If you like the notes, do like, share and comment. You can follow us on Facebook Gone Nuclear and here on Blogspot Gone Nuclear.
SOLUTIONS
INTRODUCTION
In normal life, we rarely come across pure substances. Most of these are mixtures containing two or more pure substances. Their utility or importance in life depends on their composition. The air around us is a mixture of gases primarily in O2 & N2; the water we drink contains various salts dissolved in it. Our blood is a mixture of different components. Alloys such as brass, bronze, stainless steel, etc. are also mixtures. In this unit, we will consider mostly liquid solutions and their properties.
DEFINITION
A solution is a homogeneous mixture of two or more than two components. For example, salt in water.
CLASSIFICATION
Solutions which contain two components in it are called Binary solutions.
Substances which are used to prepare a solution are its components.
The component which is present in larger quantity in the solution is called the solvent and the component less in quantity is called the solute.
The components can be in solid, liquid or in gaseous states.
STRENGTH OF SOLUTION
The amount of solute dissolved per unit solution or solvent is called the strength of a solution.
1. Mass Percentage(%w/w): It represents the mass of a component present in 100g of solution.
Mass % of a component = (Mass of a component in the soln./Total mass of soon) x 100
2. Volume Percentage(%v/v): It represents the volume of a component in 100 ml of solution.
Volume % of a component = (Volume of a component/Total volume of soln) x 100
3. Mass by Volume Percentage(%w/v): It represents the mass of solute in prams per 100 ml of solution.
Mass by Volume % = (Mass of solute in g/Volume of Soln in ml) x 100
4. Parts per Million(ppm): Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume. Its is basically 1 part in million parts, i.e., 1 part in 106 parts.
Parts per million = (No. of parts of a component/Total no. of all components of soln) x 106
5. Mole Fraction(X): It represents the moles of a solute present in one mole of solution.
Mole Fraction = (No. of moles of component/Total no. of moles of all components)
For example, in a binary mixture, if the number of moles of A and B is na and nb respectively, the mole fraction of A will be
XA = na/na+nb where X1 + X2 = 1
6. Molarity(M): It represents moles of solute present in 1L of solution.
Molarity (M) = No. of moles of solute/Volume of soln in L
Unit- mol/L
The density of a solution is the mass of solution per unit volume. So,
Density(d) = Mass of soln/Vol. of soln = m/V
7. Molality(m): It represents moles of solute present in a kg of solution.
Molality(m) = No. of moles of solute/Mass of solvent in kg
Unit- mol/kg
8. Normality(N): It represents no. of equivalents of solute present in 1L of solution.
Normality(N) = No. of equivalents of solute/Vol. of soln in L
No. of equivalents(eq) = Weight/Equivalnet weight = W/E
now, E=M/z (M is molarity and z is valency factor)
SOME IMPORTANT RELATIONSHIPS
DILUTION LAW
If a solution is diluted by adding solvents to it, then the amount of solute remains constant and we can write,
M1V1 = M2V2 and N1V1 = N2V2
Where 1 is for given solution and 2 is for solution to be prepared, i.e., 1 for initial solution and 2 for final solution.
M is molarity, N is normality and V is volume.
MOLARITY & NORMALITY
Normality = z x Molarity
Molarity = Normality/z
Z is the valency factor
Note- Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity and normality are a function of temperature. This is because volume depends on temperature whereas the mass does not.
VAPOUR PRESSURE
DEFINITION
Vapour pressure of a liquid/solution is the pressure exerted by the vapours in equilibrium with the liquid/solution at a particular temperature, i.e., pressure exerted by the vapour on the surface of the liquid/solution.
Vapour pressure α escaping tendency
VAPOUR PRESSURE OF LIQUID/SOLUTION & RAOULT'S LAW
(Raoult's law for volatile solutes)
Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each components in the solution is directly proportional to its mole fraction.
