Solubility and Solution Properties

A solution may be defined as a homogeneous mixture of two or more components that
form a single phase.
The component that determines the phase of the solution is termed the solvent and usually
constitutes the largest proportion of the system. The other components, dispersed as
molecules or ions throughout the solvent are termed solutes.
The transfer of molecules or ions from a solid state into solution is known as dissolution.
The extent to which the dissolution proceeds under a given set of experimental conditions is
referred to as the solubility of the solute in the solvent.
Solvent
Solute
Solution

1. Colligative properties
- The properties which depend on the number of particles in the solution (i.e. relative
proportions of the molecules of the solute and solvent) and not on the size or chemical nature of the
particles are called Colligative properties
Boiling Point elevation, Freezing Point depression, Vapor Pressure lowering, Osmotic Pressure
- The values are same for equal concentrations of different nonelectrolytes in solution.
- Related to each other.
2. Additive properties
- The properties of substance which depends upon the total contribution of the atoms or
functional groups within the molecules are called additive properties. e.g Mass, Molecular weight (sum of
masses of each atom in the molecule).
3. Constitutive properties
- The properties which mainly depend on the structural arrangement, number and kinds of
atoms in molecules are called constitutive properties.
- e.g. Optical rotation – depends on the chirality of the molecule which is determined by the
atomic bonding order; surface tension, viscosity etc.

Types of Solution
i.) Based on the amount of solute -
a) Saturated Solution
Saturated solution is a solution to which no more solute can be added at a particular
temperature. As temperature affects the solubility of a substance, a saturated solution
at lower temperature, may become a dilute solution at higher temperature.
b) Unsaturated Solution
It is a solution to which more solute can be added to dissolve. They are also otherwise
called dilute solutions. An unsaturated solution contains less solute than the solvent has
the capacity to dissolve at a specific temperature.
c) Supersaturated solution
A solution that contains a higher than saturation concentration of solute; slight disturbance or
seeding causes crystallization of excess solute.

Solubility
SATURATED
SOLUTION
no more solute
dissolves
UNSATURATED
SOLUTION
more solute dissolves
SUPERSATURATED
SOLUTION
becomes unstable, crystals
form
increasing concentration

Types of Solution (CONT…)
ii) Based on the physical state of the solutes and solvents
Solute Solvent Example
Gas Gas Air
Liquid Gas Water in oxygen,
Solid Gas Iodine in air
Gas Liquid Carbonated water
Liquid Liquid Alcohol in water
Solid Liquid Aqueous NaCl solution
Gas Solid Hydrogen in palladium
Liquid Solid Hg in Ag, mineral oil in paraffin
Solid Solid Gold-silver mixture

The concentration of a solution is defined as: the amount of solute present in a given
amount of solution.
Concentration is a general measurement unit, stating how much solute (solu.) is dissolved in
a given quantity of solution (soln.)
A solution containing a relatively low concentration of solute is called Dilute solution.
A solution of high concentration is called Concentrated solution.

Concentration Expression
Expression Symbol Definitions
Molarity M Moles (gm molecular weight)of solute in 1 L of
solution
Normality N Gram equivalent weights of
solute in 1 L of solution
Molality m Moles of solute in 1000 g solvent
Mole Fraction X Ratio of moles of one constituent
of a solution to the total moles of
all constituents (solute & solvent)
Mole percent -- Moles of one constituent in 100
moles of the solution. It is
obtained by multiplying X by 100
Percent by weight %w/w Gram of solute in 100 gm solution
Percent by volume %v/v ml of solute in 100 ml of solution
Percent by %w/v Gram of solute in 100 ml solution
weight-in-volume
Milligram percent --- mg of solute in 100 ml of solution

Molarity (M)
Molarity is the number of moles of solute dissolved per liter of solution.
Problems
What is the molarity of a solution prepared by dissolving 75.5 g of
pure KOH in 540 ml of solution?
Calculate the amount of HCl present in 150 ml of a 0.5 M solution?

