Henry’s Law of dissolution: How is the partial pressure of gas related to its solubility?

How much solute is dissolved in how much solvent or solution? It is one of the most important question for every solution. We have studied various ways to express the concentration of solution in the previous post of Solutions. If you want to prepare a solution, you need to understand the process of dissolution of solute and learn about the factors affecting the process of dissolution. This can help us to prepare the solution of desired properties. 

Let’s check solubility of few solutes. Why does table salt get dissolved in water but not in oil? Why does water and ethanol are miscible but water and oil are immiscible? Why is carbon dioxide more soluble in water as compared to the oxygen?

‘Like dissolves like’, you can take it as a slogan for dissolution. Table salt (Na⁺Cl^-) is an ionic compound and water is a polar molecule, that’s why water is a suitable solvent for salt. On the other hand, oil is non-polar unsaturated hydrocarbon and so it doesn't dissolve salt readily. Similarly ethanol is a polar molecule like water so they have affinity for each other in contrast to water and oil. 

Now you can guess the reason of higher solubility of CO₂in water as compared to that of O₂. Now you can predict the solubility if you know the chemistry of solute and solvent. It isn’t all about solubility though, there is some more chemistry behind it. Can you compare the solubility of O₂ and N₂ in water? Let’s see which factors decide the solubility of gases in liquids.

We have learnt the process of dissolution in the post of Solubility and Common ion Effect under chemical equilibrium. When a solute is dissolved in a solvent, an equilibrium is established between dissolved and undissolved solute molecules.

Gas molecules are much more mobile and free than the solid or liquid molecules. Therefore, when you dissolve a gas in a solvent, its molecules continually come in and out of the solvent and an equilibrium is established at the surface of the solvent between the gas molecules entering and leaving the solvent. It means that you can’t stop the exchange of molecules at the surface. There is an open gate for gas molecules. if it is crowded out there, more molecules will come inside, and if it is crowed inside, more molecules will move out, which means exchange rate will increase in either case. Is there any relation between crowding and exchange rate? Crowding shows the concentration of gas molecules and when it is outside the surface it is defined as vapour pressure, and when it is inside the surface it is defined by mole fraction. Mole fraction ultimately defines the solubility of gas. 

William Henry gave the quantitative relation between pressure and solubility of a gas. This law states that at constant temperature, solubility of a gas in a liquid is directly proportional to the pressure of the gas.

pressure = K_H. solubility

K_H is the Henry’s constant. It is a function of temperature and its unit depends on the units used for pressure and solubility. Its value is different for different gases in different solvents.

Solubility of a gas is the concentration of the gas molecules in the solvent and you can express it in terms of mole fraction or molar concentration (mol/L).

If solubility is expressed by mole fraction and pressure by partial pressure of gas then you will find a relation:

p = K_H. x

where p is the partial pressure of gas

xis the mole fraction of gas

K_H = p / x

You must have noticed in the equation that x is inversely proportional to K_H , which means that the gas which has lower value of Henry’s constant is more soluble. Value of K_H  at 298K for O₂ is 34.86kbar and for CO₂ its value is 1.67kbar. So, CO₂ is more soluble than O₂.

Partial pressure is the pressure exerted by molecules present above the surface of liquid. It gives an idea about the number of molecules present in gaseous phase. It can be defined in terms of concentration (moles/L) as well as pressure (kbar or atm).

If pressure is defined in terms of concentration and solubility is also expressed in moles/L then equation for Henry’s law will be modified as,

c_g = K_H. c_aq

where

c_g = concentration in gaseous phase

c_aq = concentration in aqueous phase

In this case the K_H will be unit less.

If pressure is defined in terms of partial pressure and solubility is expressed in moles/L, then equation for Henry’s law will be modified as,

p = K_H. c_aq

In this case, the unit of K_Hwill depend on the unit used for pressure and concentration.

The value of Henry’s constant for oxygen dissolved in water at 298K can be defined in different units like 770 atm L/mol or 3.2×10⁻² (K_H unit less) or 34.86kbar.

