Solutions preparation

Solution Preparation (2)
IntroductionIntroduction
SolutionSolution
A solution is a homogeneous mixture composed of two or more substances. In
such a mixture, a solute is dissolved in another substance, known as a solvent.
SoluteSolute
The substance which dissolves in a solution
SolventSolvent
The substance which dissolves another to form a solution
SaturationSaturation
Saturation is the point at which a solution of a substance can dissolve no more
of that substance and additional amounts of it will appear as a precipitate.
SupersaturationSupersaturation

It refers to a solution that contains more of the dissolved material than could be
dissolved by the solvent under normal circumstances.
Types of solutionsTypes of solutions
• Percentage solution
• Molar solution
• Normal solution
Percentage solutionPercentage solution
•• Weight/ Weight solution % (w/w)Weight/ Weight solution % (w/w)
This type of solution is rarely if ever prepared in the laboratory since it is easier
to measure volumes of liquids rather than weigh the liquid on an analytical
balance.
This type of percent solution is usually expressed as (w/w), where "w" denotes
weight (usually grams) in both cases.
Example:
An example of a correct designation for this type of solution is as follows: 10
g/100 g (w/w), which indicates that there are 10 grams of solute for every 100
grams total
•• Weight/volume solution % (w/v)Weight/volume solution % (w/v)
Weight-volume percentage, (sometimes referred to as mass-volume percentage
and often abbreviated as % m/v or % w/v) describes the mass of the solute in g
per 100 ml of the resulting solution.
Example:

•• Preparation of 10% (W/V) NaCl solutionPreparation of 10% (W/V) NaCl solution
CalculationsCalculations:
so10% (W/V) NaCl solution has 10 grams of sodium chloride dissolved in 100 ml
of solution.
Procedure:Procedure:
• Weigh 10g of sodium chloride.
• Pour it into a graduated cylinder containing about 80ml of water.
• Once the sodium chloride has dissolved completely add water to bring the
volume up to the final 100 ml.
Note:
Do not simply measure 100ml of water and add 10g of sodium chloride. This
will introduce error because adding the solid will change the final volume of
the solution and throw off the final percentage.
Wt of solute (g) =
C (required Conc.) X V of solvent (ml)
100
g/L unite
A gram per liter (g/L) is a unit of measurement of
concentration which shows how many grams of a
certain substance are present in one litre of liquid.
1 g/L = 1000 mg/L = 100 mg/dl = 1000000 µg/L =
0.1% (w/v) solution
ppm unite
ppm or part per million is a unit of concentration
often used and denotes one part per 1,000,000
parts, one part in 106
, and a value of 1 × 10–6
.
1 mg/L = 1 ppm = 1 µg/ml
g/L unite
A gram per liter (g/L) is a unit of measurement of
concentration which shows how many grams of a
certain substance are present in one litre of liquid.
1 g/L = 1000 mg/L = 100 mg/dl = 1000000 µg/L =
0.1% (w/v) solution
ppm unite
ppm or part per million is a unit of concentration
often used and denotes one part per 1,000,000
parts, one part in 106
, and a value of 1 × 10–6
.
1 mg/L = 1 ppm = 1 µg/ml

•• Volume /volume solution % (v/v)Volume /volume solution % (v/v)
Volume-volume percentage (abbreviated as % v/v) describes the volume of the
solute in ml per 100 ml of the resulting solution.
Example:
•• Preparation of 30% (V/V) sulfuric acidPreparation of 30% (V/V) sulfuric acid
Calculations:Calculations:
So 30% (V/V) sulfuric acid has 30 ml of sulfuric acid dissolved in 70 ml of water.
• Calculate the required volume of solute
• Subtract the volume of solute from the total solution volume
• Dissolve 30 ml sulfuric acid in a 70 ml of water to bring final volume of
solution up to 100ml.
Note:
V1 of solute (ml) =
C2 (required Conc.) X V2 of solvent (ml)
C1 (original solute Conc.)

