Lec 2 - UMDNJ

Lecture 2
Dr. Kumar
Aid-base balance-Salivary Buffering
Buffer solutions
Before you start it would be helpful to…
know that weak acids and bases are only partly
ionised in solution
• be able to calculate pH from hydrogen ion
• be able to construct an equation for the
dissociation constant of a weak acid
Buffer solutions - uses
“Solutions which resist changes in pH when
small quantities of acid or alkali are added.”
Biological Uses
In biological systems (saliva, stomach, and blood) it is essential that
the pH stays ‘constant’ in order for any metabolic processes to work
e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness
and coma
Most enzymes work best at particular pH values.
Other Uses Many household and cosmetic products need to control their
pH values.
Buffer solutions counteract the alkalinity of the soap and
prevent irritation
Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria
Washing powder, eye drops, fizzy lemonade
Buffer solutions
“Solutions which resist changes in pH when
small quantities of acid or alkali are added.”
Acidic Buffer (pH < 7)
potassium salt
made from
Alkaline Buffer (pH > 7) made from a
like ammonium chloride
a weak acid + its sodium or
weak base
Buffering biological systems (e.g. in blood)
its salt
Acid-Base Concepts and Buffer Action
Strong: Tendency to give up a proton readily---HCl
HCl + H2O Cl-- + H3O+
Weak: Weak tendency to give up a proton---Acetic Acid
CH3COO-- + H3O+
– Strong—great tendency to accept a proton--- OH ion
– Weak--- Poor tendency to accept a proton--CH3-COO- anion
Dissociation of a Weak Acid
(Weak Acid)
(conjugate Base)
Ka (acid dissociation constant) = [H+] [A--]
log Ka = log [H+] + log [A--]/[HA]
--log [H+] = --log Ka + log [A--]/[HA]
pH = pKa + log [conjugate base]/[acid]
Henderson- Hasselbalch equation
pH = -log[H+] and pKa = --log Ka
When [A-]= [HA]
pH = pKa + log [HA]/[HA] or pH = pKa + log 1
So pH = pKa
Significance of pH and pKa
pH = -log [H+]
If [H+] = 1 molar (1M) then pH = - log [1] = 0
If [H+] = 10-7 M then pH = -log [10-7]= -(-7)log10
= +7
When [H+] concentration is high pH is a low number
When [H+] concentration is low pH is a higher
pH scale is logarithmic and is usually given from 0-14
Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol/l and [A¯] of 0.1 mol/l.
Ka =
[H+(aq)] [A¯(aq)]
[H+(aq)] = [HA(aq)] x Ka
from information given
[A¯] = 0.1 mol/l
[HA] = 0.1 mol/l
If the Ka of the weak acid HA is 2 x 10-4 mol/l
[H+(aq)] = 0.1 x 2 x 10-4
pH = - log10 [H+(aq)]
= 2 x 10-4 mol/l
= 3.699
Calculating the pH of an acidic buffer solution
Calculate the pH of the solution formed when 500 mlof 0.1 mol /lof weak acid HX is
mixed with 500 ml of a 0.2 mol/l solution of its salt NaX. Ka = 4 x 10-5 mol/l
[H+(aq)] [X¯(aq)]
[H+(aq)] =
[HX(aq)] Ka
The solutions have been mixed; the volume is now 1 litre
= 0.05 mol /l
[X¯] = 0.10 mol/l
[H+(aq)] = 0.05 x 4 x 10-5
pH =
- log10 [H+(aq)]
2 x 10-5 mol/l
Dissociation of water
Water is a weak acid and a weak base
H+ + OHKw = [H+][OH-]
H2 O
Kw(H2O) = 10-14 = [H+][OH-] and pH + pOH=14
Since [H+] = [OH-]
[H+]2 = 10-14 and [H+] = 10-7
So for water, pH = 7
and pOH = 7
Titration of a weak acid
Buffering Capacity of Acid-Conjugate base
A buffer is a mixture of weak acid and its conjugate base
and its pH should not change appreciably when small
amount of acid or base is added
(Lactic acid; pKa=4.5)
At pH=4.5 which is equal to pKa, [lactate]=[lactic acid]
What will be the change in pH if we add 0.01 M base (OH) to a buffer
which contains 0.1 M each of acid and conjugate base?
pH = 4.5 + log 0.11/0.09 = 4.5 + log 1.22 = 4.5 + .086 = 4.586
What will be the change in pH if 0.01 M [H+] were added to the system?
pH =4.5 + log 0.09/0.11 = 4.5 + log 0.818 = 4.5-0.087 = 4.413
So the buffering capacity of a weak acid-conjugate base system is very
good near the pKa of the acid
Best buffering system is at +/- one pH unit above or below pKa of a weak
Salivary Buffering-Bicarbonate and pH
CO2 + H2O
HCO-3 + H +
In the oral cavity salivary conc. Of H2CO3 remains essentially
constant at 1.3 mM but pH and bicarbonate
concentrations do change:
1.When acid is produced within the dental plaque, the
reaction is driven to the left and mouth being an open
system, CO2 escapes and acid is neutralized and thus saliva
protects teeth from decay assuming there is enough
bicarbonate present in the plaque. Conversion of H2CO3 to
CO2 and water requires carbonic anhydrase VI which is
produced by acinar cells of parotid gland and appears to
form part the tooth pellicle.
Importance of salivary pH
• The normal salivary pH is ~6.3 because of
salivary bicarbonate.
• H2CO3
HCO3-1 + H+
• pH = pKa + log [HCO-13]/[H2CO3]
• After meals salivary concentration of HCO3 is
increased from ~2mM to 30 mM. Using these
values pH at 2 mM HCO-13 will be ~6.29 and at
30 mM HCO3, it should be ~7.46.
Variation of pH with bicarbonate concentration
Learning Objectives Acids and Bases
• Know the definitions of strong and weak acids
and bases?
• Know how to calculate pH and pKa.
• Know the use of Henderson-Hasselbalch
• Know the conditions for good buffering action;
which are good buffers?
• Understand salivary buffering and oral health
• Do the problems at the end of this lecture.
Problem Set
Please do these calculations
A pH of 2.5 means that the [H+] is---------------•
For an acetate buffer at pH 4.9, what will be the concentration of
acetate ion if the concentration of acetic acid is 0.06 M (pKa of
acetic acid is 4.8).
Calculate the pKa of lactic acid given that the pH is 5.28 and the
concentrations of lactic acid and lactate are 0.01 and 0.06 M
3.16 x10 –3 M
0.076 M

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