Now, consider a solution containing two volatile components 1 and 2 with mole fractions X1 and X2 respectively. Suppose at a particular temperature, their partial vapour pressure are P1 and P2 and the vapour pressure in pure state are P10 and P20.
Thus, according to Raoult's law, for components 1
P1 α X1 and P2 α X2
so, P1=P10X1 and P2=P20X2
Now, according to Dalton's law of partial pressure, the total pressure (Ptotal) over the solution phase in the container will be the sum of partical pressures of the components of the solution and it given as:
(Ptotal)=P1 + P2
Substituting the values of P1 and P2, we get
(Ptotal)=P10X1 + P20X2
as X1 + X2 = 1 so we can write X1=(1-X2)
so, (Ptotal)=(1-X2)P10 + P20X2
(Ptotal)=P10 + (P20 - P10)X2
The plot of vapour pressure and mole fraction of an ideal solution at constant temperature. The dashed line I and II represents the partial pressure of the components. It can be seen from the plot that P1 and P2 are directly proportional to X1 and X2, respectively. The total vapour pressure is given by the marked line marked III in the above figure.
Mole fraction in vapour phase,
If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phases then, using Dalton's law of partial pressure:
P1=y1Ptotal
P2=y2Ptotal
In general, Pi=yiPtotal
VAPOUR PRESSURES OF SOLUTIONS OF SOLID IN LIQUID AND RAOULT'S LAW
(Raoult's law for non volatile solutions)
If a non-volatile solute is added to a solvent to give a solution, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapour pressure is also reduced.
The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature.
Raoult's law in its general form can be stated as, for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
In a binary solution, let us denote solvent by 1 and solute by 2. When the solute is non-volatile, only the solvent molecules are present in vapour phase and contribute to vapour pressure. Let P1 be the vapour pressure of the solvent, X1 be its mole fraction, P10 be its vapour pressure in the pure state. Then according to Raoult's law,
P1 α X1
and P1=P10X1 = Ptotal
IDEAL AND NON-IDEAL SOLUTIONS
Ideal Solutions
An ideal solution is the solution in which each compoenets obeys Raoult's law under all conditions of temperature and concentration.
Properties of Ideal Solutions
ΔH(mix)=0
ΔV(mix)=0
Inremolecular attractive forces between A-A and B-B are nearly equal to those between A-B.
For example, solution of benzene and toulene; solution of n-hexane and n-heptane.
Non-Ideal Solutions
When a does not obey Raoult's law over the entire range of concentration, then it is called a non-deal solution.
Solutions showing positive deviation from Raoult's law:
Solvent-Solute(A-B) type of force is weaker than Solute-Solute(B-B) & Solvent-Solvent(A-A) forces.
Vapour pressure is higher then predicted by the law.
ΔH(mix)>0
ΔV(mix)>0
For example, solution of ethanol and acetone, solution of carbon disulphide and acetone
Pressure composition curve for solution showing positive deviation
Solutions showing negative deviation from Raoult's law
Solvent-Solute(A-B) type of force is stronger than Solute-Solute(B-B) & Solvent-Solvent(A-A) forces.
Vapour pressure is lower then predicted by the law.
ΔH(mix)<0
ΔV(mix)<0
For example, solution of phenol and aniline, solution of chloroform and acetone.
Pressure composition curves for solutions showing negative deviation
AZEOTOPES
Azeotopes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature.
MINIMUM BOILING AZEOTOPES
The solutions which show a large positive deviation from Raoult's law forms minimum boiling azeotopes at a specific composition.
For example, ethanol-water mixture containing approximately 95% of ethanol forms an azeotope with boiling point 351.15K.
MAXIMUM BOILING AZEOTOPES
The solutions which show a large negative deviation from Raoult's law forms maximum boiling azeotopes at a specific composition.
For example, Nitric acid and water mixture containing 68% nitric acid forms an azeotope with a boiling point of 393.5K.
SOLUBILITY
Solubility of a solid in a liquid
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent.
