2. BUFFER SOLUTION
► The solutions which resist the change in pH on dilution or with the
addition of small amounts of acid or alkali are called buffer
solutions.
3. BUFFER CAPACITY
► Buffer capacity (β) is defined as the moles of an acid or base necessary to
change the pH of a solution by 1, divided by the pH change and the volume of
buffer in liter
► it is a unitless number.
► A buffer resists changes in pH due to the addition of an acid or base though
consumption of the buffer.
4. BUFFER CAPACITY
► Buffer capacity quantifies the ability of a solution to resist changes
in pH by either absorbing or desorbing H+ and OH- ions.
► When an acid or base is added to a buffer system, the effect on
pH change can be large or small, depending on both the initial pH
and the capacity of the buffer to resist change in pH.
5. BUFFER CAPACITY
► The number of millimoles of acid or base to be added to a litre of
buffer solution to change the pH by one unit is the Buffer capacity of
the buffer.
Β = millimoles /(ΔpH)
6. Acidic Buffers
► These solutions are used to maintain acidic environments.
► Acid buffer has acidic pH and is prepared by mixing a weak acid and its salt with a
strong base.
► An aqueous solution of an equal concentration of acetic acid and sodium acetate has
a pH of 4.74.
► pH of these solutions is below seven (0-7)
► An example of an acidic buffer solution is a mixture of sodium acetate and acetic acid
(pH = 4.75).
7. Alkaline Buffers
► These buffer solutions are used to maintain basic conditions. Basic buffer has a basic pH
and is prepared by mixing a weak base and its salt with strong acid. The aqueous solution
of an equal concentration of ammonium hydroxide and ammonium chloride has a pH of
9.25.
► The pH of these solutions is above seven (7-14)
► An example of an alkaline buffer solution is a mixture of ammonium hydroxide and
ammonium chloride (pH = 9.25).
8. MECHANISM OF BUFFERING
► Mechanism of Buffering Action
► In solution, the salt is completely ionized and the weak acid is partly ionized.
► CH3
COONa ⇌ Na+
+ CH3
COO–
► CH3
COOH ⇌ H+
+ CH3
COO–
9. ► On Addition of Acid and Base
1. On addition of acid, the released protons of acid will be removed
by the acetate ions to form an acetic acid molecule.
H+
+ CH3
COO–
(from added acid) ⇌ CH3
COOH
(from buffer solution)
2. On addition of the base, the hydroxide released by the base will
be removed by the hydrogen ions to form water.
HO–
+ H+
(from added base) ⇌ H2
O
(from buffer solution)
10. PREPARATION OF BUFFER SOLUTION
► If the dissociation constant of the acid (pKa
) and of the base (pKb
) are
known, a buffer solution can be prepared by controlling the salt-acid or
the salt-base ratio.
► These solutions are prepared by mixing the weak bases with their
corresponding conjugate acids, or by mixing weak acids with their
corresponding conjugate bases.
► An example of this method of preparing buffer solutions can be given by
the preparation of a phosphate buffer by mixing HPO4
2-
(monohydrogen
phosphate) and H2
PO4-
(Dihydrogen phosphate)
► The pH maintained by this solution is 7.4.
11. Handerson-Hasselbalch Equation
PREPARATION OF ACID BUFFER
► Consider an acid buffer solution, containing a weak acid (HA) and its salt (KA) with
a strong base(KOH). Weak acid HA ionizes, and the equilibrium can be written as-
► HA + H2
O ⇋ H+
+ A−
► Acid dissociation constant = Ka = [H+
] [A–
]/HA
► Taking, negative log of RHS and LHS:
pH = pKa + log [Salt] / [Acid]
12. pH of acid buffer = pKa + ([salt]/[acid])
The equation is the Henderson-Hasselbalch equation, popularly
known as the Henderson equation.
13. PREPARATION OF BASE BUFFER
► Consider base buffer solution, containing a weak base (B) and its salt (BA) with
strong acid.
► pOH, can be derived as above,
► pOH of a basic buffer = pKb + log ([salt]/[base])
►
14. SIGNIFICANCE OF HENDERSON
EQUATION
► Handerson Equation can be used to:
► Calculate the pH of the buffer prepared from a mixture of the salt and weak
acid/base.
