Buffers Buffers
A buffer is a solution that resists changes in pH. The pH of a buffer changes very little when 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.
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 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 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!
strong acid ionization is complete: there is no equilibrium!
Acidic Buffers Acidic Buffers
In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal concentrations. The equilibrium is
concentrations. The equilibrium is HA
HA ↔↔ HH++ + + AA––
w
weeaak k aacciidd ccoonnjjuuggaatte e bbaassee
For a weak acid,
For a weak acid, K K aa
=
=
[[HH ++][A][A – – ]]
[HA]
[HA] so so p p K
K aa
=
=
pH pH−
−
loglog [[AA − −]] [HA] [HA]
. Thus. Thus pH pH==
p p K K aa++
loglog [A[A −−]] [HA] [HA]
:: this is thethis is the Henderso Henderson-Hassen-Hasselbalch equlbalch equationation..
These equations are useful for estimating the pH of a buffer. When [A
These equations are useful for estimating the pH of a buffer. When [A --] = [HA], it follows that] = [HA], it follows that
K
Kaa = [H = [H++] and pH = pK] and pH = pKaa..
Acidi
Acidic c BuffeBuffer r ExamplExample: Acetic acid e: Acetic acid (HC(HC22HH33OO22) , s) , sodiodium aum acetcetate ate ( ( NaCNaC22HH33OO22))
NaC
NaC22HH33OO22 is a soluble salt that dissolves in water to give Na is a soluble salt that dissolves in water to give Na++ and C and C22HH33OO22-- ions. Na ions. Na++ ions are ions are
neutral spectator ions, so can be ignored. The C
neutral spectator ions, so can be ignored. The C22HH33OO22-- ions provide the conjugate base for the ions provide the conjugate base for the
buffer. The buffer equilibrium is buffer. The buffer equilibrium is HC
HC22HH33OO22 ↔↔ HH++ + + CC22HH33OO22––
aacceettiic c aacciidd aacceettaatte e iioonn For acetic acid,
For acetic acid, K K aa(HC(HC22HH33OO22) = 1.8 x 10) = 1.8 x 10-5-5so so ppK K aa = -log = -log1010K K aa= 4.74. Thus when [HC= 4.74. Thus when [HC22HH33OO22] =] =
[C
[C22HH33OO22––], pH = p], pH = pK K aa = 4.74. = 4.74.
Basic Buffers Basic Buffers
In a basic buffer both the weak base and conjugate acid are present initially in roughly equal In a basic buffer both the weak base and conjugate acid are present initially in roughly equal concentrations. The equilibrium is
concentrations. The equilibrium is B
B + + HH22OO ↔↔ BHBH++ + + OHOH––
weak base weak base
For a weak base, K b
=
[BH +][OH – ] [B] so p K b=
pOH−
log [BH +] [B]
= p K b=
(14.00−
pH)−
log [BH +] [B]
. Thus pH=
14.00−
p K b−
log [BH +] [B]
. Since pK a = 14.00 – pK b, pH=
p K a(BH + )+
log [B] [BH+]
. This is once again the Henderson-Hasselbalch equation.These equations are useful for estimating the pH of a buffer. When [BH +] = [B],K b = [OH–].
Consequently, pOH = pK b for the buffer. It follows that pH = 14.00 – pK b = pK a(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 + H2O ↔ OH– + NH4+
ammonia ammonium ion
For ammonia, K b(NH3) = 1.8 x 10-5 so pK b = -log10K b = 4.74. Thus when [NH3] = [NH4+], pOH =
4.74 and pH = pK a(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– ↔ H+ + HPO42–
dihydrogen phosphate ion hydrogen phosphate ion
Here K a =
[H
+][HPO
42–]
[H
2PO
4−]
= 6.2 x 10–8. If [H
2PO4-] = [HPO42-], K a= [H+] and pH = pKa = 7.21.
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 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?
K a(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 HC2H3O2 H + + C 2H3O2– Initial Concentration (M) 0.240 0 0.180 Change (M) -x x x Equilibrium Concentration (M) 0.240 - x x 0.180 + x K a =
[H
+][C
2H
3O
2−]
[HC
2H
3O
2]
=1.8
×10
−5 = x(0.180
+ x)
(0.240 -
x)
≈ x(0.180)
0.240
. This approximation is OK if the %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?
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 K a =
[H
+][C
2H
3O
2−]
[HC
2H
3O
2]
=1.8
×10
−5 = x(0.170
+ x)
(0.250 -
x)
≈ x(0.170)
0.250
. Thus x = [H + ] = 2.647 x 10-5 ;pH =4.58.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
+][C
2H
3O
2−]
[HC
2H
3O
2]
=1.8
×10
−5 = x(0.190
+ x)
0.230 -
x ≈ x(0.190)
0.230
. Thus x = [H +] = 2.179 x 10-5; pH= 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?
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.