When you mix concentrated
sulfuric acid and water, you must
add acid (AA) to water in ice
bath.
Why?
Sulfuric acid reacts very
vigorously with water, in a highly
exothermic reaction. Water is
less dense than sulfuric acid, so
if you pour water on the acid, the
reaction occurs on top of the
liquid. If you add the acid to the
water, it sinks and any wild
reactions have to get through the
water.
Molar solutionMolar solution
Molecular weightMolecular weight
It is the sum of the atomic weights of all the atoms in a molecule. Also called
formula weight
MoleMole
A mole is also called gram-molecular weight and defined as number of particles
whose total mass in grams was numerically equivalent to the molecular weight
or it is the amount of substance containing the Avogadro number (6.022 × 1023
)
MolarityMolarity
Molarity or molar concentration denotes the number of moles of a given
substance per liter of solution.
Molar solutionMolar solution

It is a solution that contains 1 mole of solute in each liter of solution.
Units of molarityUnits of molarity
The units for molar concentration are mol/L. These units are often denoted by a
capital letter M (pronounced "molar").
1 mol/l = 1000 mmol/l = 1000000 µmol/l = 1000000000 nmol/l
1 mmol/ml = 1 mol/l
1 µmol/ml = 1 mmol/l
Name Abbreviation Concentration
Millimolar mM 10-3
molar
Micromolar µM 10-6
molar
Nanomolar nM 10-
9 molar
Picomolar pM 10-12
molar
Femtomolar fM 10-15
molar
•• Molarity of solid soluteMolarity of solid solute
Example:
•• Preparation of 1M NaCl solution in 1000 mlPreparation of 1M NaCl solution in 1000 ml
 M: the required molarity
 M.wt: molecular weight of the solute
 V: total volume in liters
Wt of solute (g( = M (mole/l) X M.wt (g/mole) X V (L)

Dissolve 58.5 g of NaCl in a 1000 ml (1 liter) of water to prepare 1M NaCl
solution.
•• Molarity of liquid soluteMolarity of liquid solute
To prepare 1M solutions make the following steps:
• Calculate the molecular weight of the solute
• Calculate the molarity of the solute using the following formula:
• Calculate the volume of the solute needed to prepare 1M solution using the
following formula:
• Mix both volumes of solute and solvent to reach the total required volume
Example1:
•• Preparation of 1M HCL solution in 1000 mlPreparation of 1M HCL solution in 1000 ml
% percent Density M.wt
32 1.18 36.5
HCL molarity = 10.345
Volume required from HCL = 96.66 ml
Complete to 1 liter with 903.33 ml distilled water
Example2:
•• Preparation of 1M HPreparation of 1M H22SOSO44 solution in 1000 mlsolution in 1000 ml
98 1.83 98
H2SO4 molarity = 18.3
Molarity (M( =
% X density X 10
M.wt
V1 of solute(ml( =
M2 (required molarity) X V2 (required volume) (ml)
M1 (original molarity)

Volume required from H2SO4 = 54.64 ml
Example3:
•• Preparation of 1M HPreparation of 1M H22OO22 solution in 1000 mlsolution in 1000 ml
30 1.44 34.01
H2O2 molarity = 12.70
Volume required from H2O2 = 78.74 ml
Normal solutionNormal solution
Equivalent weightEquivalent weight
An equivalent weight is equal to the molecular weight divided by the valence
(replaceable H ions).
NormalNormal
A normal is one gram equivalent of a solute per liter of solution.
NormalityNormality
Normality is the total no of gram equivalents of the solute present per
liter of the solution
Normal solutionsNormal solutions
The definition of a normal solution is a solution that contains 1 gram equivalent
weight (gEW) per liter solution.
Units of NormalityUnits of Normality

The units for normal concentration are Eq/L. These units are often denoted by a
capital letter N (pronounced "normal").
mEq/L = 0.001 Eq/L
How to calculate equivalent weightHow to calculate equivalent weight
•• Eq.wt for acids and basesEq.wt for acids and bases
Examples:
• HCL the MW= 36.5 the EW = 36.5
• H2SO4 the MW = 98 the EW = 49
• H3PO4 the MW = 98 the EW = 32.7
• NaOH the MW = 40 the EW = 40
• Ca (OH)2 the MW = 74 the EW = 37
•• Eq.wt for saltsEq.wt for salts
Eq.wt=
M.wt of acid or base
No of H+ in acids or OH- in bases
Eq.wt=
M.wt of salt
No of cations X valency or No of anions X valency

