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Osmosis and Osmotic Pressure

Hemant More — Fri, 31 Jan 2020 07:27:41 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Osmosis and Osmotic Pressure

In this article, we shall study the phenomenon of osmosis and osmotic pressure.

Semipermeable Membrane:

A semipermeable membrane is a membrane which allows the solvent molecules, but not the solute molecules through it.
Examples: Cellulosde, cellulose nitrate, animal bladder.

Osmosis:

The spontaneous and unidirectional flow of solvent molecules through a semipermeable membrane, into a solution OR flow of solvent from a solution of low concentration to a solution of higher concentration through a semipermeable membrane, is called osmosis.

Everyday Examples of Osmosis:

Raw mangoes when placed in a concentrated solution of common salt lose water through osmosis and ultimately shrivel into a pickle.
Flowers revive and regain their freshness when placed in freshwater because of osmosis.
Carrots get limed due to the loss of water to the atmosphere. But, when limped carrots are placed in water, they become firm due to the inflow of water because of osmosis.
People consuming more salt and excessive salty food suffer from edema which is swelling and puffiness produced in the body due to retention of water in tissue cells and intracellular spaces.
The preservation of meat and fishes against bacteria is done by salting it. The bacteria on meat or fishes lose water through osmosis and ultimately die.
The preservation of fruits against bacteria is done by adding sugar to it. The bacteria on meat or fishes lose water through osmosis and ultimately die.
Plants absorb water from the soil through roots due to osmosis because the root hair cells have higher osmotic pressure than that of soil water.
Red blood cells burst when kept in water due to endosmosis.
Blooming, opening, and closing of flowers are governed by osmosis.
Dead bodies swell underwater due to endosmosis.

Experiment Exhibiting Phenomenon of Osmosis OR Abbe Nollet Experiment:

A wide-mouthed thistle funnel with a narrow long stem was taken. Then pig’s bladder (semipermeable membrane) is tied tightly around the wide mouth of the funnel with the help of a thread or rubber band. Now dilute sugar solution is carefully poured into the stem of the funnel to a certain level. The wide mouth of the funnel containing sugar solution is now kept in a beaker with the help of an Iron stand. Now three-fourths of the beaker is filled with pure water. The apparatus is left undisturbed for some time.

After a few hours we find the level of sugar solution increases from its initial level. This indicates that there is a net flow of solvent molecules into the solution through the semipermeable membrane. We have to apply excess sufficient pressure from the stem side on the solution to stop this migration of solvent molecule and this excess pressure is called osmotic pressure.

Osmotic Pressure:

The excess of pressure on the side of a solution that stops the net flow of solvent into the solution through a semipermeable membrane is called osmotic pressure. The equilibrium is reached when hydrostatic pressure of the column is equal to that of osmotic pressure. Osmotic pressure is not created by the solution but it comes into existence when the solution is separated from the solvent by a semipermeable membrane. If the pressure applied to the solution is greater than the osmotic pressure of the solution then solvent starts passing from solution to solvent. This phenomenon is called reverse osmosis. This process is used for purification of seawater and hard water.

Types of Solutions on the Basis of Osmosis:

Isotonic Solutions:

Two or more solutions having the same osmotic pressure at a given temperature are called isotonic solutions. When such solutions are separated by semipermeable membrane no osmosis occurs between them.

For example, the osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/ volume) sodium chloride solution, called normal saline solution and it is safe to inject intravenously.

Hypertonic Solution:

A solution having osmatic pressure higher than that of another solution is called a hypertonic solution.

For example, the osmotic pressure associated with the fluid inside the blood cell is less than sodium chloride solution having a concentration of more than 0.9% (mass/volume). Thus the solution of sodium chloride is hypertonic. In this case, water will flow out of the cells and cells would shrink.

Hypotonic Solution:

A solution having osmatic pressure lower than that of another solution is called a hypotonic solution.

For example, the osmotic pressure associated with the fluid inside the blood cell is more than sodium chloride solution having a concentration of less than 0.9% (mass/volume). Thus the solution of sodium chloride is hypotonic. In this case, water will flow into the cells and cells would swell.

Laws of Osmotic Pressure:

van’t Hoff’s Theory of Osmotic Pressure:

He found that the solute particles in dilute solutions possess kinetic energy and move in random directions in the solutions. Thus they have similar behaviour as that of gas molecules.

