Buffer This
The Common Ion Effect and Buffers
•
1.
•
There are two common kinds of buffer solutions:
1 Solutions made from a weak acid plus a soluble ionic salt of the weak acid.
2 Solutions made from a weak base plus a soluble ionic salt of the weak base
Solutions made of weak acids plus a soluble ionic salt of the weak acid
One example of this type of buffer system is:
– The weak acid - acetic acid CH3COOH
– The soluble ionic salt - sodium acetate NaCH3COO
Buffer Solutions Weak Acids Plus
Salts of Their Conjugate Bases
One example of the type I of buffer system is:
The weak acid - acetic acid CH3COOH
CH3COOH
H3O+ + CH3COO~100%
NaCH3COO →Na+ + CH3COOThe soluble ionic salt - sodium acetate NaCH3COO
• This is an equilibrium problem with a starting concentration for both the
cation and anion. After calculating the concentration of H+ and the pH of a
solution that is 0.15 M in both acetic acid sodium acetate yields:
R CH3COOH
I
C
E
H3O+ + CH3COO-
Buffer Solutions Weak Acids Plus
Salts of Their Conjugate Bases
Alternatively you might have noticed:
[base]/[acid] = 1 = 100
log 100 = 0
pH = pKa + 0 = pKa = 4.74
[H+] = Ka = 1.8E-5
Solution
[H+]
pH
0.15 M CH3COOH
1.6 x 10-3
2.80
0.15 M CH3COOH &
0.15 M NaCH3COO buffer
1.8 x 10-5
4.74
 [H+] is ~90 times greater in pure acetic acid than in buffer solution.
Note that the pH of the buffer equals the pKa of the buffering acid.
The Common Ion Effect and Buffers
• The general expression for the ionization of a weak monoprotic acid is:
• The generalized ionization constant expression for a weak acid is:
The Common Ion Effect and Buffers
• If we solve the expression for [H+], this relationship results:
• By making the assumption that the concentrations of the weak acid and
the salt are reasonable, the expression reduces to:
The Common Ion Effect and Buffers
• The relationship developed in the previous slide is valid for buffers
containing a weak monoprotic acid and a soluble, ionic salt.
• If the salt’s cation is not univalent the relationship changes to:
The Common Ion Effect and Buffers
• Simple rearrangement of this equation and application of algebra
yields the
Henderson-Hasselbach equation.
The Henderson-Hasselbach equation is one method to calculate the pH
of a buffer given the concentrations of the salt and acid.
Henderson-Hasselbalch - Caveats and Advantages
• Henderson-Hasselbalch equation is valid for solutions whose
concentrations are at least 100 times greater than the value of
their Ka’s
• The Henderson-Hasselbach equation is one method to calculate the
pH of a buffer given the concentrations of the salt and acid.
• A special case exists for the Henderson-Hasselbalch equation
when [base]/[acid] = some power of 10, regardless of the actual
concentrations of the acid and base, where the HendersonHasselbalch equation can be interpreted without the need for
calculations:
[base]/[acid] = 10x
log 10x = x
in general pH = pKa + x
Examples:
1. when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, pH = pKa,
(corresponds to the midpoint in the titration of a weak acid or base)
2. when [base]/[acid] = 10 or 101, log 10 = 1 then pH = pKa + 1
3. when [base]/[acid] = .001 or 10-2, log 10 = -2 then pH = pKa -2
Buffer Solutions
There are two common kinds of buffer solutions:
I.
II.
Commonly, solutions made from a weak acid plus a soluble ionic salt of the
conjugate base of the weak acid.
Less common, solutions made from a weak base plus a soluble ionic salt of
the conjugate acid of the weak base.
