Ways to create a buffer--Answers
-
I. By mixing solutions of HA and NaA, where HA is a weak acid (and thus A is a weak base)
Ex. What is the pH of a buffer prepared by mixing 50.0 mL of 0.400 M HClO(aq) with 100.0 mL of
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0.700 M NaClO(aq)? Ka(HClO) = 3.5 x 10
Recognize that (due to the common ion effect), the concentrations of HClO and ClO will not change
significantly as equilibrium is reached (they will however change from the values in the "problem"
because of dilution upon mixing). Then use:
[H3O+ ]  Ka x
[HA]
moles HA
 Ka x
[A ]
moles A -
Method 1: Use dilution approach to calculate concentrations of HA and A in the buffer
 50.0 mL 
 100.0 mL 
 0.133 M ; [A-] =  0.700 M 

  0.467 M
 150.0 mL 
 150.0 mL 
 0.133 M 
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[H3O+ ]  3.5 x 10-8 
  1.0 x 10 M ; pH = 8.00
0.467
M


[HA] =  0.400 M 


Method 2: Just calculate moles of each.
-
mol HA = (0.0500 L)(0.400 M) = 0.0200 mol; mol A = (0.100 L)(0.700 M) = 0.0700 mol
 0.0200 mol 
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[H3O+ ]  3.5 x 10-8 
  1.0 x 10 M ; pH = 8.00
0.0700
mol