We have studied earlier that the enthalpy of solution is always exothermic since energy is released in this process. That’s why on increasing the temperature, the equilibrium shifts backward and more molecules start leaving the solvent. It means that the solubility of a gas decreases with the increase in temperature.

Let’s practice a few problems based on Henry’s law

 Q1. If N₂ gas is bubbled through water at 293K. How many mmoles of N₂(g) would dissolved in 1L of water? Partial pressure of N₂(g) is 0.987bar. and Henry’s law constant is for N₂ is 76.48kbar.

given,

p = 0.987 bar

K_H= 76.48 kbar = 76.48 × 10³ bar

Concentration = ? mmole/L

Partial pressure and value of K_His given in the question and units of both are the same which means that the solubility term must be unit less and that will be mole fraction. so we will use following formula:

p = K_H. x

0.987 bar = 76.48 × 10³ bar . x

x = 1.29× 10^-5

N₂(g) is dissolved in water so we need to calculate the number of mols of N_2.

Mole fraction of a component  =         Number of moles of the component in the solution 

                                                        Total number of moles of all components of the solution

let suppose number of moles of N₂ is = n

number of moles of water in 1 L = weight of 1L water/ molecular weight of water

= 1000g/18g =55.5 mol

1.29× 10^-5  =      n         

                       n+ 55.5

n+ 55.5 ⋍ 55.5

1.29× 10^-5   = n/55.5

n = 7.16 × 10^-4   mol

since 1mol = 1000ml

so 7.16 × 10^-4   mol = 7.16 × 10^-4  × 103 = 0.0716mmol

Answer is 0.0716mmol

Q2. How many grams of carbon dioxide gas are dissolved in a 1 L bottle of carbonated water if the manufacturer uses a pressure of 2.4 atm in the bottling process at 25 °C?
if: K_H of CO2 in water = 29.76 atm/(mol/L) at 25 °C

given,

p = 2.4 atm

K_H= 29.76 atm/(mol/L)

Concentration = ? g

Here partial pressure and constant both have different units.

p / K_H  = concentration

concentration =         2.4 atm               

                         29.76 atm/(mol/L)

concentration = 0.08 mol/L

1 mol of CO₂ = 44g

0.08 mole = 44 × 0.08 = 3.52g

1 L solution contains 0.08 mols of CO₂that is equal to 3.52g.

Answer is 3.52g.

Q3. The partial pressure of O2(g) in equilibrium with 6.87 × 10^-4 mol/ L O2(aq) is 16.8 kPa.

If the partial pressure of O2(g) drops to 3.46 kPa, what is the final concentration of O2(aq) in mol/ L?

given,

p initial = p₁ = 16.8 kPa

p final = p₂ = 3.46 kPa

Concentration initial = c₁=  6.87 × 10^-4 mol/ L

Concentration initial = c₂=  ? mol/ L

Henry’s law is

partial pressure = K_H . solubility

p₁   = K_H. c₁                                                                                              …..(1)

p₂   = K_H. c₂                                                                                              ….. (2)

on dividing equation 2 by equation 1 we get

p₂  = K_H. c₂                                                                                             

p₁     K_H. c₁

p₂   =  c₂                                                                                                    ….. (3)

p₁       c₁

Since our calculation is for the same gas under same temperature, the value of K_H will be the same on initial and final stages.

on putting the values in equation 3 we get the value of C₂

3.46 kPa   =  c₂                                                                                        ….. (4)

16.8 kPa    6.87 × 10^-4 mol/ L

c₂ = 1.41× 10^-4 mol/ L

Answer  is 1.41× 10^-4 mol/ L

Applications of Henry’s Law

You must have heard the popping sound on opening the bottle of a carbonated drink. As you open the cap, the pressure decreases and CO₂ molecules start escaping from the liquid due to decrease in solubility (p∝x) and if you leave the cap open, the drink doesn't remain fizzy for long because the partial pressure of CO₂in open atmosphere is very small which allows the soluble CO₂molecules to escape. 