On collision against the semipermeable membrane, the solute molecules exert osmotic pressure equal to the pressure which the solute molecules would exert if it were gas molecule at the same temperature and occupying the same volume as that of solution. Thus the gas laws are equally applicable to dilute solutions.

van’t Hoff’s Boyle’s Law of Solution:

At constant temperature, the osmotic pressure (π) of a dilute solution is directly proportional to its molar concentration (C) or inversely proportional to volume (V) of the solution.

Explanation:

van’t Hoff’s Charle’s Law of Solution:

The concentration remaining constant, the osmotic pressure (π) of a dilute solution is directly proportional to absolute temperature (T) of the solution.

Explanation:

van’t Hoff’s General Solution Equation:

By van’t Hoff Boyle’s law at a constant temperature, the osmotic pressure (π) of a dilute solution is inversely proportional to volume (V) of the solution.

By van’t Hoff Charles law, The concentration remaining constant, the osmotic pressure (π) of a dilute solution is directly proportional to absolute temperature (T) of the solution.

Where k is proportionality constant called general solution constant. van’t Hoff further proved that this constant k is equal to universal gas constant R

Equations (a) and (b) represents general solution equation.

van’t Hoff’s Avogadro’s Law of Solution:

Two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature OR equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature.

Explanation:

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Boiling Point Elevation and Freezing Point Depression

Hemant More — Fri, 31 Jan 2020 06:51:45 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Boiling Point Elevation and Freezing Point Depression

In this article, we shall study two colligative properties of solutions, namely elevation of boiling point and depression in freezing point due to addition of solute.

Elevation of Boiling Point:

Boiling Point of a Liquid:

Boiling point is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. The boiling point of a liquid is a characteristic property and can be treated as criteria for the purity of liquid. It increases with the increase in external pressure. Liquids having greater intermolecular forces have high boiling points.

Elevation of Boiling Point of a Liquid:

The vapour pressure of the solution of non-volatile solute is always less than the vapour pressure of the pure solvent.

At the boiling point of pure solvent, the solution will not boil because its vapour pressure of the solution is less than the vapour pressure of the pure solvent. Thus vapour pressure of the solution is less than the external pressure. To boil the solution we have to the increases vapour pressure of the solution to make it equal with external pressure. It is achieved by increasing the temperature of the solution. Thus there is an elevation of the boiling point of the liquid.

Let T_b0 be the boiling point of pure solvent and Tb be the boiling point of the solution. The increase in the boiling point Δ T_b = T_b – T_b0 is known as the elevation of boiling point. The elevation of boiling point (ΔTb) is directly proportional to the lowering of vapour pressure (Δp).

Thus Δ T_b α Δp.

Experiments have shown that for dilute solutions the elevation of boiling point ( Δ T_b ) is directly proportional to the molal concentration of the solute in a solution. Thus, the elevation of boiling point also depends on the number of solute molecules rather than their nature.

Δ T_b α m

Δ T_b = K_b m ………….. (1)

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant). The unit of K_b is K kg mol-1.

The molal elevation of boiling point constant is defined as the elevation of boiling point produced when one mole of solute is dissolved in 1 kg of solvent.

Now

The experimental method to determine the molecular mass of non-volatile solute by determining boiling points of pure solvent and solution of known concentration is called ebullioscopy.

Depression of Freezing Point:

Freezing Point of a Liquid:

The freezing point of a liquid is a temperature at which the vapour pressure of solid is equal to the vapour pressure of the liquid.

Depression of Freezing Point of a Liquid:

The lowering of the vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent.

We know that at the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. A solution will freeze when its vapour pressure equals the vapour pressure of the pure solid solvent as is clear from the graph.

According to Raoult’s law, 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 a lower temperature. Thus, the freezing point of the solvent decreases.

Let T_f0 be the freezing point of pure solvent and Tf be the freezing point of the solution. The increase in the freezing point Δ T_f = T_f – T_f0 is known as depression of freezing point.

The depression of freezing point (Δ T_f) is directly proportional to the lowering of vapour pressure (Δp).

Thus, Δ T_f α Δp.

