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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Qualitative analysis of Metal cations

In the last post we studied the solubility product of salts and leant how we can manipulate the solubility of any salt by applying the common ion effect. In the coming posts we will study the practical aspects of these two phenomenons.

The common metal cations are classified into five groups for systematic qualitative analysis. This classification is based on the differences of their behaviour against some reagents and solubilities of their chlorides, sulphides and carbonates. Each group of cations reacts with a particular reagent and forms precipitates. This particular reagent is known as the group reagent of corresponding group. 

Let’s see what are these groups and their characteristics, and which metal ion falls in which group.

Group I: Cations of this group are lead(II) Pb²⁺, mercury(I) Hg⁺ and silver(I) Ag⁺.

Group reagent of this group is dilute Hydrochloric acid. These cations form chloride precipitates with dilute HCl. Solubility product of Chlorides of lead, mercury and silver are lowest so they are precipitated first.

Group II: Cations of this group are divided in to two groups IIA and IIB on the basis of their solubility in ammonium polysulphide (NH₄)₂S_x. IIA group consists of mercury(II) Hg²⁺, lead(II) Pb²⁺, bismuth(III) Bi³⁺, copper(II) Cu²⁺, cadmium(II) Cd²⁺and they are insoluble in ammonium polysulphide. IIB group consists of arsenic(III) As³⁺, arsenic(V) As⁵⁺, antimony(III) Sb³⁺, antimony(V) Sb⁵⁺, tin(II) Sn²⁺ and tin(IV) Sn⁴⁺and these are soluble in ammonium polysulphide.

Group reagent: Hydrogen sulphide (gas or saturated aqueous solution). Cations of this group form precipitate in the form of sulphides on reacting with H₂S.

Did you notice? Lead(II) is common among two groups. That is because when lead reacts with 1^st group reagent dil HCl it forms lead chloride, while other cations get precipitated as chloride, it doesn’t precipitate completely because its chloride is more soluble than others. Its complete precipitation can be done as sulphide in 2^nd group.

Group III: cations of this group are cobalt(II) Co²⁺, Nickel(II) Ni²⁺, iron(II) Fe²⁺, iron(III) Fe³⁺, chromium(III) Cr³⁺, aluminium(III) Al³⁺, zinc(II) Zn²⁺, manganese(II) Mn²⁺, manganese(VII) Mn⁷⁺,  

Group reagent: ammonium sulphide solution or hydrogen sulphide gas in the presence of ammonia and ammonium chloride. Cations of this group don’t react with group reagents of 1^st or 2^ndgroup. They all precipitate with ammonium sulphide in the form of sulphides.

Group IV: calcium(II) Ca²⁺, strontium(II) Sr²⁺and barium(II) Ba²⁺

Group reagent: 1M solution of ammonium carbonate in neutral or alkaline medium. Cations of this group don’t react with previous three group reagents; they give precipitate with ammonium carbonate in the form of carbonates.

Group V: Magnesium(II) Mg²⁺, sodium(I) Na⁺, potassium(I) K⁺ and ammonium ion NH₄⁺. You might be surprised here to see ammonium with metal cations. It has similar characteristics to alkali metals. Its general properties are similar to that of potassium as the sizes of both ions are identical.

Group reagent: this group has no group reagent.  

These groups are arranged in increasing order of solubility of chloride, sulphides and carbonates of cations. For example, chlorides of I^stgroup cations have lowest solubility so they precipitate prior to others. Cations of any one group don’t react with group reagent of any other group. Let’s see how we can detect which metal cation is present in a given mixture.

Original Solution

On the basis of above information we can develop a strategy to find out which group of cations are present and then we can perform particular tests to detect that particular cation. First of all we have to dissolve the given mixture and prepare a clear solution. Take a small quantity of powdered mixture and check its solubility in different solvents to see which solvent dissolves it completely.

Water is the universal solvent so first we will try to dissolve the mixture in cold water, if it doesn’t dissolve then try hot water. If it doesn't dissolve in water then try following solvents:

a.      6M hydrochloric acid HCl

b.      Concentrated hydrochloric acid HCl

c.      8M nitric acid HNO₃

d.      Concentrated nitric acid HNO₃

e.      Auqa regia (3 part conc. HCl and 1 part conc. HNO₃)

Always check solubility in cold solvent then on warming. Once you have found the suitable solvent take 0.5-1g powdered mixture and prepare a solution for analysis. If mixture has been dissolved in conc. HCl, evaporate most of the acid. Then dilute it with water and make it up to 20-50 ml. If HNO₃or aqua regia has been used for dissolution then evaporate almost all acid then add small amount of HCl and again evaporate it to a small amount then dilute it with water. The volume of final solution must be 20-50 ml for analysis.

Scheme for the Separation of Group cations

We have prepared a solution for analysis, now we will try to find out which group cations are present in it.

Step 1: Add excess of dilute HCl (group reagent for group I) to the solution. If a white precipitate is obtained, it may contain I^stgroup cations. If there is no change on adding dilute HCl then proceed to second step.

Step 2: To the above acidified solution pass H₂S gas (group reagent of IInd group) in excess. If precipitate is obtained, it may contain IIA or IIB group cations. Filter the precipitate and wash it with dilute HCl then check the solubility of precipitate in ammonium polysulphide, if it is soluble then it may contain IIB cations otherwise IIA cations may be present. If no precipitate is formed on passing H₂S gas then follow step 3.

Step 3: Neutralize the above solution with ammonia NH₃and add ammonium polysulphide (NH₄)₂S_x (group reagent for III^rd group). If precipitate is obtained, it may contain group III cations. If you notice no changes then proceed to step 4.

Step 4: Add ammonium carbonate (NH₄)₂CO₃(group reagent for IV^th group) in excess. If precipitate is obtained, it may contain group IV cations. If no changes occur then follow step 5.

Step 5: Add disodium hydrogen phosphate Na₂HPO₄solution in excess. If precipitate formed, it may contain Mg²⁺cation. If no precipitate is formed then test for Na⁺ and K⁺.

Now you are able to detect groups of cations in a given mixture. In the coming posts we will discuss each group separately and learn how to perform confirmatory tests for particular cations.

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