Chang, 7th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2002
Buffers
A buffer is a solution that resists changes in pH. The pH of a buffer changes very little when
small amounts of a strong acid or strong base are added to the buffer.
A buffer consists of approximately equal amounts of a conjugate weak acid/weak base pair in
equilibrium with each other. Strong acids and their conjugate bases don’t produce a buffer since
strong acid ionization is complete: there is no equilibrium!
Acidic Buffers
In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal
concentrations. The equilibrium is
HA

weak acid
H+
+
A–
conjugate base
[A ] 
[A ] 
[H  ][A– ]


: this is
For a weak acid, K a 
so pK a  pH  log
. Thus pH  pKa  log 
[HA]
[HA]
[HA] 
the Henderson-Hasselbalch equation.
These equations are useful for estimating the pH of a buffer. When [A-] = [HA], it follows that
Ka = [H+] and pH = pKa.
Acidic Buffer Example: Acetic acid (HC2H3O2), sodium acetate (NaC2H3O2)
NaC2H3O2 is a soluble salt that dissolves in water to give Na+ and C2H3O2- ions. Na+ ions are
neutral spectator ions, so can be ignored. The C2H3O2- ions provide the conjugate base for the
buffer. The buffer equilibrium is
HC2H3O2
acetic acid

H+ +
C2H3O2–
acetate ion
For acetic acid, Ka(HC2H3O2) = 1.8 x 10-5 so pKa = -log10Ka = 4.74. Thus when [HC2H3O2] =
[C2H3O2–], pH = pKa = 4.74.
Basic Buffers
In a basic buffer both the weak base and conjugate acid are present initially in roughly equal
concentrations. The equilibrium is
B
+
weak base
H2 O


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BH+ + OH–
conjugate acid
Chang, 7th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2002
[BH  ]
[BH  ][OH– ]
 =
so pK b  pOH  log 
[B]
 [B] 

[BH ] 

. Thus pH  14.00  pK b  log [BH ]
. Since pKa = 14.00 –
pK b  (14.00  pH)  log
 [B] 
 [B] 
 [B] 
pKb, pH  pKa (BH )  log 
. This is once again the Henderson-Hasselbalch equation.
[BH  ]
For a weak base, K b 
These equations are useful for estimating the pH of a buffer. When [BH+] = [B], Kb = [OH–].
Consequently, pOH = pKb for the buffer. It follows that pH = 14.00 – pKb = pKa(BH+) for the
buffer.
Basic Buffer Example: Ammonia (NH3), ammonium chloride (NH4Cl)
NH4Cl is a soluble salt that dissolves in water to give NH4+ and Cl– ions. Cl– ions are neutral
spectator ions, so can be ignored. The NH4+ ions provide the conjugate acid for the buffer. The
buffer equilibrium is
NH3 +
ammonia

H2O
OH–
+ NH4+
ammonium ion
For ammonia, Kb(NH3) = 1.8 x 10-5 so pKb = -log10Kb = 4.74. Thus when [NH3] = [NH4+], pOH
= 4.74 and pH = pKa(NH4+) = 14.00 – 4.74 = 9.26.
Neutral Buffers
Neutral buffers have a pH close to 7.00. A good example is a NaH2PO4/Na2HPO4 buffer. Since
Na+ ions are neutral spectator ions, this is a dihydrogen phosphate/hydrogen phosphate (H2PO4–
/HPO42–) buffer. The buffer equilibrium is
H2PO4–

dihydrogen phosphate ion
H+
+
HPO42–
hydrogen phosphate ion
[H  ][HPO 4 ]
Here K a 
= 6.2 x 10–8. If [H2PO4-] = [HPO42-], Ka = [H+] and pH = pKa = 7.21.

