General Biochemistry MODULE

Biochemistry

1
Notes
GENERAL
BIOCHEMISTRY

1.1 INTRODUCTION
Solutions of chemical reagents are a big part of biochemistry, biological and
chemical based work. For a beginner of experimental procedure making
solutions can also be the most frustrating part. Preparation and handling
solutions are essential part of experimental biochemistry. Thus any of new
science graduates should be clear in preparing reagents, buffers, and accuracy
in pipeting.

The concentration of a dissolved salt in water refers to the amount of salt (solute)
that is dissolved in water (solvent). Solutes are the substance of interest to be
dissolved and the term solvent denotes the material in which the solute is
dissolved.

Solution is a mixture that contains solute and a solvent. Solute can be denoted
as the component of a solution present in the lesser amount and the solvent is
the component of a solution present in the greater amount. Concentration can
be written as the amount of a solute present in a solution per amount of solvent.

OBJECTIVES
After reading this lesson, you will be able to:
z describe the importance of solution preparation in biochemistry
z explain different concentration units
z describe different terms of percent solutions

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Biochemistry
1.2 UNITS OF CONCENTRATION
z There are many ways to express concentrations. Concentration may be
expressed several different ways and some of the more common concentration
units are:
1. Equivalent weight
2. Molarity
Notes 3. Molality
4. Normality
5. Percent solution (weight/weight)
6. Percent solution (weight/volume)
7. Percent solution (volume/volume)

1.2.1 Equivalent Weight

The equivalent weight is determined by dividing the atomic or molecular weight
by the valence. A major use of the concept of equivalents is that one equivalent
of an ion or molecule is chemically equivalent to one equivalent of a different
ion or molecule. The mass of a substance especially in grams is chemically
equivalent to eight grams of oxygen or one gram of hydrogen : the atomic or
molecular weight divided by the valence
Valance could be determined as
1. The absolute value of ion charge
2. The number of H+ or OH– that a species can react with
3. The absolute value of change in charge on a species when undergoing a
chemical reaction.

1.2.1.1 Preparation of NaOH

Solutions of NaOH can be prepared by either dissolving solid NaOH pellets in
water or by diluting a concentrated solution of NaOH. However, the exact
concentration of the solution prepared by these methods cannot be calculated
from the weighed mass or using the dilution equation for two reasons:
1. Solid sodium hydroxide is hygroscopic (“water-loving”). Pellets of NaOH
exposed to air will increase in mass as they become hydrated so the actual
mass of pure NaOH is not accurately known.
2. Sodium hydroxide in solution reacts with carbonic acid and its concentration
decreases over time. The acid is formed when small amounts of CO2 gas
(which is always present in air) dissolves in solution.

ZZX H2O + Na+(aq) + HCO3–(aq)

H2CO3(aq) + NaOH(aq) YZZ

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The water used to make the NaOH solution can be boiled to expel the dissolved Biochemistry
CO2 gas but this time-consuming procedure is often not possible in a short
laboratory period. A stock solution of NaOH can be made in advance with boiled
water but will re-absorb CO2 over a period of time unless stored in airtight
containers. Therefore, if we want to know the exact concentration of a freshly
made NaOH solution, we need to “standardize” it. That is, determine its exact
concentration by titrating it with a known mass of a primary standard acid.
A “primary standard” is a substance that is used to determine the concentration Notes
of a solution. A primary standard should have the following properties: It should
be available in very pure form at reasonable cost and should have a high
equivalent weight to minimize weighing errors. It should be stable at room
temperature, easy to dry, and should not easily absorb water when exposed to
air (hygrophobic).
Potassium hydrogen phthalate (“KHP”) is the primary standard reagent commonly
used to standardize NaOH. It is a monoprotic acid whose formula is KHC8H4O4
and molecular weight is 204.22 g/mol.
KHC8H4O4(aq) + NaOH(aq) ⎯→ H2O + Na+(aq) + K+(aq) + C8H4O4(aq)
The white powdery acid is normally heated at 110°C for one hour to remove any
loosely bound waters of hydration and then cooled in a desiccator before use.
The exact mass (and number of moles of acid) is determined by weighing the
dried acid on an analytical balance. The acid is then dissolved in water and NaOH
is added until an endpoint (the point at which an indicator changes color) is
reached. The phenophthalein indicator used in this experiment is colorless in
acid and pink in base. Therefore, the solution containing KHP will remain
colorless as long as some KHP is still present. Once the last of the KHP has
reacted, the solution will turn pink with one excess drop of base.
The exact concentration of NaOH is calculated by using the stoichiometry from
reaction to convert the number of moles of KHP used to moles of NaOH and
then dividing by the volume of NaOH used to reach the endpoint of the reaction.