At higher altitude climbers have trouble in breathing and sometimes they have anoxia, which is a clinical condition when brain becomes unable to work properly due to lack of oxygen. At high altitude, the atmospheric pressure decreases and so does the concentration of oxygen. The decrease in partial pressure of oxygen decreases the solubility of it in blood.

In contrast to higher altitude, atmospheric pressure increases in deep sea and this increased pressure also increases solubility of available gases in the blood. When divers come towards the surface, pressure suddenly decreases and the dissolved gases try to escape from the blood due to decreased solubility. Escaping gases form bubbles which can migrate to any part or joint of the body to find the way out. This causes severe pain in different joints. This condition is known as diver’s disease or Decompression sickness or bends or caisson disease. You might have read about this under the applications of boyels law.

You know that life on the earth was first started in the sea. You might be interested to find out why? Let’s try to calculate the concentration of oxygen in sea water which supports life under the sea.

By the Dalton's law of partial pressures we know that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. if we know the mole fraction or percent composition of gas we can calculate the partial pressure of particular gas if total pressure is given.   

Total pressure = p₁ + p₂ + p₃+ p₄ ….

Atmospheric pressure = p_oxygen + p_nitrogen+ p_{carbondioxide} + p_hydrogen ….

Partial pressure of gas 1 (p₁ ) = pressure × mole fraction x₁

or,

Partial pressure of gas 1 (p₁ ) = pressure × percentage composition of gas₁

Atmospheric pressure at sea level is 760mm Hg = 1atm = 1.01325 bar

Percentage of oxygen in air is 21%

So, the partial pressure of oxygen will be = (21 × 1)/ 100 = 0.21atm

Henry’s constant for O₂ in water at 298K is 770 atm L/mol.

p = K_H . Concentration

0.21atm = 770 atm L/mol . concentration

concentration of O₂ = 2.7 × 10^-4 mol/L

Pressure in the sea increases by approximately 1 atm for every 10m depth. Higher pressure and lower temperature at deep sea allow more oxygen to stay in water which supports giant aquatic creatures like sharks.

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What is Solution?

Solution! Even though you might not know its proper definition, but you use it frequently in your daily life. Chocolate syrup, sugar syrup, deodorants, perfumes, paints, nail paints, milk, tea, coffee, air, blood, and sweat are some examples of solutions. Did you notice any similarity among them? Yes, these are all mixed so well that we cannot separate their ingredients. It means that solutions are homogeneous mixture of two or more substances. There is one more similarity among above examples, they have one main substance and other substances are dissolved in it, for example, in sugar syrup, sugar is dissolved in water. The substance which is present in larger quantity is called the solvent and other substances which are present in lesser quantities are known as solutes. There are three states of any substance, solid, liquid and gas. Solvent and solute both can be solid, liquid or gas, there are no restrictions. It means we can have 9 combinations or solutions. Let’s have a look.

Solution	Solvent	Solute	Example
Solid solution	Gold(solid)	Copper(solid)	This solution is used for jewelry making
	Silver(s)	Mercury(l)	Used in dental filling
	Palladium(s)	Hydrogen(g)	Use as catalyst
Liquid solution	Water(l)	Sugar(s)	Sugar syrup
	Water(l)	Ethanol(l)	Alcoholic beverages
	Water(l)	CO2(g)	Carbonated drinks
Gaseous solution	Air(g)	Particles(s)	Smoke
	Air(g)	Water(l)	Fog you have seen in the morning
	Nitrogen(g)	O2(g)	Air is the solution of all gases

Every solution has its unique properties such as taste, texture, viscosity, smell and so on. For example, beer, whisky, wine, and vodka are all alcoholic beverages, but every one of them is different from others. Each one has different alcohol content. Sugar syrups used for different purposes have different sugar content. It means that the ratio of solute and solvent plays an important role and there are a number of ways to represent this ratio or the concentration of solution. "How much solute is dissolved in how much solvent?", let’s see how many different ways and units are there to answer this question:

Weight percentage (w/w)

Weight percentage of a solute =  Weight of solute in the solution  × 100

      Total weight of solution

If it is said that it’s a 10% glucose solution. That means 10g glucose is dissolved in 90g of water.