Experiments have shown that for dilute solutions the depression of freezing point (Δ T_f) is directly proportional to the molal concentration of the solute in a solution. Thus, the depression of freezing point also depends on the number of solute molecules rather than their nature.

Δ T_f α m

Δ T_f = K_f m ………….. (1)

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, K_f is called Freezing Point Elevation Constant or Molal Elevation Constant (cryoscopic Constant). The unit of K_f is K kg mol-1.

The molal elevation of freezing point constant is defined as the depression of freezing point produced when one mole of solute is dissolved in 1 kg of solvent.

Now,

The experimental method to determine the molecular mass of non-volatile solute by determining freezing points of pure solvent and solution of known concentration is called cryoscopy.

Problems:

Which of the following aqueous solutions will have maximum depression in freezing point. a) 0.5 M Li2SO4 b) 1 M NaCl c) 0.5 M Al2(SO4)3 d) 0.5 M BaCl2
A solution containing 0.73 g of camphor (molar mass 152 g mol-1) in 36.8 g of acetone (boiling point 56.3° C) boils at 56.55° C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46° C. calculate molar mass of the unknown compound.
1.0 x 10-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases the freezing point of the solvent by 0.211 K. 1.6 x 10-3 kg of another non-electrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of another solute.
Which of the following aqueous solutions will have minimum elevation in boiling point. a) 0.1 M KCl b) 0.05 M NaCl c) 1 M AlPO4 d) 0.1 M MgSO4
The boiling point of a solvent is 80.2° C. When 0.419 g of the solute of molar mass 252.4 g mol-1, is dissolved in 75 g of above solvent, the boiling point of the solution is found to be 80.26° C. Find molal elevation constant.

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Numerical Problems on Lowering of Vapour Pressure

Hemant More — Fri, 31 Jan 2020 05:24:53 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Lowering of Vapour Pressure

In this article, we shall study to solve problems based on relative lowering of vapour pressure and to calculate the molecular mass of a solute.

Example – 01:

The vapour pressure of a pure liquid at 298K is 4 x 10⁴ N/m². When a non-volatile solute is dissolved the vapour pressure becomes 3.65 x 10⁴ N/m². Calculate relative vapour pressure, lowering of vapour pressure and relative lowering of vapour pressure

Given: Vapour pressure of pure liquid = p^o = 4 x 10⁴ N/m², vapour pressure of solution = p = 3.65 x 10⁴ N/m², temperature = T = 298 K,

To Find: Relative vapour pressure = p/p^o =? Lowering of vapour pressure = p^o – p =? and relative lowering of pressure = (p^o – p)/p^o = ?

Solution:

Relative vapour pressure = p/p^o = (3.65 x 10⁴ N/m²)/(4 x 10⁴ N/m²) = 0.9125

Lowering of vapour pressure = p^o – p = 4 x 10⁴ N/m²– 3.65 x 10⁴ N/m²

Lowering of vapour pressure = 0.35 x 10⁴ N/m²= 3.5 x 10³ N/m²

Relative lowering of pressure = (p^o – p)/p^o = (3.5 x 10³ N/m²)/(4 x 10⁴ N/m²) = 0.0875

Ans: Relative vapour pressure = 0.9125, Lowering of vapour pressure = 3.5 x 10³ N/m², Relative lowering of pressure = 0.0875

Example – 02:

The vapour pressure of a solution containing 13 × 10^-3 kg of solute in 0.1 kg of water at 298 K is 27.371 mm Hg. calculate the molar mass of the solute. Given that the vapour pressure of water at 298 K is 28.065 mm Hg.

Given: mass of solute W₂ = 13 × 10^-3 kg, mass of solvent (water) = W₁ = 0.1 kg, vapour pressure of pure solvent (water) = p^o = 28.065 mm of Hg, vapour pressure of solution = p = 27.371 mm of Hg, temperature = T = 298 K, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 94.63 g mol^-1

Example – 03:

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile solute of a mass 2.175 × 10^-3 kg is added to 39.0 × 10^-3 kg of benzene. The vapour pressure of a solution is 600 mm Hg. What is the molar mass of the solute? Given C= 12, H = 1.

Given: mass of solute W₂ = 2.175 × 10^-3 kg, mass of solvent = W₁ = 39.0 × 10^-3kg, vapour pressure of pure solvent (benzene) = p^o = 640 mm of Hg, vapour pressure of solution = p = 600 mm of Hg,

To Find: Molecular mass of solute = M₂ =?