[H2 PO4 ]
2–
Response of a Buffer to the Addition of a Strong Acid or a Strong Base
Added Acid. When a small amount of a strong acid such as HCl is added to a buffer, the H+ from
the acid reacts with the basic part of the buffer to give more of the acidic part of the buffer. The
reaction is assumed to go 100%. The new concentration of the acidic part of the buffer
(increased from the initial value) and the new concentration of the basic part of the buffer
(decreased from the initial value) are then used to calculate the pH of the buffer.
Added Base. When a small amount of a strong base is added to a buffer, the OH– reacts with the
acidic part of the buffer to give more of the basic part of the buffer. The reaction is assumed to
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Chang, 7th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2002
go 100%. The new concentration of the acidic part of the buffer (decreased from the initial
value) and the new concentration of the basic part of the buffer (increased from the initial value)
are then used to calculate the pH of the buffer.
Exercises
1. Identify which of the following mixed systems could function as a buffer solution. For each
system that can function as a buffer, write the equilibrium equation for the conjugate acid/base
pair in the buffer system
(a) KF/HF
(b) NH3/NH4Br
(c)KNO3/HNO3
(d) Na2CO3/NaHCO3
Answer:
(a) Buffer: HF  H+ + F–
(b) Buffer: NH3 + H2O  NH4+ + OH– (or NH4+  H+ + NH3)
(c) Not a buffer since HNO3 is a strong acid. It is 100% ionized.
(d) Buffer: HCO3–  H+ + CO32–
2. What is the pH of a 1 L solution containing 0.240 mol HC2H3O2 and 0.180 mol NaC2H3O2?
Ka(HC2H3O2) = 1.8 x 10-5. HINT: Begin by filling out the equilibrium table below.
Balanced Equation
HC2H3O2 
H+
+
C2H3O2–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
Answer:
Balanced Equation
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
H+
HC2H3O2 
0.240
-x
0.240 - x
0
x
x
+
C2H3O2–
0.180
x
0.180 + x

Ka 
[H ][C2 H3 O2 ]
x (0.180  x) x (0.180)
 1.8 105 

. This approximation is OK if the %
[HC 2 H3 O2 ]
(0.240 - x)
0.240
ionization is < 5%; it is in this case. Thus x = 2.4 x 10-5 = [H+]. pH = 4.62.
3. What would be the pH of the buffer from #2 above if 0.010 mol of HCl were added per liter of
the buffer?
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Chang, 7th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2002
Answer:
The new initial concentrations are:
[C2H3O2-] = (0.180 - 0.010) M = 0.170 M (basic part decreases)
[HC2H3O2] = (0.240 + 0.010) M = 0.250 M (acidic part increases)

Thus
Ka 
[H ][C2 H3 O2 ]
x (0.170  x) x (0.170)
 1.8 105 

. Thus x = [H+] = 2.647 x 10-5; pH = 4.58.
[HC 2 H3 O2 ]
(0.250 - x)
0.250
Note that the pH has decreased very little!!
4. What would be the pH of the buffer from #2 above if 0.010 mol NaOH were added per liter of
the buffer?
Answer:
The new initial concentrations are:
[C2H3O2-] = (0.180 + 0.010) M = 0.190 M (basic part increases)
[HC2H3O2] = (0.240 - .010) M = 0.230 M (acidic part decreases)

Thus K a 
[H ][C2 H3 O2 ]
x (0.190  x) x (0.190)
 1.8 105 

. Thus x = [H+] = 2.179 x 10-5; pH =
[HC 2 H3 O2 ]
0.230 - x
0.230
4.66. Note that the pH has increased very little!!
5. What would be the pH of 0.010 M HCl without the buffer?
Answer:
HCl is a strong acid and is therefore 100% ionized. Thus [H+] = 0.010M, so the pH = 2.00.
Note how much lower the pH is than it was with the buffer in #3 above (pH = 2.00 here versus 4.58 in #3 above)!
6. What would be the pH of 0.010 M NaOH without the buffer?
Answer:
NaOH is a strong base and is therefore 100% ionized. Thus [OH –] = 0.010M, so the pOH = 2.00. Thus pH = 14.00 –
pOH = 12.00.
Note how much higher the pH is than it was with the buffer in #4 above (pH = 12.00 here versus 4.66 in #4 above)!
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