Biochemistry 17/9
SHEET FOR LECTURE #3
We ended our last lecture with the statement that scientists are lazy, they tend to
transform big numbers into small ones. Thus they started using Pka
As you know:
PKa = -log(Ka)
Q: if the PKa of formic acid is 3.8 and acetic acid is 4.8 which one is more acidic?
Formic acid is more acidic.
because the less Pka is the more it's acidic. Pka is diversely proportional to the
concentration of the H+ thus with the acidity
Henderson and Hasselbalch have reached to an equation that explains the
relationship between PKa with PH which is:
[A-] is the concentration of the conjugate base which is capable of accepting a proton
[HA] is the concentration of the acid which donate a proton
Hint:
When the acidic acid lose a proton we transform the "ic" into " ate", for example
Acid
acetic
formic
lactic
CH3COOH
HCOOH
CH3CH(OH)COOH
Conjugate base
Acetate
CH3COOformate
HCOOLactate
CH3CH(OH)COO-
From Henderson and Hasselbalch equation we can redefine PKa that:
PKa is the PH when the concentration of the conjugate base [A-] equals to the
concentration of
[HA]. *so that is equals to 1 so the log will equals to
zero*
In another meaning it's when %50 of the acid dissociate into conjugate base .
So when we define for example the PKa of acetic acid 4.8 : it's when %50 of it is in
the acidic form and the other %50 is in the conjugate base form the PH of the acid
will be equaled to 4.8.
Buffer
What's a buffer? it means in English a region or a zone of no action
In Biochemistry it means a solution that resist changes in PH
For example if we added water to HCl the PH will immediately drop to 1 *roughly*
Also water to NaOH the PH will increase to 13 *roughly*
But the acetic acid PH is 4.8 if we added HCl to it ,it may drop to 4.7 maybe.
Also if we tried in the other way around by adding NaOH to acetic acid the pH will
increase from 4.8 to 4.9.
As you see in both situations the change in the pH was minor negligible.
So HOW DOES IT WORK ?
It's based in the principle that solutions are in equilibrium balanced.
So there is a constant amount of products and a constant amount of reactants. Thus
Ka is constanst.
If the reactants increase the products will increase and vise versa. Because Ka
remain constant.
So based on that principle if we had acetate as a buffer which exist in pH 4.8 and
added HCl to it which dissociate immediately completely into H+ and Cl- as showing:
CH3COOH  H+ + CH3COOHCl  H+ + ClUpon this dissociation the number of protons will increase and the equation would be
imbalanced so the reaction would move towards the reactants direction to get rid of
the access protons be removed from the solution.
In order to achieve that the conjugate base "which in this case is CH3COO-" would
react with the access protons that came from the added acid "HCl" to produce (the
reactant) the acid "CH3COOH". As a result we will maintain the equilibrium and the
balance and more importantly the concentration of H+. Because at the end the pH
should remain constant.
BUT what would happen if we added more HCl?
Of course the concentration of the excess H+ will increase that will react with the
conjugate base. And this reaction will result more acid and less conjugate base while
the maintenance of the concentration of H+ thus the value of pH of the solution.
What if we kept on adding more and more HCl? What will happen to the pH ?
The pH will drop suddenly. Because the conjugate base has been consumed
completely, there is no more left to react with the access protons. So the
concentration of the H+ will increase and thus the pH will decrease
Lets go the other way around
What if we added NaOH to acetic acid?
NaOH is a strong base which dissociate completely into the conjugate acid Na+ and
OH- as following:
NaOH Na+ + OHCH3COOH  CH3COO- + H+
The added OH- will react with H+ which already exist in the soloution producing
water molecules thus decreasing the concentration of H+ and imbalance the system:
OH- + H+  H2O
So the system will try to return to its balance point and move it towards the products
by the dissociation of the acid ( CH3COOH ) into more products (conjugate base and
protons) to compensate the loss of the protons and maintaining the equilibrium.
So at the end we'll have more conjugate base, less acid and constant concentration
of protons and the pH stays almost the same.
And if we added more NaOH the acid will dissociate more.
What if we added more and more of NaOH to the solution?
The acid will continue dissociating until there is no more left of the acidic form of
acetic acid to dissociate and compensate the loss of protons so the pH goes up really
high.