1.2.1.2 Equivalent Weight of an Acid

In an acid base titration, the equivalence point is the volume of added base where
the moles of –OH added (from the base) equal the moles of H+ initially present
(from the acid). (i.e) moles of H+ initially present = moles –OH added (at
equivalence point)
To approximate the equivalence point, an indicator with an endpoint close to the
equivalence point is added to the analyte solution. A balanced equation can be
written describing the chemical reaction occurring between the titrant (the base
in this experiment) and analyte (the acid in this experiment) if the identity of
both is known.

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Biochemistry For example, the titration of hydrochloric acid with potassium hydroxide can be
written:
HCl(aq) + KOH(aq) ⎯→ H2O + KCl(aq)
In the case if the identity of the acid is unknown, but the number of acidic
hydrogens (H+) carried by the acid is known, a balanced equation can still be
written.
Notes For example, the titration of a triprotic acid (an acid with 3 H+) with sodium
hydroxide can be written:
H3X(aq) + 3KOH(aq) ⎯→ 3H2O(l) + K3X(aq)
(X : the unknown anion of the acid)
The formula weight of this unknown acid can be calculated by using dimensional
analysis. First, the base’s concentration is used to convert the base’s volume at
the endpoint to moles. Then, multiplying by the mole ratio between acid and base
from the balanced chemical equation allows for the calculation of the moles of
the acid. Now the mass of acid titrated must be divided by the moles of acid
calculated giving a result with the units of g/mol.

1.2.1.3 Equivalent Weight of an Oxidizing Agent

The concept of equivalents and equivalent mass is not restricted to acid-base
reactions alone. Unlike acid-base reactions in redox reactions, the electrons are
the active units (the equivalents) and the equivalent weights are the masses of
oxidizing or reducing agent that deliver or accept 1 mole of electrons. But in
case of acid and base the hydrogen or hydroxide ions plays key role in
determination of equivalent weight.

INTEXT QUESTIONS 1.1

1 The equivalent weight is determined by dividing the atomic or molecular
weight by the ...................
2 The mass of a substance especially in grams is chemically equivalent to
................... grams of oxygen
(a) 2 (b) 5 (c) 8 (d) 10
3 ................... is the primary standard reagent commonly used to standardize
NaOH3 Molarity

1.2.2 Molarity
Molarity is based on the volume of solution containing the solute. Since density
is a temperature dependent property a solution’s volume, and thus its molar

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concentration, changes with temperature. By using the solvent’s mass in place Biochemistry
of the solution’s volume, the resulting concentration becomes independent of
temperature.
Molarity is the common way of referring to concentrations of solutions. The goal
of most basic molarity problems shall be to get the moles from grams by dividing
the molecular weight and then dividing by the total number of liters or by given
the molarity find the number of grams of the solution by multiplying the volume Notes
then the molecular weight. Molarity might give you the density of the solution,
from which you can obtain the mass by multiplying the density by the volume.
Although there are several ways in which the concentration of a solution can be
quantified, molarity is one of the most basic and widely used. Molarity (M) is
defined as the number of moles of solute dissolved in one liter of solution. The
higher the molarity, the more concentrated or strong the solution is. For example,
a 12 M (which is said “twelve molar”) solution of HCl (ie. hydrochloric acid)
is much more concentrated than a 0.10 M solution! The basic formula for
calculating molarity is:
Molarity (M) = moles of solute (mol) per liters of solution (L)
To solve for moles of solvent, we can use algebra to manipulate the above
equation producing the following derived formulas:
moles of solute (mol) = Molarity (M) × liters of solution (L)
In simple terms, the following formula could be used for preparation of molar
solutions for lab solutions preparation
For preparation of molar solution

Molecular weight of the compound (A)

1000
× Required morality (B) × Required volume of solution (C) = D gram
In the above equation for preparation of solution of ‘B’ molarity, ‘D’ grams of
the solute could be dissolved in ‘C’ ml of solvent.