Calculating Molarity
• What is the molarity of a solution containing 18.
0 g of NaOH in 0.100 L of solution?
• We also need to convert grams NaOH to moles Na
OH (MM = 40.00 g/mol).
= 4.50 M NaOH×18.0 g NaOH
0.100 L solution
1 mol NaOH
40.00 g NaOH

Molar Concentration Problem
• How many grams of K2Cr2O7 are in 250.0 mL of
0.100 M K2Cr2O7?
• We want mass K2Cr2O7, we have mL solution.
= 7.36 g K2Cr2O7
0.100 mol K2Cr2O7
1000 mL solution250.0 mL solution × ×
294.2 g K2Cr2O7
1 mol K2Cr2O7

Molar Concentration Problem
• What volume of 12.0 M HCl contains 7.30 g of H
Cl solute (MM = 36.46 g/mol)?
• We want volume, we have grams HCl.
= 16.7 mL solution
1 mol HCl
36.46 g HCl7.30 g HCl × × 1000 mL solution
12.0 mol HCl

Normality (N)
Number of gram equivalent (eq.) weights of solute in 1L of solution.
Relationship between molarity and normality for the same solute in the same solution:
So
Thus: N=nM n ≥ 1 therefore N ≥ M

Equivalent mass (Eq. M.): equivalent mass of a substance can be calculated by
dividing its molar mass with the number of active units in one molecule of this
substance (n) thus:
The active unit in the acid-base reactions is the number of hydrogen ions liberated by a
single molecule of an acid or reacted with a single molecule of a base.
Acid: one definition of acid is any substance that ionizes in water to form H+, includes
monoprotic and polyprotic. Monoprotic acid like hydrochloric acid (HCl) which includes
only one proton (hydrogen ion), and polyprotic acid like sulfuric acid (H2SO4) that
contains two protons (hydrogen ion).
The active unit in the oxidation-reduction (redox) reactions is the number of electrons (e)
transferred from one reactant to another during the reaction.
A 3M H2SO4 solution is the same as a 6N H2SO4 solution.
For a basic solution, n is the number of OH- ions provided by a formula unit of base.
A 1M Ca(OH)2 solution is the same as a 2N Ca(OH)2 solution

Problems for using Molarity and Normality
1. Volume of solution change with temperature. So it should not be used when one wishes to study t
he properties of solutions at various temperatures.
2. Difficult to study the properties of solutions such as vapor pressure and osmotic pressure which
are related to the concentration of the solvent.
Problem
Calculate the amount required to produce 250 ml, 1N HCl solution. The purity of HCl 37.0 % & the
density of the reagent is 1.19 g/mL.?
Find the Normality of 0.67 g. KOH in 120 mL of water?
Molality (m)
Number of moles of solute in 1000 g solvent. It does not change with temperature.
A solution obtained by dissolving one mole of the solute in 1000 g of solvent is called one molal or
1m solution.
Problem
What is the molality of a solution prepared by dissolving 50.0 g of HCl in 225 g of water?

Mole Fraction(X)
Mole fraction is the ratio of the number of moles of one constituent (solute/Solvent) in
solution to the total number of moles of solute and solvent. If n represents moles of solute
and N number of moles of solvent,
Problem: Calculate the mole fraction of HCl in a solution of hydrochloric acid in water,
containing 36 per cent HCl by weight
Mole Percent
Moles of one constituent in 100 moles of the solution. It is obtained by multiplying ‘X’ by 100.
Percent by weight in weight (%w/w)
It is the weight of the solute as a per cent of the total weight of the solution. That is,
For example, if a solution of HCl contains 36 per cent HCl by weight, it has 36 g of HCl for
100g of solution.
Problem:What is the per cent by weight of NaCl if 2.75 g of NaCl is dissolved in 9.85 g of
water?

Super problem
50 g of NaCl is dissolved in 1000 g of water.
If the density of the resulting solution is 0.997 g per
ml, calculate the molality, molarity, normality and
mole fraction of the solute, assuming volume of the
solution is equal to that of solvent.

Percent by volume in volume (%v/v)
ml of solute in 100 ml of solution
Problem
Make 1000ml of a 5% by volume solution of ethylene glycol in water.
Percent by weight-in-volume (%w/v)
Gram of solute in 100 ml solution
Problem
15mL of an aqueous solution of sucrose contains 750mg sucrose. What is the
weight/volume percentage concentration of this solution in g/100mL?

Dilution
Dilution is the addition of solvent to decrease the concentration of solute. The solution volume
changes, but the amount of solute is constant.
moles of solute (mol) = molarity (M) x volume (V)
M1V1 = M2V2
Initial values final values
Problem
Prepare 250 mL of 0.100 M NaCl solution from a 2.00 M NaCl solution.