Examples:
• KCL the MW= 74.55 the EW = (74.55 / 1 X 1) = 74.55
• CaCl2 the MW= 110.98 the EW = (110.98 / 1 X 2) or (110.98 / 2 X 1) = 55.49
• Na2CO3 the MW= 105.98 the EW = (105.98 / 1 X 2) or (105.98 / 2 X 1) =
52.99
•• Normality of solid soluteNormality of solid solute
Examples:
•• Preparation of 1N NaOH solutionPreparation of 1N NaOH solution
•• Preparation of 1N Ca(OH)Preparation of 1N Ca(OH)22 solutionsolution
•• Preparation of 1N KCL solutionPreparation of 1N KCL solution
 N: the required normality
 Eq.wt: equivalent weight of the solute
 V: total volume in liters
• Dissolve 40 g of NaOH in a 1000 ml (1 liter) of water to prepare 1N NaOH
solution.
• Dissolve 37 g of Ca (OH)2 in a 1000 ml (1 liter) of water to prepare 1N Ca
(OH)2 solution.
Wt of solute (g) = N (Eq/l) X Eq.wt X V (L)

• Dissolve 74.55 g of KCl in a 1000 ml (1 liter) of water to prepare 1N KCl
solution.
•• Normality of liquid soluteNormality of liquid solute
To prepare 1N solutions make the following steps:
• Calculate the equivalent weight of the solute
• Calculate the normality of the solute using the following formula:
• Calculate the volume of the solute needed to prepare 1N solution using the
following formula:
• Mix both volumes of solute and solvent to reach the total required volume
Example1:
•• Preparation of 1N HCL solutionPreparation of 1N HCL solution
32 1.18 36.5
HCL normality = 10.345
Volume required from HCL = 96.66 ml
Example2:
Normality (N) =
% X density X 10
Eq.wt
V1 of solute(ml) =
N2 (required normality) X V2 (required volume)
(ml)N1 (original normality)

•• Preparation of 1N HPreparation of 1N H22SOSO44 solutionsolution
98 1.83 98
H2SO4 normality = 36.6
Volume required from H2SO4 = 27.32 ml
Dilution of solutionsDilution of solutions
•• Simple DilutionSimple Dilution
A simple dilution is one in which a unit volume of a liquid material of interest is
combined with an appropriate volume of a solvent liquid to achieve the desired
concentration.
The dilution factor is the total number of unit volumes in which your material will
be dissolved. The diluted material must then be thoroughly mixed to achieve the
true dilution.
Example:
Dilute a serum sample 1:5 dilution as follow:
Combining 1 unit volume of serum (for example 200 l) + 4 unit volumes of the
saline (for example 800 l)
In this case the dilution factor is 5 (1 + 4 = 5 = dilution factor).
•• Serial DilutionSerial Dilution

A serial dilution is simply a series of simple dilutions which amplifies the dilution
factor quickly beginning with a small initial quantity of material.
The source of dilution material for each step comes from the diluted material of
the previous.
In a serial dilution the total dilution factor at any point is the product of the
individual dilution factors in each step up to it.
Final dilution factor (DF) = DF1 * DF2 * DF3 etc
Example:
Using a primary stock solution (for example 2 mmol/l) prepares a series of
different concentrations using serial dilution as follow:
The initial step combines 1 unit volume of stock solution (100 ml) with 9 unit
volumes of distilled water (900 ml) = 1:10 dilution.
Prepare another 6 tubes each one contains 500 ml distilled water
Combines 1 unit volume of the first tube (500 ml) with the 500 ml distilled water
in the second tube
Combines 1 unit volume of the second tube (500 ml) with the 500 ml distilled
water in the third tube and so on until the last tube
The total dilution would be: 1:10 X 2 X 2 X 2 X 2 X 2 X 2 = 1:640