Mass percentage

Mass percentage of a solute =  Mass of solute in the solution  × 100

       Total mass of solution

let’s try to understand it by taking an example of water. Find out the mass percentage of Oxygen in the water molecule.

mass of oxygen is 16 and molecular mass of water is 18 so,

Mass percentage of a Oxygen =  16 × 100

    18

           = 88.88%

Volume percentage (v/v)

 Volume percentage of a solute =  Volume of solute in the solution  × 100

       Total volume of solution

In laboratory we often use this unit of concentration. For example, 3% (v/v) hydrogen peroxide H₂O₂solution means 3ml H₂O₂ is dissolved in water such that the total volume of solution is 100ml.

Weight by Volume percentage (w/v)

w/v % of a solute = weight of solute in the solution  × 100

Total volume of solution

This unit commonly used in pharmacy. Here we measure the weight of solute in grams dissolved in 100ml of solution. For example: 5% (w/v) saline solution is means 5g NaCl is dissolved in water to make 100ml of solution.

Parts per million (ppm)

You must have heard this term when talking about pollutants in air or water. When a solute is present in very little quantity then its concentration is expressed in ppm.

Parts per million solute =          Number of parts of the solute in the solution     × 10⁶

 Total number of parts of all components of the solution

For example, 6×10^-3g of oxygen O₂ is dissolved in 1liter of sea water ( which weighs 1030g), let's express its concentration in ppm.

  Parts per million O₂  =   6×10^-3g × 10⁶

     1030

                                   = 5.8ppm

Mole fraction(x)

Mole fraction of a component  =         Number of moles of the component in the solution 

                    Total number of moles of all components of the solution

you can calculate the number of moles by dividing the given weight by its molecular weight.

number of moles = given weight in g / molecular weight

let’s take an example: suppose there is a solution of A, B and C and the number of moles of A, B and C are n_A, n_B and n_Crespectively.

Mole fraction of component A (x_A)  =         n_A     

                                                            ( n_A + n_B + n_C)

Similarly you can calculate the mole fraction of B and C. The sum of the mole fractions of all the components of a solution is always unity. let’s check it:

x_A + x_B + x_C = ?

On putting the mole fractions in the above equation

       n_A                           +        n_B            +        n_C               

 ( n_A + n_B + n_C)       ( n_A + n_B + n_C)    ( n_A + n_B + n_C)

    n_A  +  n_B + n_C   

   ( n_A + n_B + n_C)

= 1

So, x_A + x_B + x_C = 1

Mole itself is an important measuring unit and mole fraction is useful when we talk about gaseous mixture or solutions of different components. You will find its applications when we discuss the properties of solutions.

Molarity (M)

In laboratory you must have seen the sign ‘M’ on reagent bottles. It is the number of moles of solute dissolved in one liter of solution.

Molarity M  =      Number of moles of solute  

    Volume of solution liter

or,

Molarity M  =      Number of moles of solute ×1000

        Volume of solution ml

For example 2M HCl solution means 2moles of HCl is dissolved in 1liter. Let’s calculate the molarity of a solution containing 5g NaOH in 500ml solution.

Molecular weight of NaOH = 40

40g NaOH contains 1 mole

so 5g NaOH will contain = 5g/ 40g = 1/8 mole

and given volume of the solution is 500ml

M = 1 ×1000

        8 × 500

M=1×2/8 =1/4

M = 0.25M

Molarity of a solution depends on temperature because volume of the solution may vary with temperature.

Normality (N)

It is the number of equivalent weight present in 1 liter of solution. Equivalent weight is the amount of substance which gives 1mol of H⁺ or 1mol of electrons. Let's try to understand it in a simpler way, 1 mol of HCl can give 1mol H⁺ so its equivalent weight will be same as its molecular weight. 1 mol H₂SO₄ can give 2mol H⁺ so its equivalent weight will be half of its molecular weight.

Normality (N)  =    Number of equivalents of solute  

      Volume of solvent in L

Number of equivalents  = given weight / equivalent weight

For example 1 N solution of HCl means that 1equivalent of HCl is dissolved in 1L solution.