Titration:
Let's look at the work of a buffer in terms of a titration experiment:
We began the experiment when pH equals to 4.8. at that point pH equals to pKa
which means it's the pH of the acid when half of it in the acidic form and half of it in
the conjugate base form. And this point is called the midpoint.
At the left side of the graph we added more and more HCl when the pH was still
almost constant until we reach a point when anymore HCl is added the pH goes
down dramatically.
Let's say we added NaOH thee pH goes up slowly until we reach a point goes up
really high.
At the red lighted point all the acid has dissociated into conjugate base and at the
blue lighted point all acid is in its acidic form because all the conjugate base has
been consumed.
Buffering capacity
Any buffer has a range works in it, this range is called buffering capacity a range
(region) in which a buffer acts in. and it's equals to pKa (plus minus) +-1.
For example:
1. the pKa of acetic acid is 4.8 what's its buffering capacity? (3.8-5.8)
2. the pKa of formic acid is 3.8 what's its buffering capacity? (2.8-4.8)
Q: at the end of a buffering capacity calculate without using calculator the ratio of the
conjugate base to acid knowing that the pKa is 4.8 at the point showing on the
graph?
The answer is 10 to 1 meaning that when we have 10 molecules of the conjugate
base we have 1 molecule of the acid.
How?
by Henderson and Hasselbalch equation we will find that:
Q: can I use formic or acetic acid buffer to maintain pH at 4.2 ?
Yes
Why? Because 4.2 falls within the buffering capacity of both acetic and formic acid.
Now if I wanted to keep the pH of the solution at 3.9 would I use the acetate buffer?
No, because it's at the end of the buffering capacity when any little change in the
concentration would change the pH dramatically. So it's better to choose a buffer
where the wanted pH falls in the middle of the box of the buffering capacity.
Q: Predict without using the calculation the value of pH when the
is 5 to 1.
ratio
It logically should be 5.3 depending on this graph:
Because logically the
points
ratio 5 to 1 will fall between the blue and red
meaning at the middle of the distace between the pH (4.8-5.8)
preparation of acetate buffer what should we do ? a combination of the acid and its
conjugated base Sodium acetate and so on with the other acids.
Acid
Conjugate base
CH3COOH
CH3COONa
(NaCH3COO)
H3PO4
NaH2PO4
H2PO4- (or NaH2PO4)
Na2HPO4
H2CO3
NaHCO3
How do we choose a buffer?
Based on the buffer pKa according to the pH I want to maintain to fall within the
buffering capacity of the solution.
For example:
1. if we wanted the solution to be alkaline we should use ammonia as our buffer
which pKa is around 9
2. if we wanted to use a neutral the pH to be round 7 we use H2PO4- because
its pka is around 7
3. if we wanted the solution to be acidic we use acetic acid as our buffer
because it hase a low pKa equals to 4.8
*Hint: do not memorize the numbers just understand the concept.
So I choose the buffer based on its pKa that it has to fall within its buffering
capacity.
•
Q:Problems and solutions A solution of 0.1 M acetic acid and 0.2 M acetate
ion. The pKa of acetic acid is 4.8. Hence, the pH of the solution is given by
•
•
Similarly, the pH of an acid can be calculated, just plug in the numbers.
Exercises:
•
What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 104)
The answer is 3.45 simply because the concentration of the acid equals to the
conjugate base so pH equals to pKa
Or we can solve it directly by Henderson and Hasselbalch equation:
•
What is the pH of a solution containing 0.1M HF and 0.1M NaF, when 0.02M
NaOH is added to the solution?
When we add OH- to the solution we subtract it's concentration from the acid and
add it to the conjugated base so:
[HF] = 0.08 , [F-] = 0.12
And Henderson and Hasselbalch equation:
Equivalent point: Neutralization to all the acidic form or consuming of all the
conjugated base.
Exercise:
What is the concentration of 5 ml of acetic acid knowing that 44.5 ml of 0.1 N of
NaOH are needed to reach the end of the titration of acetic acid? Also, calculate
the normality of acetic acid
Solution
M NaOH = N/n = 0.1/1 = 0.1 M
MV n = MV n
0.1 x 0.0445 x 1 = M x 0.005 x 1
M = 0.89 M
N = M n = 0.89 N