Dilution Problem
• What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L
if 0.10 M NaOH?
• We want final volume and we have our final volume and concentra
tion.
M1 × V1 = M2 × V2
(6.0 M) × V1 = (0.10 M) × (5.00 L)
V1 = = 0.083 L
(0.10 M) × (5.00 L)
6.0 M

Solutions of Solids in Liquid
Like Dissolves Like
Polar molecules are soluble in polar solvents
e.g. Polar compounds, like table sugar
(C12H22O11), are soluble in polar solvents and insoluble
in nonpolar solvents; C2H5OH in H2O .
Non-polar molecules are soluble in non-polar solvents
e.g. naphthalene (C10H8), are soluble in nonpolar
solvents and insoluble in polar solvents;CCl4in C6H6
This it the like dissolves like rule.
Will sugar dissolve in water? (Homework) Will petroleum dissolve in water?
The mechanism of solution of Ionic compounds, like sodium chloride (Homework)

Miscible & Immiscible
Two liquids that completely dissolve in each other are miscible liquids.
Two liquids that are not miscible in each other are immiscible liquids.
Polar water and nonpolar oil are immiscible liquids and do not mix to form
a solution.
Partition Coefficient
A partition-coefficient (P)is the ratio of concentrations of a compound in a
mixture of two immiscible phases at equilibrium.Partition coefficient is generally
defined as the fraction of a compound (drug) in an oil phase to that of an adjacent aqueous
phase.
P o/w = (C oil / C water) equilibrium
Accordingly compounds with relatively high partition coefficient are predominantly lipid
soluble and consequently have very low aqueous solubility.
Compounds with very low partition coefficients will have difficulty in penetrating
membranes resulting poor bioavailability.

Factors Affecting Solubility of a Solute
Temperature
Generally in many cases solubility increases with the rise in temperature
and decreases with the fall of temperature but it is not necessary in all
cases. However we must follow two behaviours:
In endothermic process solubility increases with the increase in
temperature and vice versa.
For example: solubility of potassium nitrate increases with the increase in
temperature.
In exothermic process solubility decrease with the increase in temperature.
For example: solubility of calcium oxide decreases with the increase in
temperature.
Effect of pressure
The effect of pressure is observed only in the case of gases.
An increase in pressure increases of solubility of a gas in a liquid.
For example carbon di oxide is filled in cold drink bottles (such as coca cola, Pepsi
7up etc.) under pressure.

Chemical natures of the solute and solvent
A polar solute will dissolve in a polar solvent but not in a nonpolar solvent. The adage
"like dissolves like" is very useful.
Example: Alcohol (polar substance) dissolves in water (polar substance). Water (polar
substance) does not dissolve in oil (nonpolar substance)
Stirring: With liquid and solid solute, stirring brings fresh portions of the solvent in contact with
the solute, thereby increasing the rate of dissolution.
Amount of solute already dissolved
When there is little solute already in solution, dissolving takes place relatively rapidly. As
the solution approaches the point where no solute can be dissolved, dissolving takes
place more slowly.
Molecular size
The larger the molecules of the solute are, the larger is their molecular weight and their size.
If the pressure and temperature are the same than out of two solutes of the same polarity,
the one with smaller particles is usually more soluble.

Nernst gave a generalization governing the distribution of a solute between
. This is called law or
or simply or
If a solute X distributes itself between two
immiscible solvents A and B at constant temperature and X is in the same
molecular condition in both solvents,
Then
If C1 denotes the concentration of the solute (X) in solvent, A and
C2 the concentration in solvent B, then Nernst’s Distribution law can be expressed
as
The constant KD (or simply K) is called the Distribution coefficient or Partition
coefficient or Distribution ratio.
DK
BinXofionConcentrat
AinXofionConcentrat

DK
C
C

2
1

Explanation
If we take two immiscible solvents A and B in a beaker, they form separate layers. When a
solute X which is soluble in both solvents is added, it gets distributed or partitioned between
them. Molecules of X pass from solvent A to B and from solvent B to A. Finally a dynamic
equilibrium is set up. At equilibrium, the rate, at which molecules of X pass from one solvent
to the other is balanced.
Figure: Distribution of solute X between solvent A and B.
Now according to Nernst’s distribution law the distribution of solute X between solvent A
and B will be
Constanta
BinXofionConcentrat
AinXofionConcentrat


Derivation of distribution law
Let us take two immiscible solvents, A and B in a beaker. Now add a solute X which is
soluble in both of the solvents, thus it gets distributed or partitioned between them.
Molecules of X pass from solvent A to B and from solvent B to A. Finally a dynamic
equilibrium is set up.
Fig. At equilibrium, the number of molecules of X passing from solvent A into B is
proportional to its concentration in A and vice versa. Also, the rate of migration of
solute molecules from A to B and B to A is equal (Illustration).