•• Specific DilutionSpecific Dilution
It is used when we need to make a specific volume of known concentration from
stock solutions
To do this we use the following formula:
V1C1=V2C2
• V1: the volume of stock we start with. (Unknown)
• C1: the concentration of stock solution
• V2: total volume needed at the new concentration
• C2: the new concentration
Example:

Suppose we have 3 ml of a stock solution of 100 mg/ml and we want to
make 200 ml of solution having 25 mg/ ml.
V1 = (V2 x C2) / C1
V1 = (0.2 ml x 25 mg/ml) / 100 mg/ml
V1 = 0.05 ml, or 50 ml
So, we would take 0.05 ml stock solution and dilute it with 150 ml of solvent to
get the 200 ml of 25 mg/ ml solution needed
Conversion of Conc. unitsConversion of Conc. units
•• Conversion of Wt% Percent to molarityConversion of Wt% Percent to molarity
Molarity (M) = (% X 10)/Molecular wt
•• Conversion of molarity to Wt% PercentConversion of molarity to Wt% Percent
Wt% Percent = (M X Molecular wt)/ 10
•• Conversion of Wt% Percent to normalityConversion of Wt% Percent to normality
Normality (N) = (% X 10)/Equivalent wt
•• Conversion of normality to Wt% PercentConversion of normality to Wt% Percent
"X" units
Stock solutions of stable compounds are routinely maintained in labs as
more concentrated solutions that can be diluted to working strength
when used in typical applications. The usual working concentration is
denoted as 1X. A solution 20 times more concentrated would be denoted
as 20X and would require a 1:20 dilution to restore the typical working
concentration.
"X" units
Stock solutions of stable compounds are routinely maintained in labs as
more concentrated solutions that can be diluted to working strength
when used in typical applications. The usual working concentration is
denoted as 1X. A solution 20 times more concentrated would be denoted
as 20X and would require a 1:20 dilution to restore the typical working
concentration.

Wt% Percent = (N X Equivalent wt)/ 10
•• Conversion of Normality to MolarityConversion of Normality to Molarity
Normality (N) = Molarity (M) X n
•• Conversion of Molarity to NormalityConversion of Molarity to Normality
Molarity (M) = Normality (N) / n
Where n =
number of (H+
) in acids or (OH-
) in bases or valency in salts
•• Conversion of Wt% Percent to g/lConversion of Wt% Percent to g/l
Wt% Percent = g/l / 10
•• Conversion of g/l to Wt% PercentConversion of g/l to Wt% Percent
g/l = Wt% Percent X 10
•• Conversion of Molarity to g/lConversion of Molarity to g/l
g/l = M X Molecular wt
•• Conversion of g/l to MolarityConversion of g/l to Molarity
M = Molecular wt / g/l
•• Conversion of Normality to g/lConversion of Normality to g/l
g/l = M X Equivalent wt

•• Conversion of g/l to NormalityConversion of g/l to Normality
M = Equivalent wt / g/l
Examples:
1M NaCl1M NaCl 5.85%5.85% NaClNaCl 58.5 g/l58.5 g/l NaClNaCl 11NN NaClNaCl
1M HCL1M HCL 3.09%3.09% HCLHCL --------------- 1N HCL1N HCL
0.50.5M HM H22SOSO44 2.67%2.67% HH22SOSO44 --------------- 1N H1N H22SOSO44
11MM NaOHNaOH 4%4% NaOHNaOH 40 g/l40 g/l NaOHNaOH 1N NaOH1N NaOH
0.5M0.5M Ca(OH)Ca(OH)22 3.7%3.7% Ca(OH)Ca(OH)22 37 g/l37 g/l Ca(OH)Ca(OH)22 1N Ca(OH)1N Ca(OH)22
0.5M0.5M NaNa22COCO33 5.29%5.29% NaNa22COCO33 52.99 g/lg/l NaNa22COCO33 11NN NaNa22COCO33

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