II. By conversion of some of one conjugate into the other (i.e., by "reaction")
A. By converting some of a weak acid into its conjugate base (by adding some NaOH)
Ex. Start with 100. mL of 0.10 M HClO (0.010 mol). Add 4.0 mL of 1 M NaOH (0.004 mol)
0.100 L x 0.10 M HClO = 0.010 mol HClO
+
0.0040 L x 1 M NaOH = 0.004 mol NaOH (= 0.004 mol OH ; Na irrelevant)
“ICF” reaction (goes until someone runs out): HClO + OH  ClO + H2O
-
-
-
-
in moles
HClO
OH
I
0.010
0.004 (LR)
0
C
-0.004
-0.004
+0.004
F
0.006
0
0.004
ClO
Decent amounts of HClO and ClO are present after this reaction  a buffer!
-
[H3O ]  Ka x
+
moles HClO
0.006
 3.5 x 10 -8 x
 5.25 x 10 -8 M
0.004
moles ClO
 pH  -log(5.25 x 10 )  7.279..  7.3
-8
Answer is reasonable. There is a bit more
HClO than ClO- here. Thus, the pH should be a
bit lower than pKa (= 7.46) and it is. (Was a bit
silly to have only 1 SF though, in moles.)
B. By converting some of a weak base into its conjugate acid (by adding some HCl)
Ex. Start with 100. mL of 0.10 M NaClO (0.010 mol). Add 4.0 mL of 1 M HCl (0.004 mol)
Ways to create a buffer--Answers
“ICF” reaction (goes until someone runs out): ClO + HCl  HClO + Cl
-
in moles
ClO
I
-
-
HCl
HClO
0.010
0.004 (LR)
0
C
-0.004
-0.004
+0.004
F
0.006
0
0.004
Decent amounts of HClO and ClO are present after this reaction  a buffer!
-
[H3O ]  Ka x
+
moles HClO
0.004
 3.5 x 10 -8 x
 2.33.. x 10 -8 M
0.006
moles ClO
Answer is reasonable. There is a bit more ClOthan HClO here. Thus, the pH should be a bit
higher than pKa (= 7.46) and it is. (Was a bit
silly to have only 1 SF though, in moles.)
 pH  -log(2.33.. x 10 )  7.63..  7.6
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----------------------------------------------------Which of the following give a buffer solution when equal volumes are combined? **Must consider RXNS**
Shortest Answers (see below for full explanation:
-
a) 0.10 M HF and 0.10 M NaF
(a) Buffer. Mixed some HF (WA) with F (its CB; sodium salt). Both
“remain” at equilibrium (opposite sides of acid ionization rxn)
b) 0.10 M HF and 0.10 M NaOH
(b) Not. The OH reacts will all of the HF. This solution will be the
equivalent of a solution of NaF. (F only; no HF left)
c) 0.20 M HF and 0.10 M NaOH
(c) Buffer. The OH reacts will some but not all of the HF. So some F
(created) and some HF (remaining) is present after rxn.
d) 0.10 M HCl and 0.20 M NaF
e) 0.30 M HCl and 0.10 M NaF
-
-
-
-
(d) Buffer. The HCl reacts with some but not all of the F . So some HF
(created) and some F (remaining) is present after rxn.
(e) Buffer. Same as (d). [Note: One does not need to have equal
amounts of HA and A to have a buffer. Just a “reasonable” amount
of each (no more than 10-100 x more of one vs. the other)
Strategy:
NOTE: A buffer must contain some of a weak acid and some of its conjugate base. So if those
two components are mixed together directly, clearly a buffer will result. However, one must
realize that if a strong acid or strong base is one of the substances being mixed a reaction
may occur to generate one of the components of the buffer that is not “there” initially. So you
must be on the lookout for this kind of mixture. As will be described below, in such an
instance, a calculation needs to be done to see whether or not a buffer will result.
1) Look to see what two kinds of substances are being mixed. If the following substances are
mixed, you can make the conclusion right away without doing a calculation:
a) If two acids are mixed, it won’t be a buffer.
(You would neither have or generate the weak
base component.)
b) If two bases are mixed, it won’t be a buffer.
(You would neither have or generate the weak
base component.)
c) If a weak acid and a Gp I salt of its conjugate base are mixed (e.g., HA and NaA or KA), it
will be a buffer.
d) If a weak molecular base and a salt of its conjugate acid with a negligible base anion (e.g.,
RNH2 and RNH3Cl, where “R” represents C (with other atoms bound to it) or H), it will be a
buffer.
2) In the following two cases, a calculation (an “ICF” type) will be needed:
Ways to create a buffer--Answers
d) If a weak acid and strong base are mixed, it may be a buffer because they will react with
one another until one of them runs out, producing some of the conjugate base of
the weak acid. The question becomes, “What will be left over?” So do an ICF type
calculation (with moles) and assess the result:
i) If there is excess weak acid (i.e., the SB is limiting), the resulting solution is a buffer.
ii) If there is excess strong base (i.e., the WA is limiting), the resulting solution will not be a
buffer (in effect, a buffer was initially created but then so much SB was around that the buffer capacity was
exceeded).
iii) If there is no excess of either, the resulting solution will not be a buffer. The solution
contains only the conjugate base of the weak acid. (Solution of a weak base.)
e) If a weak base salt (i.e., Gp I salt of a base anion) and a strong acid are mixed, it may be a
buffer because they will react with one another until one of them runs out, producing
some of the conjugate acid of the weak base. The question becomes, “What will be left
over?” So do an ICF type calculation (with moles) and assess the result:
i) If there is excess weak base (i.e., the SA is limiting), the resulting solution is a buffer.
ii) If there is excess strong acid (i.e., the WB is limiting), the resulting solution will not be a
buffer (in effect, a buffer was initially created but then so much SA was around that the buffer capacity was
exceeded).
iii) If there is no excess of either, the resulting solution will not be a buffer. The solution
contains only the conjugate acid of the weak base. (Solution of a weak acid.)
Execution of Strategy:
-
(a) HF and NaF are mixed. HF and F will both be present. This will make a buffer.
The amounts are basically irrelevant since neither one is ridiculously small.
~100%
(b) HF is mixed with NaOH. They will react to form F : HF + OH 
 F + H2O
-
-
-
 Do an ICF type calculation to see who’s limiting. For ease of calculation, assume 1 L
(exactly) of each solution.
moles HF: 1 L x 0.10 mol/L  0.10 mol HF
moles OH : 1 L x 0.10 mol/L  0.10 mol OH
-
-
All of both will react  Will not make a buffer.
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to
“prove” it to you (and for completeness), I will show the results of that here:
in moles
OH-
F-
HF
I
0.10
0.10
0
C
- 0.10
- 0.10
+ 0.10
F
0
0
0.10
A solution of only F
-
~100%
 F + H2O
(c) HF is mixed with NaOH. They will react to form F : HF + OH 
-
-
-
 Do an ICF type calculation to see who’s limiting. For ease of calculation, assume 1 L
(exactly) of each solution.
moles HF: 1 L x 0.20 mol/L  0.20 mol HF
moles OH : 1 L x 0.10 mol/L  0.10 mol OH
-
-
L.R.  This will make a buffer.
Ways to create a buffer--Answers
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to
“prove” it to you (and for completeness), I will show the results of that here:
in moles
OH-
HF
F-
I
0.10
0.20
0
C
- 0.10
- 0.10
+ 0.10
F
0
0.10
0.10
Both HF and F  buffer
-
(d) NaF is mixed with HCl. They will react to form HF:
~100%
F + HCl 
 HF + Cl
-
-
 Do an ICF type calculation to see who’s limiting.
moles F : 1 L x 0.20 mol/L  0.20 mol F
-
-
moles HCl: 1 L x 0.10 mol/L  0.10 mol HCl
L.R.  This will make a buffer.
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to
“prove” it to you (and for completeness), I will show the results of that here:
in moles
F-
HCl
HF
I
0.20
0.10
0
C
- 0.10
- 0.10
+ 0.10
F
0.10
0
0.10
A buffer!
(e) NaF is mixed with HCl. They will react to form HF:
~100%
F + HCl 
 HF + Cl
-
-
 Do an ICF type calculation to see who’s limiting.
moles F : 1 L x 0.30 mol/L  0.30 mol F
-
-
moles HCl: 1 L x 0.10 mol/L  0.10 mol HCl
L.R.  This will make a buffer.
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to
“prove” it to you (and for completeness), I will show the results of that here:
in moles
F-
HCl
HF
I
0.30
0.10
0
C
- 0.10
- 0.10
+ 0.10
F
0.20
0
0.10
A buffer!