Both of the above may also be prepared by starting with a weak acid (or
weak base) and add half as many moles of strong base (acid)
Buffer Solutions: Weak Bases Plus
Salts of Their Conjugate Acids
One example of the type II of buffer system is:
The weak base – ammonia NH3
NH3
The soluble ionic salt – ammonium nitrate NH4NO3
NH4NO3
NH4 ++ OH ~100%
→ NH
4
++ NO 3
• This is an equilibrium problem with a starting concentration for both the
cation and anion. After calculating the concentration of OH- and the pOH of
the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in
ammonium nitrate, NH4NO3 yeilds:
R
NH3
NH4+
+
OH-
I
C
E
• Substitute the quantities determined in the previous relationship into
the ionization expression for ammonia.
The Common Ion Effect and Buffers
• We can derive a general relationship for buffer solutions that contain a
weak base plus a salt of a weak base similar to the acid buffer
relationship.
– The general ionization equation for weak bases is:
• Simple rearrangement of this equation and application of algebra
yields the
Henderson-Hasselbach equation.
The Common Ion Effect and Buffers
• A comparison of the aqueous ammonia concentration to that of the buffer
described above shows the buffering effect.
Solution
[OH-]
pH
0.15 M NH3
1.6 x 10-3 M
11.20
0.15 M NH3 &
0.15 M NH4NO3 buffer
9.0 x 10-6 M
8.95
 The [OH-] in aqueous ammonia is 180 times greater than in the buffer.
Buffering Action
• If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is
0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how
much does the pH change? Assume no volume change due to addition of
the HCl.
1 Calculate the pH of the original
buffer solution.
~100%
NH4NO3
NH3
R
NH3
→ NH
4
++ NO 3
NH4 ++ OH NH4+
+
OH-
I
C
E
• Substitute the quantities determined in the previous relationship into
the ionization expression for ammonia.
Buffering Action
2 Next, calculate the concentration of all species after the addition of
the gaseous HCl.
– The HCl will react with some of the ammonia and change the
concentrations of the species.
– This is another limiting reactant problem.
HCl
NH3 + H+
R
I
C
E
NH3
~100%
→ H + Cl
+
-
NH4 +
NH4+
+
OH-
Buffering Action
3 Using the concentrations of the salt and base and the HendersonHassselbach equation, the pH can be calculated.
4 Finally, calculate the change in pH.
Buffering Action
1. If 0.020 mole of NaOH is added to 1.00 liter of
solution that is 0.100 M in aqueous ammonia
and 0.200 M in ammonium chloride, how much
does the pH change?
Assume
no
volume
~100%
→ Na +of
OH the solid NaOH.
change due to NaOH
addition
+
NH4 + + OH-
R
I
C
E
NH3
-
NH3
NH4+
+
OH-
Buffering Action
2.
3.
Using the concentrations of the salt and base and the HendersonHassselbach equation, the pH can be calculated.
Finally, calculate the change in pH.
Buffering Action
Original Solution
1.00 L of solution
containing
0.100 M NH3 and
0.200 M NH4Cl
Original
pH
Acid or base
added
New
pH
pH
0.020 mol NaOH 9.08
+0.13
0.020 mol HCl
-0.14
8.95
8.81
• Notice that the pH changes only slightly in each case.
Preparation of Buffer Solutions
• Calculate the concentration of H+ and the pH of the solution prepared by
mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium
hydroxide solutions.
• Determine the amounts of acetic acid and sodium hydroxide prior to the
acid-base reaction.
• NaOH and CH3COOH react
in a 1:1 mole ratio.
• After the two solutions
are mixed, Calculate
total volume.
• The concentrations of
the acid and base are:
• Substitution of these
values into the ionization
constant expression (or
the HendersonHasselbach equation)
permits calculation of the
Preparation of Buffer Solutions
•
For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Starting with a solution that is 0.100M in
aqueous ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using ammonium chloride as the source of the soluble ionic salt of
the conjugate weak acid.
•
The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair
•
•
•
•
pOH can be determined from the pH:
pKb can be looked up in a table:
[base] concentration is provided:
Solve for [acid]:
~100%
•
Does this result make sense?
NH4Cl
+
NH4 +
Cl-
NH3
H2 O
NH4++ OH-