INTEXT QUESTIONS 1.2

1. .................... is based on the volume of solution containing the solute
(a) Normality (b) Molarity (c) Molality (d) Percent
2. Molarity is the common way of referring to .................... of solutions

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Biochemistry 3. .................... is defined as the number of moles of solute dissolved in one

liter of solution
4. The higher the ...................., the more concentration of the solution

1.2.3 Molality
The molal unit is not used nearly as frequently as the molar unit and is used in
Notes thermodynamic calculations where a temperature independent unit of
concentration is needed. A molality is the number of moles of solute dissolved
in one kilogram of solvent. The term molality and molarity should not be
confused. While expressing the Molality it is represented by a small “m,”
whereas molarity is represented by an upper case “M.”
In case of preparation of molar solution except water all other solvent must be
weighed. The water is exempted from weighing because; one liter of water has
a specific gravity of 1.0 and weighs one kilogram. So one can measure out one
liter of water and the solute could be directly added to it. But other solvents might
have a specific gravity greater than or less than one. Therefore, one liter of any
solvent other than water is not likely to occupy a liter of space.
For example to make a one molal aqueous (water) solution of sodium chloride
(NaCl), measure out one kilogram of water and add one mole of the solute, NaCl
to it. The atomic weight of sodium is 23 and the atomic weight of chlorine is
35. Therefore the formula weight for NaCl is 58. So 58 grams of NaCl could
be dissolved in 1kg water for preparation of 1 molal solution of NaCl.

INTEXT QUESTIONS 1.3

1. .................. is the unit used in thermodynamic calculations where a
temperature independent unit of concentration is needed.
2. A molality is the number of moles of solute dissolved in ..................
kilogram of solvent
(a) 3 (b) 2 (c) 1 (d) 6
3. Molality it is represented by a ..................
(a) Small ‘m’ (b) Capital ‘M’ (c) Small ‘n’ (d) Capital ‘N’

1.2.4 Normality
The concentration of a solution could also be expressed in terms of Normality.
It is based on an alternate chemical unit of mass called the equivalent weight.
The normality of a solution is the concentration expressed as the number of

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equivalent weights (equivalents) of solute per liter of solution. In a chemical Biochemistry
mixture 1 normal (1 N) solution contains 1 equivalent weight of solute per liter
of solution. Since normality simplifies the calculations required for chemical
concentration, it is widely used in analytical chemistry.
Every substance may be assigned an equivalent weight. The equivalent weight
may be equal to the formula weight (molecular weight, mole weight) of the
substance or equal to an integral fraction of the formula weight (i.e., molecular Notes
weight divided by 2, 3, 4, and so on).
The above phenomenon could be better explained with the following example
to gain an understanding of the meaning of equivalent weight:
HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O
1 mole 1 mole
(36.5 grams) (40.0 grams)
H2SO4(aq) + 2NaOH(aq) ⎯→ Na2SO4(aq) + 2H2O
1 mole 1 mole
(98.1 grams) (80.0 grams)
In the above chemical reaction 1 mole of hydrochloric acid (HCl) reacts with
1 mole of sodium hydroxide (NaOH) and 1 mole of sulfuric acid (H2SO4) reacts
with 2 moles of NaOH. If you made 1 molar solutions of these substances, 1
liter of 1 M HCl will react with 1 liter of 1 M NaOH and 1 liter of 1 M H2SO4
will react with 2 liters of 1 M NaOH. Therefore, H2SO4 has twice the chemical
capacity of HCl when reacting with NaOH. The equivalent weight of HCl is
equal to its molecular weight, but that of H2SO4 is ½ its molecular weight.
Expressions for normality are shown below. Notice the similarity to molar
solution definition.

Number of equivalents of solute Equivalents

Normality (N) = =
1 liter of solution liter
grams of solute
where Number of equivalents of solute =
equivalent weight of solute

grams of solute grams

finally N = =
eq wt solute × L solution eq wt × L
So, 1 liter of solution containing 36.5 grams of HCl would be 1 N, and 1 liter
of solution containing 49.0 grams of H2SO4 would also be 1 N. A solution
containing 98.1 grams of H2SO4 (1 mole) per liter would be 2 N.

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Biochemistry

INTEXT QUESTIONS 1.4

1. ...................... based on an alternate chemical unit of mass called the
equivalent weight
(a) Normality (b) Molarity (c) Percent (d) Molality
Notes 2. One mole of ...................... reacts with 1 mole of sodium hydroxide (NaOH)
3. One mole of sulfuric acid (H2SO4) reacts with ...................... moles of
NaOH
4. The equivalent weight of HCl is equal to its ......................