When the distribution of the solute X has reached dynamic equilibrium, the rate (R1) at which
molecules of X pass from solvent A to B is proportional to its concentration (C1) in A.
That is, R1 ∝ C1
or R1= k1C1 where k1 is a constant
The rate (R2) at which molecules of X pass from solvent B to A is proportional to its concentr
ation (C2) in B.
R2 ∝ C2
or R2 = k2C2 where k2 is a constant
Also, at equilibrium, the rate of migration of solute from one solvent to the other is equal.
Thus we have, R1= R2
or, k1C1 = k2C2
If temperature is fixed the distribution coefficient KD is also constant, since k1 and k2 are cons
tants at the same temperature.
This is the Nernst’s Distribution law equation.
D
D
K
C
C
or
K
k
k
C
C
or


2
1
2
1
2
1
,
,

Limitations of distribution law
The conditions to be satisfied for the application of the Nernst’s Distribution law
are :
1. Constant temperature. The temperature should be kept constant throughout
the experiment.
2. Same molecular state. The molecular state of the solute has to be the same
in the two solvents. The law does not hold if there is association or dissociation of
the solute in one of the solvents.
3. Equilibrium concentrations. The concentrations of the solute requires to be
noted after the equilibrium has been established.
4. Dilute solutions. Suitable if solute concentration is low in two solvents.
The law does not hold when the concentrations are high.
5. Non-miscibility of solvents. The two solvents should be non-miscible or only s
lightly soluble in each other. The extent of mutual solubility of the solvents
remains unaltered by the addition of solute to them.

The relationship between pressure and solubility of a gas in a particular solvent was
investigated by William Henry (1803). He gave a generalisation which is known as
Henry’s Law. It may be stated as
Henry’s law states: at a constant temperature the solubility of a gas in a liquid is
proportional to the pressure of the gas above it.
Mathematically, Henry’s Law may be expressed as
C ∝ P
Or C = k P
Where P = pressure of the gas;
C = concentration of the gas in solution; and
k = proportionality constant known as Henry’s Law Constant.
The value of k depends on the nature of the gas and solvent, and the units of P and
C used.

Derivation
Henry’s law is, is a form of Distribution law. If a vessel containing a liquid and a gas is
shaken, at equilibrium the gas can be regarded as distributed between the liquid
(Phase A) and the space above (Phase B).
Let C1 be the molar concentration of the gas in phase B
C be the concentration of the gas in phase A
Applying the Distribution law, C1/C2=KD (a constant)
We know that molar concentration of gas is proportional to
its pressure, P.
C/P =K (a constant)
or C = kP
This is Henry’s Law equation.
Like distribution law, Henry’s law holds good for dilute solutions of gases which do not react
with the solvent.

At a certain temperature, the Henry’s law cons
tant for N2 is 6.0  104 M/atm. If N2 is presen
t at 3.0 atm, what is the solubility of N2?
1. 6.0  104 M
2. 1.8  103 M
3. 2.0  104 M
4. 5.0  105 M

Correct Answer:
Henry’s law, Sg
= kPg
Sg = (6.0  104 M/atm)(3.0 atm)
Sg = 1.8  103 M
1. 6.0  104 M
2. 1.8  103 M
3. 2.0  104 M
4. 5.0  105 M

Limitations of Henry’s Law
It applies closely to gases with nearly ideal behaviour.
(1) At moderate temperature and pressure.
(2) If the solubility of the gas in the solvent is low.
(3) If the gas does not react with the solvent to form a new
species. Thus ammonia (or HCl) which react with water
does not obey Henry’s Law
NH3 + H2O ⇌NH4
+ + OH–
(4) If the gas does not associate or dissociate on dissolving
in the solvent.

Use
The influence of partial pressure on solubility is utilized in making carbonated beverages like
beer, champagne, and many soft drinks. So called ‘soda water’ is bottled under a carbon
dioxide pressure of about 4 atm. When the bottle is opened to the air, the partial pressure of
CO2 above the solution is decreased (about 0.001 atm), and CO2 bubbles out.
a ) CO2 gas at 4 atm in equilibrium with dissolved CO2 resulting in high solubility of CO2 ;
(b ) In opened bottle pressure is released to 1 atm and hence equilibrium shifted upward, gas
bubbles evolved causing brisk effervescence; ( c) Partial pressure of CO2 in air being 0.001 atm,
practically the whole of CO2 is removed from solution, leaving the soft drink flat as the
equilibrium is established.

