# [H 2 PO 4 - ] = 0.800 M.

```Chapter 8 – pH Buffers
Lect 2 Oct 18
Chapter 8 – pH Buffers
Buffers are extremely important in the regulation
of a variety of biochemical reactions. Often the
rate of the reaction is dependent on the pH. In
certain cases the enzyme is degraded and no longer
functions correctly outside a narrow pH range. The
next slide shows this for the rate of -chymotrysin
as a catalyst for the cleavage of a C – N bond.
Rate of cleavage of
C – N bond with chymotrysin (catalyst)
As a function of pH.
Chapter 8 – pH Buffers
This last expression may be rearranged to solve for
[H+ ] as
[H+ ] = Ka x {[HX] / [X- ]}
(Note that the last term is the ratio of the acid : base
forms.)
Henderson and Hasselbalch first rewrote this last
equation by taking the – log 10 of each term to obtain
pH = pKa + log 10{[base] / [acid]}
This last equation bear Henderson and Hasselbalch
names and is widely used in biochemistry.
Chapter 8 – pH Buffers
pH = pKa + log 10{[base] / [acid]}
It is important to note form the Henderson-Hasselbalch
equation that the pH depends on
1) The pKa of the conjugate acid form of the acid-base
pair
Chapter 8 – pH Buffers
pH = pKa + log 10{[base] / [acid]}
It is important to note form the Henderson-Hasselbalch
equation that the pH depends on
1) The pKa of the conjugate acid form of the acid-base
pair
2) The ratio of Base : Acid
Chapter 8 – pH Buffers
pH = pKa + log 10{[base] / [acid]}
Also note that whenever [Base] = [Acid],
log 10 1.00 = 0, so the pH = pKa of the conjugate acid.
•
This provides us with a handle for selecting a suitable
buffer;
Choose a system whose pKa of the conjugate acid is
close to the desired region that you want to control the
pH.
Chapter 8 – pH Buffers
pH = pKa + log 10{[base] / [acid]}
Farther, note that a 10-fold change in the
ratio of [Base] : [Acid] leads to a change
of 1.00 pH unit in the buffer.
Chapter 8 – pH Buffers
pH = pKa + log 10{[base] / [acid]}
Generally, the pH range that the buffer
will work effectively is
pH = pKa ± 1.00
Problem 1 - Calculate the pH in a solution prepared
by adding 0.200 mol of acetic acid and 0.200 mol of
sodium acetate to one liter.
 Ka = 1.8 X 10-5 ; pKa = 4.74
 [HC2H3O2] = 0.200 mol/L = 0.200 M
 [C2H3O2 -] = 0.200 mol/L = 0.200 M




pH = pKa + log 10{[base] / [acid]}
pH = 4.74 + log 10{0.200M/ 0.200M}
pH = 4.74 + log 10{1.00}
pH = 4.74
Problem 2 - Choose a system (conjugate pair) to
give a buffer whose pH = 7.0
 As mentioned above, the pH of the buffer is roughly equal
to pKa of the weak acid. From Appendix B, pages 546ff
there are several system whose pKa values are close to 7.0 .
 I am going to choose the phosphate buffer with pKa =
7.199.
 The ratio of ([HPO4 2-] / [H2PO4 -]) is calculated from the
Henderson-Hasselbalch expression.
 pH = pKa + log 10{[HPO4 2-] / [H2PO4 -]}
2 7.00 = 7.199 + log 10{[HPO4 ] / [H2PO4 ]}
 log 10{[HPO4 2-] / [H2PO4 -]}= 7.00 – 7.199 = - 0.199
 {[HPO4 2-] / [H2PO4 -]}= 10 -0.199 = 0.632
 This means that any ratio of {[base] / [acid]} = 0.632 will
have a pH of 7.00
Effect of adding strong acid or
base to a buffer
Whenever a strong acid or a strong base is
added to a buffer the following reactions
occur:
• Addition of strong base (OH-)
HB(aq) + OH-(aq)  H2O + B- (aq)
Effect of adding strong acid or
base to a buffer
Whenever a strong acid or a strong base is
added to a buffer the following reactions
occur:
• Addition of strong base (OH-)
HB(aq) + OH-(aq)  H2O + B- (aq)
• Addition of strong acid (H+ or H3O+)
B-(aq) + H+(aq)  HB(aq)
Effect of adding strong acid or
base to a buffer
So long as the system has plenty of HB and
B– to consume the H+ or OH- ions that have
been added there is not a drastic change in
the pH. The actual pH will depend on the
ratio of the base form : acid form as shown in
the Henderson-Hasselbalch equation.
What is the new pH whenever 0.100 mol of HCl is added to 1
liter of the pH 7.0 phosphate buffer chosen earlier if the
[H2PO4-] = 0.800 M.
Earlier we calculate that Base : Acid ratio needed to be
0.632, so if the [acid] = 0.800 M,
the [base] = 0.632 x 0.800M = 0.506M
What is the new pH whenever 0.100 mol of HCl is
added to 1 liter of the pH 7.0 phosphate buffer chosen
earlier if the [H2PO4-] = 0.800 M.
The addition of 0.100 mol of HCl (H+) will cause
H2PO4- to increase by 0.100 mol and the HPO4-2 to
decrease by 0.100 mol; the reaction is
HPO4-2 + H+ → H2PO4-
What is the new pH whenever 0.100 mol of HCl is
added to 1 liter of the pH 7.0 phosphate buffer chosen
earlier if the [H2PO4-] = 0.800 M.
HPO4-2 + H+ → H2PO4The new mol of HPO4-2 = (1.00L)(0.506) – 0.100
= 0.406 mol; since in 1.00 L, [HPO4-2] = 0.406 M
The new mol of H2PO4- = (1.00)(0.800) + 0.100
= 0.900 mol; since in 1.00 L, [H2PO4-] = 0.900M
What is the new pH whenever 0.100 mol of HCl is
added to 1 liter of the pH 7.0 phosphate buffer chosen
earlier if the [H2PO4-] = 0.800 M.
The new pH is found by substituting the new
concentration values into the H-H equation:
pH = pKa + log 10{[base] / [acid]}
pH = 7.199 + log 10{0.406 / 0.900}
pH = 7.199 + (- 0.346) = 6.853
What is the new pH whenever 0.100 mol of HCl is
added to 1 liter of the pH 7.0 phosphate buffer chosen
earlier if the [H2PO4-] = 0.800 M.
pH = 7.199 + (- 0.346) = 6.853
{Note that the addition of strong acid causes the pH of
the buffer to become more acidic (lower pH).
Conversely, the addition of a strong base would cause
the pH of the buffer to become more basic (higher pH)}
Buffer capacity
Buffer capacity () is the number of moles of OH–
or H + that 1.00 Liter of a buffer can absorb without
the pH change exceeding 1 pH unit.
The buffer capacity depends on the concentrations of
the weak acid and its conjugate base.
For the addition of base: nOH- = nHB originally present
nH+ = nB- originally present
In practice, pH starts to change drastically as nHB or
nB→ 0, as is shown in the next slide.
The Buffer capacity
of the 0.500 M
lactic acid/lactate
buffer. Note the
middle of the buffer
range occurring at
pH of = 3.85, the
pKa of this system.
Next time, finish Buffers, Chapter
7, start Chapter 12 – EDTA
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