1.2.5 Percentage Solutions

The percentage solution could be expressed in terms of weight percent (% w/
w), volume percent (% v/v) and weight-to-volume percent (% w/v) units of
solute present in 100 units of solution. For example a solution of 1.5% w/v
NH4NO3, contains 1.5 gram of NH4NO3 in 100 mL of solution.

1.2.5.1 Percent by weight (% w/w)

In case of preparing a solution based on percentage by weight, one would simply
determine what percentage was required (for example, a 20% by weight aqueous
solution of sodium chloride) and the total quantity to be prepared.
If the total quantity needed is 1 kg, then it would simply be a matter of calculating
20% of 1 kg which, of course is:
20 /100 * 1000 g/kg = 200 g NaCl/kg.
Thus finally to bring the total quantity to 1 kg, it would be necessary to add 800g
water.

1.2.5.2 Percent by volume (% w/v)

Preparation of solutions based on percent by volume it requires the calculation
same as for percent by weight, except that calculations are based on volume. In
simple terms one should plan that what percentage was desired (for example,
a 20% by volume aqueous solution of sodium chloride) and the total quantity
to be prepared in terms of volume.
For example if 20 % is to be prepared for a total quantity of 1 liter, then it would
simply be a matter of calculating 20% of NaCl in 1 liter, the formula can be
written as:
20/100 * 1000 ml/l = 200 g NaCl/l.

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1.2.5.3 Percent by volume (% v/v) Biochemistry

Volume percent or volume/volume percent most often is used when preparing

solutions of liquids. This is typically only used for mixtures of liquids. Volume
percent is relative to volume of solution, not volume of solvent. The advantage
of volume/volume units is that gaseous concentrations reported in these units
do not change as a gas is compressed or expanded.
For example, 70% v/v rubbing alcohol may be prepared by taking 700 ml of Notes
isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution
(which will not be 300 ml).
Table 1.1 Common Units for Reporting Concentration
Sl.No Name Units Symbol
1. Molarity moles solute M
liters solution
2. Normality equivalents solute N
liters solution
3. Molality moles solute m
kilograms solvent
4. Weight percent grams solute % w/w
100 grams solution
5. Volume percent mL solute % v/v
100 mL solution
6. weight-to-volume percent grams solute % w/v
100 mL solution

INTEXT QUESTIONS 1.5

1. Percent by weight could be expressed as ................
2. For preparation of 20 % NaCl by (w/v) ................ grams of NaCl is to be
dissolved in 1 L of water.
(a) 200 (b) 100 (c) 400 (d) 50
3. volume/volume percent most often is used when preparing solutions of
................
4. Match the following:
(i) Molarity (a) N
(ii) Normality (b) m
(iii) Molality (c) M

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Biochemistry

WHAT HAVE YOU LEARNT

z Preparation solutions of chemical reagents are very important task for the
beginners of biochemistry.
z Concentration of a biological solution could be expressed in terms of
equivalent weight, normality, molarity, molality, percent solution (weight/
Notes
weight, weight/ volume, volume /volume).
z The equivalent weight is determined by dividing the atomic or molecular
weight by the valence.
z Molarity could be simply termed as “M” and are units of moles of solute
per liter of solution.
z Molality is units of moles of solute per kilogram of solution and is termed
as “m”. While the normality is termed as “N” and are units of equivalent
of solute per liter of solution.
z The percentage solution could be expressed in terms of weight percent (%
w/w), volume percent (% v/v) and weight-to-volume percent (% w/v) units
of solute present in 100 units of solution.

TERMINAL QUESTIONS
1. Write a brief note about the importance of solution preparation in
biochemistry.
2. Write a note on Equivalent weight.
3. Describe about molarity and molality.
4. Write a short note on normality.
5. Write about different terms of percent solutions.
6. Tabulate the units and symbols of different modes of concentrations.

ANSWERS TO INTEXT QUESTIONS

1.1
1. Valence
2. 8
3. Potassium hydrogen phthalate

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1.2 Biochemistry

1. Molarity
2. Concentrations
3. Molarity
4. Molarity
Notes
1.3
1. Molality
2. 1
3. Small ‘n’

1.4
1. Normality
2. Hydrochloric acid
3. Two
4. Molecular weight

1.5
1. (% w/w)
2. 200
3. Liquids
4. Match the following:
(i) (b)
(ii) (a)
(iii) (c)

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