Solution:

Molecular mass of solvent (benzene C₆H₆) = M₁ = 12 x 6 + 1 x 6 = 78 g mol^-1

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 69.6 g mol^-1

Example – 04:

In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.

Given: mass of solute (mannitol) W₂ = 18.04 g, mass of solvent (water) = W₁ = 100g, vapour pressure of pure solvent (water) = p^o = 17.535 mm of Hg, decrease in vapour pressure of solution = p^o – p = 0.309 mm of Hg, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute (mannitol) = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 184.3 g mol^-1.

Example – 05:

A solution is prepared from 26.2 × 10^-3 kg of an unknown substance and 112.0 × 10^-3 kg acetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. Calculate the vapour pressure of solution if the molar mass of a substance is 273.52 × 10^-3 kg mol^-1. Given C = 12, H = 1, O = 16.

Given: mass of solute W₂ = 26.2 × 10^-3 kg, mass of solvent = W₁ = 112.0 × 10^-3kg, vapour pressure of pure solvent (acetone = p^o = 0.526 atm, Molecular mass of solute = M₂ = 273.52 × 10^-3 kg

To Find: vapour pressure of solution = p =?

Solution:

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 12 x 3 + 1 x 6 + 16 x 1

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 58 g mol^-1= 58 × 10^-3 kg

For dilute solution relative lowering of vapour pressure is given by

Ans: Vapour pressure of solution is 0.500 atm.

Example – 06:

The vapour pressure of water at 20 °C is 17 mm Hg. Calculate the vapour pressure of a solution containing 2.8 g of urea (NH₂CONH₂) in 50 g of water.

Given: Temperature of water = 20 °C = 20 + 273 = 293 K, vapour pressure of pure solvent (water) = p^o = 17 mm of Hg, mass of urea (solute)(NH₂CONH₂) = W₂ = 2.8 g, mass of water (solvent) = W₂= 50 g

To Find: Vapour pressure of solution = p =?

Solution:

The molecular mass of solute urea (NH₂CONH₂) = M₂

M₂ = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1 = 60 g mol^-1.

The molecular mass of water M₁ = 18 g mol^-1.

Ans: Vapour pressure of solution = 16.714 mm of Hg.

Example – 07:

At 300 K the vapour pressure of water is 1.2 x 10⁴ Pa. 0.8 x 10^-2 kg of oxalic acid (molecular mass = 126) is dissolved in 700 cm³ of water at the same temperature, find the vapour pressure of the solution.

Given: Temperature of water = 300 K, vapour pressure of pure solvent (water) = p^o = 1.2 x 10⁴ Pa, mass of solute (oxalic acid) = W₂ = 0.8 x 10^-2 kg, volume of water (solvent) = 700 cm³, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute (oxalic acid) = M₂ = 126 g mol^-1.

To Find: Vapour pressure of solution = p =?

Solution:

Volume of water = 700 cm³

Mass of water = W₁ = 700 cm³ x 1 g cm^-3 = 700 g = 0.7 kg

Ans: Vapour pressure of solution = 1.198 x 10⁴ Pa.

Example – 08:

Calculate the decrease in the vapour pressure when 1.81 g x 10^-2 kg of a solute (molecular mass = 57) is dissolved in 0.1 kg of water. Vapour pressure of water is 1.223 x 10⁴ Pa.

Given: Vapour pressure of pure solvent (water) = p^o = 1.223 x 10⁴ Pa, mass of solute = W₂ = 1.81 x 10^-2 kg, mass of water (solvent) = 0.1 kg, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute = M₂ = 57 g mol^-1.

To Find: Decrease in vapour pressure of solution = p^o – p =?

Solution:

Ans: Decrease in vapour pressure is 699 Pa

Example – 09:

A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25 ^oC. Further 18 g of water is then added to the solution. The new vapour pressure becomes 22.15 mm of Hg at 25 ^oC. Calculate the molecular mass of the solute and the vapour pressure of water at 25 ^oC.

