Solution Preparation (1)

Introduction Introduction

Solution Preparation (2)

Solution Solution

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 solutions Types of solutions

Percentage solution

Molar solution

Normal solution

Percentage solutionPercentage solution

Solution Preparation (3)

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 solution Preparation 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.

Wt of solute (g)= C (required Conc.) X V of solvent (ml)100

Solution Preparation (4)

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.

Volume /volume solution % (v/v)Volume /volume solution % (v/v)

g/L uniteA 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 g/L = 0.1% (w/v) solution

ppm uniteppm 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 = g/ml

Solution Preparation (5)

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.

Procedure:Procedure: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: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

V1 of solute (ml)= C2 (required Conc.) X V2 of solvent (ml)

C1 (original solute Conc.)

Solution Preparation (6)

Molecular weightMolecular weight

It is the sum of the atomic weights of all the atoms in a molecule. Also called

formula weight

Mole Mole

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 solution Molar 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 mol/l = 1000000000 nmol/l 1 mmol/ml = 1 mol/l

1 mol/ml = 1 mmol/l

Name Abbreviation Concentration

Solution Preparation (7)

Millimolar mM 10-3 molar

Micromolar M 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

Calculations: Calculations:

M: the required molarity M.wt: molecular weight of the solute V: total volume in liters

Procedure:Procedure: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:

Wt of solute (g)= M (mole/l) X M.wt (g/mole) X V (L)

Molarity (M)= % X density X 10

M.wt

V1 of solute(ml)= M2 (required molarity) X V2 (required volume) (ml)

M1 (original molarity)

Solution Preparation (8)

Subtract the volume of solute from the total solution volume

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 = 10.345Volume required from HCL = 96.66 ml

Complete to 1 liter with 903.33 ml distilled water

Example2: Preparation of 1M HPreparation of 1M H 22SOSO44 solution in 1000 ml solution in 1000 ml

% percent Density M.wt

98 1.83 98

H2SO4 molarity = 18.3Volume required from H2SO4 = 54.64 ml

Complete to 1 liter with 945.36 ml distilled water

Example3: Preparation of 1M HPreparation of 1M H 22OO22 solution in 1000 ml solution in 1000 ml

% percent Density M.wt

30 1.44 34.01

H2O2 molarity = 12.70Volume required from H2O2 = 78.74 ml

Complete to 1 liter with 921.25 ml distilled water

Solution Preparation (9)

Normal solutionNormal solutionEquivalent 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 solutions Normal 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

Solution Preparation (10)

How to calculate equivalent weightHow to calculate equivalent weight

Eq.wt for acids and bases Eq.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 salts Eq.wt for salts

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

Eq.wt= M.wt of acid or base

No of H+ in acids or OH- in bases

Eq.wt= M.wt of saltNo of cations X valency or No of anions X valency

Solution Preparation (11)

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 solution solution Preparation of 1N KCL solutionPreparation of 1N KCL solution

Calculations: Calculations:

N: the required normality Eq.wt: equivalent weight of the solute V: total volume in liters

Procedure:Procedure: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.

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:

Wt of solute (g)= N (Eq/l) X Eq.wt X V (L)

Normality (N)= % X density X 10

Eq.wt

Solution Preparation (12)

Calculate the volume of the solute needed to prepare 1N solution using the

following formula:

Subtract the volume of solute from the total solution volume

Mix both volumes of solute and solvent to reach the total required volume

Example1:

Preparation of 1NPreparation of 1N HCL solution

% percent Density M.wt

32 1.18 36.5

HCL normality = 10.345Volume required from HCL = 96.66 ml

Complete to 1 liter with 903.33 ml distilled water

Example2: Preparation of 1N HPreparation of 1N H 22SOSO44 solution

% percent Density M.wt

98 1.83 98

H2SO4 normality = 36.6Volume required from H2SO4 = 27.32 ml

Complete to 1 liter with 972.67 ml distilled water

V1 of solute(ml)= N2 (required normality) X V2 (required volume) (ml)

N1 (original normality)

Solution Preparation (13)

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

Solution Preparation (14)

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

Solution Preparation (15)

SS pecific Dilutionpecific 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=V2C2V1: 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) / C1V1 = (0.2 ml x 25 mg/ml) / 100 mg/ml

V1 = 0.05 ml, or 50 mlSo, 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

"X" unitsStock 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.

Solution Preparation (16)

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

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) / nWhere n =

number of (H+) in acids or (OH-) in bases or valency in salts

Solution Preparation (17)

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/lExamples:

1M NaCl 1M NaCl 5.85% 5.85% NaCl NaCl 58.5 g/l 58.5 g/l NaCl NaCl 11 NN NaCl NaCl

1M HCL 1M HCL 3.09%3.09% HCL HCL --------------- 1N HCL 1N HCL

0.50.5 M HM H 22SOSO44 2.67%2.67% H H 22SOSO44 --------------- 1N H1N H22SOSO44

11 MM NaOH NaOH 4%4% NaOH NaOH 40 g/l40 g/l NaOH NaOH 1N NaOH 1N 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 Na Na 22COCO33 5.29%5.29% Na Na 22COCO33 52.99 g/l g/l NaNa 22COCO33 11 NN Na Na 22COCO33