► Calculate the pKa value.
► Prepare buffer solution of needed pH.
16. USES OF BUFFER SOLUTIONS
► There exists a few alternate names that are used to refer buffer solutions, such as
pH buffers or hydrogen ion buffers.
► An example of the use of buffers in pH regulation is the use of bicarbonate and
carbonic acid buffer system in order to regulate the pH of animal blood.
► Buffer solutions are also used to maintain an optimum pH for enzyme activity in
many organisms.
► The absence of these buffers may lead to the slowing of the enzyme action, loss
in enzyme properties, or even denature of the enzymes. This denaturation
process can even permanently deactivate the catalytic action of the enzymes.
17. Density, Specific Gravity, and Specific
Volume
Density (d) is mass per unit volume of a substance. It is usually expressed as
grams per cubic centimeter (g/cc).
the gram is defined as the mass of 1 cc of water at 4o
C, the density of water is
1 g/cc.
The United States Pharmacopeia states that 1 mL may be used as the
equivalent of 1 cc, the density of water may be expressed as 1 g/mL
18. ► Thus, if 10 mL of sulfuric acid weighs 18 g, its density is:
Density may be calculated by dividing mass by volume, that
is:
19. Specific gravity
► Specific gravity (sp gr)
► Specific gravity is a ratio, expressed decimally, of the weight of a substance to
the weight of an equal volume of a substance chosen as a standard, both
substances are needed to be at the same temperature
► Water is used as the standard for the specific gravities of liquids and solids; the
most useful standard for gases is hydrogen.
► Specific gravity may be calculated by dividing the weight of a given substance
by the weight of an equal volume of water, that is:
20. ► Thus, if 10 mL of sulfuric acid weighs 18 g, and 10 mL of water, under similar conditions, weighs
10 g, the specific gravity of the acid is:
► Substances that have a specific gravity less than 1 are lighter than water.
► Substances that have a specific gravity greater than 1 are heavier than water
21. * If 54.96 mL of oil weighs 52.78 g, what is the specific gravity of the oil?
54.96 mL of water weighs 54.96 g
Specific gravity of oil= 52.78 (g)/ 54.96 (g)
= 0.9603, answer.
* If a pint of a certain liquid weighs 601 g, what is the specific gravity of the liquid?
1 pint =473 mL.
473 mL of water weighs 473 g
Specific gravity of liquid= 601 (g) /473 (g)
= 1.27, answer.
22. Pycnometer or Specific Gravity Bottle
► A pycnometer is a special glass bottle used to
determine specific gravity.
► Pycnometers are generally available for laboratory
use in volumes ranging from 1 mL to 50 mL.
► Pycnometers have fitted glass stoppers with a
capillary opening to allow trapped air and excess fluid
to escape.
► Some pycnometers have thermometers affixed,
because temperature is a factor in specific gravity
determinations.
23. Example:
A 50 mL pycnometer is found to weigh 120 g when empty, 171 g when filled with water, and
160 g when filled with an unknown liquid. Calculate the specific gravity of the unknown
liquid.
Weight of water: 171 g - 120 g = 51 g
Weight of unknown liquid: 160 g - 120 g = 40 g
Specific gravity= Weight of substance/ Weight of equal volume of water
Specific gravity of unknown liquid = 40 (g)/ 51 (g)
= 0.78, answer.
24. Grams = Milliliters × Specific gravity
Grams (other liquid)= Grams (of equal volume of water) × Specific gravity (other liquid)
Example:
What is the weight, in grams, of 2 fl. oz. of a liquid having a specific gravity of 1.118?
2 × 29.57 mL =59.14 mL
59.14 mL of water weigh 59.14 g
59.14 g × 1.118 =66.12 g, answer.
What is the cost of 1000 mL of glycerin, specific gravity 1.25, bought at $54.25 per pound?
1000 mL of water weigh 1000 g
Weight of 1000 mL of glycerin =1000 g × 1.25 =1250 g
1 lb =454 g
454 (g) /1250 (g) = ($) 54.25 ($) /x
x = $149.37, answer