Factors affecting the solubility of a solid in liquid
1. Nature of solute & solvent
Like dissolves like. That is polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents.
For example, While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene readily dissolve in benzene, but sodium chloride and sugar do not.
2. Temperature
In a nearly saturated solution,
If (ΔsolH>0), the solubility increases with rise in temperature and
If (ΔsolH<0), the solubility decreases with rise in temperature.
3. Pressure
Pressure does not have any significant effect as solids and liquids are highly incompressible.
HENRY'S LAW
It states that at a constant temperature, the solubility of gas in a liquid is directly proportional to the pressure of the gas.
The most commonly used form of Henry's law states that the partial pressure of the gas in vapour phase (P) is proportional to the mole fraction of the gas (X) in the solution. This is expressed as
P=KHX
Here KH is the Henry's law constant.
Characteristics of KH
KH is a function of the nature of the gas.
Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.
KH value increases with increase of temperature indicating that the solubility of gases increases with decrease of temperature.
Raoult's law as a special case of Henry's law
According to Raoult's law,
Pi=Pi0Xi
In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Its solubility according to Henry's law,
P=KHX
Thus, Raoult's law becomes a special case of Henry's law in which KH becomes equal to Pi0
COLLIGATIVE PROPERTIES
The properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are called colligative properties.
There are four colligative properties
1. Relative lowering of vapour pressure
2. Elevation in boiling point
3. Depression in freezing point
4. Osmotic pressure
1. RELATIVE LOWERING OF VAPOUR PRESSURE
When a non-volatile solute is added to a solvent, the vapour pressure decreases.
The lowering of apour pressure with respect to the vapour pressure of the pure solvent is called Relative lowering of vapour pressure.
According to Raoult's law
P1=P10X1
The reduction in the vapour pressure of solvent (ΔP1) is given as
ΔP1 = P10-P1 = P10-P10X1 = P10 (1-X1)
Know that X2=1-X1
ΔP1 = P10X2
Equation can be written as
The expression on the left hand side of the equation as mentioned earlier is called realtive lowering of vapour pressure and is equal to the mole preaction of the solute. The above equation can be written as
Here, n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions, n2<<n1, hence neglecting n2 in the denominator we have
or 
where; R = gas constant (8.314 JK/mol)
Here w1 and w2 are the masses and M1 and M2 are the molar masses of the solvent and solute respectively.
2. ELEVATION IN BOILING POINT
Boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.
On addition of non-volatile solute the vapour pressure of the solvent decreases and therefore, to boil the solution the required temperature will be higher. So, there will be a rise in the boiling point of the solution.
The increase in boiling point ΔTb = Tb - Tb0
where Tb0 is the boiling point of the pure solvent and Tb is the boiling point of solution, this is known as elevation in boiling point.
ΔTb = Kb m
Kb is called boiling point elevation constant or molal elevation constant (Ebullioscopic constant).
Calculation of molar mass of solute
Substituting the value of molality in this equation, we get
Kb
It is defined as the elevation in boiling point when the molality of the solution is unity.
Unit- K kg/mol
Determination of Kb
Tb = boiling temperature in K
M1 = Molar mass of solvent in kg/mol
ΔvapH = enthalpy of vaporisation of solvent in J/mol
The vapour pressure curve for solution lies below the curve for pure water. The diagram shows that ΔTb denotes the elevation of boiling point of a solvent in solution.
3. DEPRESSION IN FREEZING POINT
The freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
When a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at low temperature. Thus, the freezing point of the solvent decreases.
ΔTf = Tf0 - Tf where Tf0 is the freezing point of pure solvent and Tf is its freezing point when non-volatile solute is dissolved, this is known as depression in freezing point.
ΔTf = Kf m
Kf is called freezing point depression constant or molal depression constant (Cryoscopic constant).