Applications of Distribution Law
There are numerous applications of distribution law in the laboratory as well as in industry.
Here we will discuss some more important ones by way of recapitulation.
(1) Solvent Extraction
The removal of a substance(s) {solute} by a solvent of an organic substance from an
aqueous solution is called extraction.
(2) Partition Chromatography
This is a modern technique of separating a mixture of small amounts of organic materials.
(3) Desilverization of Lead (Parke’s Process)
(4) Confirmatory Test for Bromide and Iodide
(5) Determination of Association
(6) Determination of Dissociation
(7) Determination of Solubility
(8) Deducing the Formula of a Complex Ion (I3 )
(9) Distribution Indicators

(1) Solvent Extraction
The removal of a substance(s) {solute} by a solvent of an organic
substance from an aqueous solution is called extraction.
Classification
Simple extraction: when the process of extraction is carried out with the
total amount of the given solvent in a single operation, is referred to as
simple extraction.
Multiple extraction or multi-step extraction: To recover the maximum
amount of the substance from aqueous solution, when the extraction is
made in two or more successive operations using small portions of the
provided solvent, then it is called multiple extraction or multi-step
extraction.

Simple extraction
The process of Simple extraction of organic substances from aqueous solutions is
carried out by shaking the aqueous solution with a immiscible organic solvent, say
ether (ethoxyethane), in a separatory funnel.
The distribution ratio being in favour of ether, most of the organic substance passes
into ether layer.
On standing, the aqueous and ether layers separate in the funnel. The lower aqueous
layer is run out, leaving the ether layer behind. This is then transferred to a distillation
flask. Ether is distilled over while the organic substance is left as residue in the flask.
If desired, the process may be repeated with aqueous layer left after the first extraction
with a fresh quantity of the solvent.
The other common solvents used for extraction are hexane, benzene, chloroform,
acetone, carbon disulphide, etc.
The greater the distribution ratio is in favour of the organic solvent, the greater will be
the amount extracted in any one operation.

Multiple extractions
In Multiple extractions process the aqueous solution is first extracted with a
portion of the solvent in a separatory funnel. The aqueous layer from which some
substance has been removed is then transferred to another funnel. This is shaken
with another ( second) portion of the solvent. Similarly, the aqueous layer from the
second extraction is treated with a third portion of solvent, and so on.
Superiority of multiple extraction over simple extraction
(** For better extraction, multiple extraction is more efficient than single extraction.)
Generally it is more efficient to use a specified volume of solvent in small portions
rather than in one whole.
Suppose we have 100 ml of an aqueous solution containing A grams of an organic
substance.
We can extract the substance with ether (ethoxyethane), its distribution ratio being
twice in favour of ether. Provided that 100 ml of ether which may be used in one
lot or in two portions of 50 ml each.

Using all the ether in one lot.
Let x grams be the weight of the substance extracted in the solvent layer.
Then the amount of substance left in the water layer = (A – x) grams.
Therefore,
Concentration in ether layer = x/100
Concentration in water layer = (A-x)/100
According to the Distribution law, Cether/Cwater=K
Thus
Hence, x = 2/3A
Thus 66% of substance is extracted.

Using two 50 ml portions of ether
Let x1 grams of substance be extracted in the first operation with 50 ml ether.
Thus,
Concentration in ether layer= x1 /50
Concentration in water layer = (A- x1 )/100
Thus
Hence, x =1/2A
The substance left in water layer is =1/2A

Now,
Let x2 grams of substance be extracted in the first operation with 50 ml ether.
Thus,
Concentration in ether layer= x2 /50
Concentration in water layer = (A/2- x2 )/100
Thus
Hence, x =1/4A
Now the total percent of substance is extracted by using two 50 ml portions of the solvent is
= (50+25) %
=75%
It is greater than 66 percent obtained by using 100 ml solvent in one lot.
Hence multiple extraction is more efficient than simple extraction-proved.

Solubility and Solution Properties

Solubility and Solution Properties

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Solubility and Solution Properties

Similar to Solubility and Solution Properties (20)

More from Imran Nur Manik

More from Imran Nur Manik (20)

Recently uploaded

Recently uploaded (20)

Solubility and Solution Properties