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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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.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

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.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.