pH = pKa + log [base]
[acid]
pH = 4.30
pH = - log(1.8 x 10-5) + log (0.08)
0.22
Adding a base
Use the base reaction of the salt
X- + H2O  HX + OHExample – A buffer is made from 0.1 moles of NaF and 0.1
moles of HF (Ka = 6.8 x 10-4) in 1.0 L of solution. What is the
pH? What is the pH if we add 0.050 moles of NaOH?
1. Use the buffer formula to calculate the pH
pH = pKa + log [base]
[acid]
pH = 3.16
pH = - log(6.8 x 10-4) + log (0.1)
0.1
Now set up the base equation to determine the
concentrations after adding base
Before 0.1
0.1
0.050
F- + H2O  HF + OHAfter
0.150
0.050
0
Force the equilibrium to shift to use up all the OH-
Now calculate the new pH
pH = pKa + log [base]
[acid]
pH = 3.64
pH = - log(6.8 x 10-4) + log (0.15)
0.050
NOTE – If you use too much added acid or base, calculate
the pH based on the leftover H+ or OH-!
Before 0.1
0.1
0.15
F- + H2O  HF + OHAfter
0.20
0
0.05
[OH-] = 0.05 M
pOH = 1.30
pH = 12.7