Given: Temperature of water = 25 ^oC = 25 + 273 = 298 K

For solution 1: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g, solution vapour pressure = p = 21.85 mm of Hg

For solution 2: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g + 18 g = 108 g, solution vapour pressure = p = 22.15 mm of Hg

To Find: Molecular mass of solute = M₂ =? vapour pressure of pure water solution = p^o =?

Solution:

5p^o – 109.25 = 6p^o – 132.9

p^o = 132.9 -109.25 = 23.65 mm of Hg

Substituting in equation (1)

Ans: Molecular mass of solute is 72.83 u and vapour pressure of pure water solution 23.65 mm

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Lowering of Vapour Pressure of Solution

Hemant More — Fri, 31 Jan 2020 04:55:13 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Colligative Properties

In this article, we shall learn the meaning of colligative properties, colligative properties of solutions, and the lowering of vapour pressure due to the addition of solute in a solvent.

Colligative Properties:

Colligative properties are those properties of dilute solutions that depend only on the number of solute particles (atoms or molecules, ions or aggregates of molecules) in the solution and not on the nature of solute particles. These properties are important because they are used to determine molar masses of non-electrolyte solutes. Similarly, they are related to one another. If one is measured then other properties can be calculated. These properties can be observed clearly if the solution is very dilute, the solute is non-volatile and the solute does not dissociate or associate in the solution. There are four colligative properties of solutions

Lowering of vapour pressure.
Elevation of the boiling point of the
solvent in the solution.
Depression in the freezing point of
the solvent in the solution.
Osmotic pressure

The Concept of Vapour Pressure:

If a container is partially filled with a liquid, a portion of the liquid evaporates to fill the remaining volume of the container with vapour. The molecules of liquid evaporated are in continuous random motion, they collide with the walls of the container and with each other. Thus they create pressure on the walls of the container and on the liquid. At the same time some molecules which have left liquid return back to the liquid, the process is called condensation. After some interval of time, an equilibrium is established between the two phases of substance. At this stage, the rate of evaporation is equal to the rate of condensation.

The pressure exerted by the vapours of the liquid on the surface of the liquid when equilibrium is established between liquid and its vapour is called vapour pressure of the liquid. The temperature at which vapour pressure of the liquid is equal to the external pressure is called boiling temperature at that pressure.

Relation Between Vapour Pressure and Temperature:

Vapour pressure of liquid increases with the increase in the temperature.

Clausius Clapeyron Equation:

Using this relation we can find vapour pressure at some another temperature when its value at some temperature is known.

Lowering of Vapour Pressure:

Liquids at a given temperature vapourize and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure [Fig (a)].

In a pure liquid, the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. (b)], the vapour pressure of the solution is solely from the solvent alone. This vapour pressure of the solution at a given temperature is found to be lower than that of the pure solvent at the same temperature.

In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced. Consequently, 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.

Reasons for Lowering of Vapour Pressure:

The evaporation of liquid and evaporation of the liquid is a surface phenomenon. It is directly proportional to the surface area available for evaporation. Due to the addition of non-volatile solute the surface area of liquid decreases. It results in a decrease in the rate of evaporation and vapour pressure decreases.

By Graham’s law, the rate of evaporation is inversely proportional to the square root of the density of the liquid. When a solute is added to a solvent, the density of the resulting solution is more than the pure solvent. Hence the rate of evaporation decreases which results in the decrease of the vapour pressure of the solution.

Relative Lowering of Vapour Pressure:

The relative lowering of vapour pressure for the given solution is the ratio of vapour pressure lowering of solvent from the solution to the vapour pressure of the pure solvent.
Mathematically, the relative lowering of vapour pressure is given by

Where,
p₁⁰= Vapour pressure of the pure solvent
Δp= Lowering of vapour pressure
p= Vapour pressure of the solution

Raoult’s Law:

The partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

Explanation:

Let us consider a solution containing two volatile components say A₁ and A₂, with mole fractions x₁ and x₂ respectively. Let p₁^o and p₂^obe the vapour pressures of the pure components A₁ and A₂ respectively, then by Raoult’s law

p₁ = p₁^ox₁ and p₂ = p₂^o x₂

The total vapour pressure of the solutions of two volatile components is the sum of partial vapour pressures of the two components

p_T = p₁ + p₂

p_T = p₁^ox₁ + p₂^o x₂

But x₁ + x₂ = 1

∴ x₂ = 1 – x₁

∴ p_T = p₁^ox₁ + p₂^o (1 – x₁)

∴ p_T = p₁^ox₁ + p₂^o – p₂^ox₁

∴ p_T = p₂^o + ( p₁^o – p₂^o)x₁

The solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoul’s law are called non-ideal solutions.