Q. calculates the normality of solution containing 7.88g of HNO₃ per liter solution.

Equivalent weight of HNO₃ = molecular weight of HNO₃ = 63.02

number of equivalents in 7.88g = 7.88 / 63.02 = 0.125 eq

Normality (N)  =    Number of equivalents of solute  

           Volume of solvent in L

N = 0.125 / 1L

Answer = 0.125N

We will practice more problems on normality when we study redox reactions.

Molality (m)

It is defined as the number of moles of solute present in 1kg of solvent.

Molality (m)  =    Number of moles of solute  

       Mass of solvent in kg

for example 2m KI solution means that 2 moles of KI dissolved in 1kg of water. let’s calculate the molality of a solution containing 120g acetic acid in 100g water.

1 mole of acetic acid CH₃COOH =  60g

number of moles in 120g = 120/60 = 2moles

Molality  = 2 mole

                  0.1kg

answer:  20m

Molality is independent of temperature since mass doesn’t vary with temperature.

Let’s practice a few examples:

Q1. The density of a 2.03 M solution of acetic acid in water is 1.017 g/mL. Calculate the molality (m) of the solution.

To calculate the molality, first we will calculate the mass of acetic acid dissolved in 2.03M solution and then we will calculate the mass of solvent that is water.

Given solution contains 2.03 moles of acetic acid.

since molecular mass of acetic acid is 60g

then 2.03 moles = 2.03 × 60g = 121.80 g

In the given 2.03M solution 121.80 g acetic acid is dissolved in 1 L of solution. So we have to find out the weight of 1 L solution.

density of the solution is given 1.017 g/mL

Density = weight / volume

1.017 = wt / 1ml

So, wt = 1.017

1ml of solution weighs 1.017g

so 1000ml will weigh = 1.017× 1000 = 1017g

Now we know the weight of solution. But to calculate the molality we need to know the weight of solvent.

weight of solvent = weight of solution - weight of solute

                            =1017- 121.80

                            =895.2g = 0.8952kg

now we can calculate the molality of solution

molality = 2.03moles

                0.8952kg

Answer = 2.26m

Q2. Calculate the molarity of a sulphuric acid H₂SO₄ solution of density 1.198g/cm³, containing 27% sulphuric acid by weight.

Hint: To calculate M, we need to find out the weight of H₂SO₄ present in given solution, then we can calculate the moles and then we calculate M.

it is given that,

1cm³solution weighs 1.198 g and it contains 27% H₂SO₄

= 27 × 1.198  = 0.32g H₂SO₄

         100

molecular weight of H₂SO₄ = 98.1

number of moles present in 0.32g = 0.32/ 98.1 = 0.0033 moles

1cm³ = 1ml = 1× 10^-3 L

M = number of moles of solute

            volume in L

M = 0.0033  

       1 × 10^-3

M = 3.3 mol/L = 3.3M

Q3. How many g of CaCl₂ should be added to 300mL water to make 2.46m solution?

molecular weight of CaCl₂ = 111

Molality (m)  =    Number of moles of solute  

     Mass of solvent in kg

density of water is 1 g/mL

so the weight of 300ml water will be = 300g = 0.3kg

Molality (m)  =    (weight/ molecular weight of solute)  

             Mass of solvent in kg

2.46m  =    (weight/ 111)  

  0.3kg

2.46m × 0.3kg × 111 =    weight

Answer: 81.91g of CaCl₂ should be added.

Q4. what is the mole fraction of the solute in 1m aqueous solution?

1m aqueous solution means 1mole of solute is dissolved in 1kg of water.

molecular weight of water = 18

number of moles of water in 1000 g = 1000/18 = 55.5 moles

Mole fraction (x) of a component  =         Number of moles of the component in the solution 

                          Total number of moles of all components of the solution

x  =    1  

       1+ 55.5

answer:  0.017

In the next post we will discuss more about solutions.