Calculation of molar mass of solute
Kf
It is defined as the elevation in boiling point when the molality of the solution is unity.
Unit- K kg/mol
Determination of Kf
Tf = freezing temperature in K
M1 = Molar mass of solvent in kg/mol
ΔfusH = enthalpy of fusion of solvent in J/kg
4. OSMOTIC PRESSURE
Osmosis
When a pure solvent and solution are kept with a semi-permeable between them then the solvent particles passes through the membrane from the solvent side to the solution side. This phenomenon is called 'Osmosis'.
The semi-permeable membrane is a membrane that allows only one small molecule to pass through and blocks the larger solute molecules.
The osmotic pressure (π) of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semi-permeable membrane into the solution.
Expression
For dilute solutions, osmotic pressure is proportional to the concentration that is molarity, C of the solution at a given temperature T. Thus,
π = C R T
Here, R is gas constant.
Calculation of molar mass
OSMOTIC SOLUTIONS
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
The solution with lower concentration or lower osmotic pressure is called hypotonic with respect to more concentrated solutions.
The solution with higher concentration or higher osmotic pressure is known as hypertonic with respect to dilute solution.
REVERSE OSMOSIS
If a pressure larger than the osmotic pressure is applied to the solution side, the solvent will flow from the solution into the pure solvent through semi-permeable membrane. This phenomenon is called reverse osmosis.
Application of Reverse Osmosis
Desalination of sea water: When pressure more than osmotic pressure is applied, pure water is squeezed out of sea water through the membrane.
ABNORMAL MOLAR MASS
When molar mass of a substance determined by studying any of the colligative properties come out to be different than the theoretically expected value, the is said to show abnormal molar mass. They are observed when,
1. the solute undergoes association in solution.
2. the solute undergoes dissociation in solution.
VAN'T HOFF FACTOR
To calculate extent of association or dissociation, van't Hoff introduced a factor i, known was van't Hoff factor.
Association
Number of particles will always decrease due to association therefore i<1.
Dissociation
Number of particles will always increase due to association therefore i>1.
MODIFIED EXPRESSIONS OF COLLIGATIVE PROPERTIES
VAPOUR PRESSURE
On increasing the temperature of the liquid the escaping tendency of the molecule increases and the vapour pressure increases.
Distribution of molecular kinetic energies in a liquid
Claussius-Clapeyron Equation
In P2/P1 = (ΔHvap/R)(1/T1 - 1/T2)
Where ΔHvap represents the enthalpy of vaporisation of liquid.
THERMODYNAMICS OF DISSOLUTION
If the interaction grow stronger the process is exothermic and if they go weaker during the formation of solution the process becomes endothermic. In general ΔS is positive in dissolution process. If the mixing process is spontaneous then ΔG has to be negative.
Elevation in boiling point
ΔGvap = 0
Tb = ΔHvap/ΔSvap
The non-volatile solute increases the randomness of the solution phase and the entropy of the vapour remains the same. Due to this ΔSvap decreases thus giving rise to the boiling point.
The higher boiling point of a solution relative to that of a pure solvent is due to a difference in their entropies of vapourisation, ΔSvap.
Depression in freezing point
ΔGfus = 0
Tf = ΔHfus/ΔSfus
The entropy difference will increase in this case due to the increase in the entropy of solution. This increase in entropy will result decrease of the freezing point according to the above reaction.
The lower freezing point of a solution relative to that of a pure solvent is due to a difference in their entropies of fusion, ΔSfus.
OSTWALD WALKER METHOD
w1 and w2 represent the decrease in weight of the vessels and w3 represents the increase in weight of the third vessel due to absorption.
If you like the notes, do like, share and comment. You can follow us on Facebook Gone Nuclear and here on Blogspot Gone Nuclear.
Atish J. Bain
HOD (Dept. of Chemistry)
St. Thomas High School (10+2)
Dhanbad, Jharkhand
No comments:
Post a Comment
Thank you. Will get back to you.