Raoult’s Law for a Solution of Non-Volatile Solute:

Let us consider a solution containing two volatile component A₁ and non-volatile component A₂, with mole fractionsx₁ and x₁ respectively.

Let p₁⁰ and p₂⁰be the vapour pressures of the pure components A₁ and A₂ respectively. Now component A₂ is non-volatile, hence it will not contribute to vapour pressure. Thus p₂⁰= 0. We have

p_T = p₂^o + ( p₁^o – p₂^o)x₁

∴ p_T = 0 + ( p₁^o -0)x₁

∴ p_T = p₁^ox₁

Thus vapour pressure of a solution of non-volatile solute is the product of vapour pressure p₁⁰ of pure solvent and mole fraction x₁ of the solvent, which is Raoult’s law.

The equation shows that vapour pressure of the solution p < p₁⁰ , i.e. there is a lowering of the vapour pressure of the solution mathematically, it is given by

Δp = p₁^o – p

∴ Δp = p₁^o – p₁^ox₁

∴ Δp = p₁^o( 1 – x₁)

∴ Δp = p₁^ox₂

In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Now, the relative lowering of vapour pressure is given by

This relation proves that the lowering of vapour pressure is colligative property because it depends on the concentration of non-volatile solute.

Raoult’s Law as a Special Case of Henry’s Law:

By Raoult’s law, we have

p = p₁^ox …………… (1)

By Henry’s law, we have

p = K_Hx …………… (2)

If we compare the two equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction of solute in the solution. Only the proportionality constant K_H differs from p₁⁰. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H is equal to p₁⁰.

Relation Between Molar Mass of Solute and Lowering of Vapour Pressure:

Let W₂ g of the solute of molar mass M₂ be dissolved in W₁ g of the solvent of molar mass M₁. The numbers of moles of solvent and solute are given by n₁ = W₁/M₁ and n2 = W₂/M₂ respectively

The mole fraction of solute is given by

Now for dilute solutions n₂ << n₁. Hence n₂ can be neglected.

This is the relation between the molar mass of solute and the lowering of vapour pressure.

Measurement of Lowering of Vapour Pressure:

Barometric Method:

Raoult introduced the liquid or the solution into the Torricellian vacuum of a barometer tube and measured the depression of the mercury level.

This method is neither practicable nor accurate as the lowering of vapour pressure is too small.

Manometric Method:

This method is generally used for aqueous solutions.

The bulb of the apparatus is charged with the liquid or solution. The air in the connecting tube is then removed with a vacuum pump. When the stopcock is closed, the pressure inside is due only to the vapour evaporating from the solution or liquid. The manometric liquid can be mercury or n-butyl phthalate which has a low density and low volatility.

Ostwald and Walker’s Dynamic Method (Gas Saturation Method):

In this method, the relative lowering of vapour pressure can be determined directly. The measurement of the individual vapour pressures of a solution and solvent is thus eliminated.

Apparatus: The apparatus used is shown in the figure. It consists of two sets of bulbs. The first set of three bulbs is filled with the solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like P₂O₅, concentrated H₂SO₄etc. The bulbs of the solution and pure solvent are kept in a thermostat maintained at a constant temperature.

Procedure: A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up the amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of the solution, i.e. p₀ – p. The two sets of bulbs are weighed again. The guard tubes are also weighed.

Calculation:

The loss in the mass in the solution bulbs ∝ p

The loss in the mass in the solvent bulbs ∝ (p₀ – p)

The total loss of the mass in both sets of bulbs ∝ [p_s + (p₀ – p)] ∝ p₀

The total loss in the mass of both sets of bulbs is equal to gain in mass of guard tubes.

Thus measuring quantities on R.H.S. relative lowering of vapour pressure can be found.

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Solutions of Gases in Liquids

Hemant More — Fri, 31 Jan 2020 04:04:06 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Solutions of Gases in Liquids

In this article, we shall study solutions of gases in liquids and Henry’s law of solubility.

