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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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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Qualitative analysis of V group cations

Group 5 cations are magnesium (II) Mg²⁺, potassium (I) K⁺ and sodium (I) Na⁺. These cations do not react with hydrochloric acid HCl, hydrogen sulphide H₂S, ammonium sulphide and ammonium carbonate (NH₄)₂CO₃. 

Magnesium does show similar reactions to IV^th group cations; it forms basic magnesium carbonate MgCO₃.Mg(OH)­₂.5H₂O with IVth group reagent ammonium carbonate (NH₄)₂CO₃. But this basic magnesium carbonate is soluble in presence of ammonium salts and therefore it doesn’t precipitate with IV group cations.

5Mg²⁺ + 6CO₃^2- + 7H₂O ⟶ 4 MgCO₃.Mg(OH)­₂.5H₂O ↓ + 2HCO₃^-

NH₄⁺ + CO₃² ⇌ NH₃ + HCO₃^-

Magnesium carbonate doesn’t precipitate in the presence of ammonium salts due to common ion effect. Solubility product of magnesium carbonate is quite high and cannot be achieved with lower concentration of carbonate ions. High concentration of ammonium ions shifts the equilibrium in forward direction which decreases the concentration of carbonate ions causing the basic magnesium carbonate to remain soluble.

Take the filtrate of IV group in a porcelain dish and evaporate it to a pasty mass. Add 3ml concentrated nitric acid HNO₃ to dissolve it, evaporate again to dryness and heat until white fumes of ammonium salts cease to evolve. If you get a white residue, it means group V is present.

Add 4ml water to the residue, stir and warm it up for 1minute and then filter. We will test for Mg(II) in the residue, and for K(I) and Na(I) in the filtrate. 

Confirmatory test for Mg²⁺

Dissolve the residue in a few drops of dil HCl and add 2-3ml water. Divide the solution in two parts.

Part 1

Add a little ammonium chloride NH₃Cl solution followed by ammonical oxine reagent (take 1ml 2% 8-hydroxyquinoline solution and add 2M acetic acid followed by 5ml 2M ammonia solution, warm to dissolve any precipitated oxine) and heat to boiling for 1-2minutes or till the odour of ammonia becomes noticeable. You will get a pale yellow precipitate of magnesium oxine Mg(C₉H₆ON)₂.4H₂O which confirms the Mg²⁺ ion.

Part 2

Take 1-2 drops of test solution in a spot plate and add 2-3 drops of magneson I reagent (4-(4-Nitrophenylazo)-resorcinol) and add 1 drop of 2M sodium hydroxide NaOH to make it alkaline. Blue colouration or blue precipitate is formed depending on the concentration of magnesium.

We will test for K(I) and Na(I) in the filtrate we got above. If the residue of V^th group dissolves completely, dilute it up to 6ml and filter if necessary. Divide this solution into three equal parts to test magnesium (II) Mg²⁺, potassium (I) K⁺ and sodium (I) Na⁺. In first part, directly apply magneson test for the confirmation of Mg(II) ion and in other two parts test for K(I) and Na(I).

Confirmatory test for Na⁺

Add a little uranyl magnesium acetate reagent, shake and allow to stand for few minutes. Yellow crystalline precipitate of sodium magnesium uranyl acetate is formed. If precipitation doesn’t occur, add 1/3^rdvolume of ethanol; it helps in precipitation.

Na⁺ + Mg²⁺ + 3UO₂²⁺+ 9CH₃COO^- ⟶NaMg(UO₂)₃(CH₃COO)₉ ↓

If you perform flame test persistent yellow flame confirms Na⁺.

Confirmatory test for K⁺

Add a little sodium hexanitritocobaltate(III) solution and a few drops of 2M acetic acid. Stir and then allow it to stand for 1-2 minutes. Yellow precipitate of potassium hexanitritocobaltate(III) is obtained. If precipitation doesn’t occur immediately, warm it a little; it will accelerate the precipitation.

3K⁺ + [Co(NO₂)₆]^3- ⟶  K₃[Co(NO₂)₆] ↓

The precipitate is insoluble in dilute acetic acid. If larger amount of sodium is present or you have added excess of reagent, then a mixed salt K₂Na[Co(NO₂)₆] is formed.

We have successfully separated the metal cations. Unlike cations, there is no well-defined system for analysis of anions. In the coming posts of analytical chemistry we will discuss the tests for anions.

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