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chapter
Buffers
>> Parts of this equation/concept include:
Buffers are also examples of the common ion effect since they are
mixtures where both substances produce the same ion. One substance
reacts completely; the other, in equilibrium. Always do the complete
reaction first to determine the initial concentrations for the ICE
table. What makes the mixture a buffer is that the concentration
of both components of a conjugate acid–base pair is significant.
>> Example 1
What is the pH of a solution of 0.12 M HOBr and 0.53 M NaOBr?
Solution:
NaOBr contains always soluble Na+, so it completely
ionizes.
NaOBr 
Na+ + OBr–
The common ion is OBr–. [OBr–]
= 0.53 M. Since OBr– is the conjugate base
of HOBr, the system is a buffer.
The equilibrium is
HOBr + H2O 
H3O+ + OBr–
The ICE table is
| |
HOBr |
H3O+ |
OBr– |
| Initial |
0.12 |
0 |
0.53 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.12 – x |
x |
0.53 + x |
Assuming x is small,
4.5 x 10–10 = x
pH = 9.34
| A. Henderson–Hasselbalch Equation |
Because of the common ion effect, changes in concentration will
be small. Thus the assumption that "x is small" is more likely
to be true. This assumption is made in the Henderson–Hasselbalch
equation. It can be used instead of Ka, although
Ka is an equally acceptable way to solve the problem.
The Henderson–Hasselbalch equation
can be used for any buffer, even if the buffer is based on a weak
base rather than a weak acid. Because a conjugate acid–base
pair is used, it is always possible to use the pKa
of the acid. Kb is not used because Kb
is proportional to the concentration of OH– not
H3O+.
>> Example 2
What is the pH of a solution that is 0.13 M HCOOH and 0.24 M KHCO2?
Solution:
Since potassium salts are always soluble
KHCO2
K+ + HCO2–
[K+] = [HCO2–] = 0.24 M
And the weak acid reacts as
HCOOH + H2O
H3O+ + HCO2–
In the equilibrium, HCOOH is the acid and HCO2–
is its conjugate base. The Ka of the acid, HCOOH,
is 1.8 x 10–4. The pKa = –log
Ka = –log(1.8 x 10–4) =
3.74
So using the Henderson–Hasselbalch equation,
pH = 3.74 + 0.27
pH = 4.01
>> Example 3
What is the pH of a buffer made of 0.21 M NH3 and 0.48 M NH4Cl?
Solution:
NH3 is a weak base. Its reaction is
NH3 + H2O
NH4+ + OH–
NH4Cl is a soluble salt. Its reaction is
NH4Cl
NH4+ + Cl–
The conjugate acid–base pair is NH3 and NH4+,
and NH4+ is the acid.
The Kb for NH3 is the value listed
in the table, 1.8 x 10–5. Therefore the Ka
of NH4+ can be determined from
KaKb = 1.0 x 10–14
Ka(1.8 x 10–5) = 1.0 x 10–14
Ka = 5.6 x 10–10
pKa = –log(5.6 x 10–10)
= 9.25
Using the Henderson–Hasselbalch equation
pH = 8.89
>> Example 4
What ratio of sodium oxalate (Na2C2O4)
and sodium hydrogen oxalate (NaHC2O4) is
required for a buffer with a pH of 4.5?
Solution:
Na2C2O4
2 Na+ + C2O42–
NaHC2O4
Na+ + HC2O4–
The conjugate acid–base pair is acid = HC2O4–
and base = C2O42–. The Ka
of the acid is Ka2 of oxalic acid (H2C2O4)
= 6.4 x 10–5. Consequently, the pKa
= 4.19. Using Henderson–Hasselbalch and solving for the acid–base
ratio,
Therefore, you need twice as much base as acid to create this
buffer.
>> back
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The buffer capacity is the amount of acid or base that can be added
to a buffer before a significant pH change. Significance is in the
eye of the beholder, however, so it is normally defined as part
of the question.
Recall that the addition of acid or base does two things. First,
it reduces the concentration of the substance it reacts with. Second,
it increases the amount of the conjugate. Stoichiometry is used
to do this, so changes are calculated in terms of moles.
Good news! Because the amount of acid and base is a ratio, you
can work in either moles or molarity of acid and conjugate base
when using the Henderson–Hasselbalch equation.
>> Example 5
How many moles of NaOH are required to change the pH of 100 mL
of buffer made from 0.10 M NaNO2 and 0.22 M
HNO2 to pH = 4.00?
Solution:
For the conjugate acid–base pair of HNO2/NO2–,
the pKa of the acid (HNO2) is
pKa = –log(4.5 x 10–4)
= 3.35.
For the solution to have a pH = 4, the ratio of acid to base
must be
Adding base will remove HNO2 and create NO2–.
For every mole of HNO2 removed, 1 mole of NO2–
will be created.
HNO2 + NaOH
NaNO2 + H2O
To work in moles, the moles of NO2–
are (0.10 mol/L)(0.1L) = 0.01 mol NO2–
The moles of HNO2 are (0.22 mol/L)(0.1L) = 0.022 mol
HNO2.
To get the appropriate mole ratio,
0.09834 – 4.47x = 0.01 + x
0.08834 = 5.47x
0.016 = x
That's it. 0.016 mole NaOH will remove 0.016 mole of HNO2
(leaving 0.204 mole) and will add 0.016 mole NO2–
(making 0.116 mole).
Although the question wasn't asked, for perspective—the original
pH of the solution was 3.0. The 0.016 mol of HaOH could come from
16 mL of 1 M NaOH or 160 mL of 0.1 M NaOH or 0.64
g of solid NaOH.
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