Almost all the gases are soluble in water as well as in other liquids to a greater or lesser extent. The solubility of gases in the liquid is expressed in terms of the absorption coefficient. The absorption coefficient is defined as the volume of gas in mL that can be dissolved by 1 mL of a liquid solvent at the temperature of the experiment at one atmospheric pressure. The volume of the gas is measured at STP. Mathematically

α = v/VP

Where, α = absorption coefficient of a gas

v = volume of gas dissolved in V volume of liquid solvent at atmospheric pressure P

Factors Affecting Solubility of Gases in Liquids:

Nature of Gas and Nature of Solvent:

The gases which are easily
liquefiable (e.g. CO₂) are relatively more soluble than dihydrogen
and dioxygen in common solvent.
The gases which are capable of
undergoing a chemical reaction with the water (e.g. HCl, NH₃) are
relatively more soluble in water than other solvents.
Oxygen, nitrogen and carbon dioxide
are more soluble in ethyl alcohol than in water at the same temperature and
pressure.

Effect of Temperature:

The solubility of gases in liquids decreases with the rise in temperature. When dissolved, the gas molecules are present in the liquid phase and the process of dissolution can be considered similar to condensation and heat is evolved in this process.

We know that the dissolution process involves dynamic equilibrium and thus must follow Le Chatelier’s Principle. As dissolution is an exothermic process, the solubility should decrease with the increase of temperature.

From the graph, we can see that the solubility of oxygen in water decreases rapidly with the increase in temperature. This results in a condition called thermal pollution. Thermal pollution is defined as a sudden increase or decrease in temperature of a natural body of water which may be ocean, lake, river or pond by human influence. It is harmful to its living inhabitants as the oxygen content of water decreases.

Effect of pressure:

The solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of gases increases with the increase in pressure.

For a solution of gases in a solvent, consider a system as shown in the following fig (a). The lower part is the solution and the upper part is the gaseous system at pressure P and temperature T. Assume this system to be in a state of dynamic equilibrium, i.e., under these conditions rate of gaseous particles entering and leaving the solution phase is the same.

Now increase the pressure over the solution phase by compressing the gas to a smaller volume. This will increase the number of gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of the solution to enter it. The solubility of the gas will increase until a new equilibrium is reached resulting in an increase in the pressure of a gas above the solution and thus its solubility increases.

It is to be noted that the pressure does not have any significant effect on so the ability of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure. The effect of pressure on the solubility of a gas in a liquid is given by Henry’s law.

Henry’s Law of Solubility:

Henry was the first to give a quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry’s law. The law states that at a constant temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas.

S α P

∴ S = KP

Where K = Henry’s constant

When P = 1, then S = K

Thus the solubility of gas at unit pressure is equal to Henry’s constant.

Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of the partial pressure of the gas.

If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that the mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution. The most commonly used form of Henry’s law states that “the partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as:

p = K_Hx

Where K_H is Henry’s law constant.

If we draw a graph of the partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown.

Different gases have different K_H values at the same temperature. This suggests that K_H is a function of the nature of the gas. From the above equation, we can conclude that the higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid.

Applications of Henry’s Law of Solubility:

Henry’s law finds several applications in industry and explains some biological phenomena.

To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.
Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in the blood. When the divers come towards the surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

Limitations of Henry’s Law:

Henry’s law is applicable only under the following conditions.

The pressure of the gas is not too high.
The temperature is not too low.
The gas should not undergo any chemical reaction with the solvent.
The gas should not undergo dissociation in the solvent.

Henry’s Law is not strictly followed.

Higher Solubility than expected:

Ammonia not only dissolves in water but also reacts with it. The reaction is as follows.

NH₃ + H₂O →NH₄OH ⇌ NH₄⁺ + OH^–

CO₂not only dissolves in water but also reacts with it. The reaction is as follows.

CO₂ + H₂O →H₂CO₃

In both cases, higher solubilities result than expected. Water is sparingly soluble in water but is highly soluble in the blood to the presence of haemoglobin of the blood.

Salting out effect:

Salting out is an effect based on the electrolyte-non electrolyte interaction, in which the non-electrolyte could be less soluble at high salt concentrations. Gases are less soluble in aqueous solutions of electrolytes. Similarly, sugar the non-electrolyte decreases the solubility of a gas in water.

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