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Collision Theory

Max Trautz and William Cudmore McCullgh Lewis gave collision theory to interpret the exponential factor in the Arrhenius equation. Their theory was based on kinetic theory of gases. They assumed that the molecules are hard spheres and when they collide, the product is formed. It means that the rate depends on total number of collisions per unit volume, which is termed as collision number Z.

You can understand it by taking the example of Pool or Billiards. To sink a pool ball into the pocket, it is necessary to hit it by cue ball with desired kinetic energy. For a successful shot, the orientation of cue ball is as important as the kinetic energy.

Let’s take an example of a real reaction, the formation of methanol from methyl bromide.

CH₃Br + OH^- ⟶ CH₃OH +  Br ^–

As you know, the hybridization of methylbromide is sp³ and its shape is tetrahedral, but it gains inverted umbrella shape for this reaction. Formation of methanol depends on the orientation of methylbromide when OH^-collides with it.   If OH^- attacks from the methyl side, it leads to the formation of methanol. If OH^-attacks from the bromide side, it bounces back and no product is formed.

Right now I am trying to explain the importance of orientation in any reaction. We will discuss the detailed reaction mechanisms in our future posts of organic reaction mechanisms.

Trautz and Lewis first took a simple reaction between similar molecules to calculate the collision factor Z.

2A(g) ⟶ Product

Collision between A-A is given by equation:

Z_A-A  = ½ √2π  σ²ū N²_A              .......(1)

where π= 3.14,

σ is the distance,

N_Ais the number of molecules per unit volume,

ū is molecular mean velocity = (8k_BT/πm)^1/2

where ,

k_Bis the Boltzmann constant,

T is the temperature in Kelvin,

m is the molecular mass,

On putting the value of ū in equation 1:

Z_A-A  = ½ √2 πσ²(8k_BT/π m)^1/2 N²_A

Z_A-A  = 2 σ² N²_A(π k_B T/m)^1/2                  .......(2)

This quantity is known as collision number Z and its unit is m^-3 s^-1. This equation is restricted to the reactions between same molecules. They modified the equation of collision number Z for the reactions which involve two different molecules like A and B.

For the collision between two different molecules like:

A + B ⟶ Product

Z_A-B  = d ² N_A N_B (8π k_BT)^1/2                 .......(3)

                                    µ

Where d is the distance between the centres of A and B when collision occurs or it is the sum of the radii of A and B.

µ is the reduced mass =  m_A m_B

                                       m_A + m_B

On putting the value of µ in equation 3, we get:

Z_A-B  = d ² N_A N_B (8π k_BT m_A + m_B)^1/2    

                                               m_A m_B

Trautz and Lewis suggested that multiplying collision number with Arrhenius Factor gives the rate of formation of product (ν) in terms of number of molecules formed per unit volume and per unit time.

ν  =  d ² N_A N_B (8π k_BT)^1/2  e^{-Ea / RT}         

                               µ

And to get the rate constant, they divided the above equation by N_A N_B and multiplied it by Avogadro constant L,

k = L d ² (8π k_BT)^1/2  e^{-Ea / RT}      

                     µ

k = Z_AB e^{-Ea / RT}

The pre-exponential factor in this equation is called the collision frequency factor and is denoted by symbols Z_AA (for same molecules) or Z_AB(for different molecules). They applied this theory to the following reaction:

2HI ⟶ H₂ + I₂

And the calculated value of Z was quite closer to the experimental value. This theory gained confidence but when it was applied to other reactions, the difference between experimental and calculated values was quite considerable. By a lucky coincidence,  Lewis and Trautz picked the reaction which gave them accurate results. In order to resolve the discrepancies, they introduced another pre-exponential factor known as the steric factor P, which represents the fraction of the total number of collisions that are effective and from correct orientation.

 k = PZ_ABe^{-Ea / RT}

The introduction of P did improve the collision theory, but it is quite complicated to estimate the value of P as it involves the orientation.

The hard sphere collision theory may give satisfactory explanation to the reactions which involve gaseous reactants and products but it fails to explain the reactions between ions or in solutions. In our next post of chemical kinetics, we will discuss yet another theory to decipher the chemical reactions, the Transition State Theory.

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