Form 3 Chemistry – QUANTITATIVE ANALYSIS AND VOLUMETRIC ANALYSIS

Definition;

Volumetric analysis,  is the process of determining the amount of substance in term of volume and concentration of reacting solutions.

Concentration;

This is the amount of substance per one dm³ (litre) of solution

Units are g/dm³, mole/dm³

Types of concentration

1. Molar concentration (morality) is the amount of substance in moles per one dm³ of solution (mol dm ^-3) it is denoted by M

For example; 0.1M H₂SO₄, 2M H₂SO₄ etc

Mass concentration is the amount of substance in games per one dm³ of solution (g/dm³)

Question:

a) Define (I) concentration (ii) molarity

b) Calculate mass concentration of the following

i) 4g of NaOH in 400cm³ of solution

ii) 10g of H₂SO₄ in 2dm³

iii) 2g of HCl in 4l

 Solution

a) Give the meaning and SI units of molarity

b) What is the molarity of a solution containing?

     i) 0.05mole in 80 cm³

    ii) 0.25mole in 2dm³

    iii) 0.5mole in 4l

Solution

2b). given n = 0.05mol                       

= 0.625mol/dm³

STANDARD VOLUMETRIC APPARATUS

The apparatus needed in volumetric analysis include

• Retort stand and clamps
  
• White tiles
  
• Burette – This is calibrated from 0 to 50 cm³ it measures volume of an acid
  
• Pipette – common capacity volumes are 25 cm³ or 20cm³. It measures volume of a base.
  
• Conical flask
  
• Filter funnel
  
• Reagent bottles
  
• Volumetric  flask
  
• Wash bottle
  
• Measuring cylinder
  
• Beakers
  
• Watch glass
  

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STANDARD SOLUTION

A standard solution is the one whose concentration is exactly known

For example; 10g of NaOH in 1dm³ of solution

         10g of KOH in 2dm³ of solution

The molarity of the above substance can be calculated because their concentration (in gdm^-3) and formulae are known

           

MOLAR SOLUTION

A molar solution is the solution which contains one mole of a solute dissolve in one dm³ of solution

Example; 4Og of NaOH dissolved in 1dm³ of solution

            98g of H₂SO₄ dissolved in 1 litre of solution

            132g of (NH₄) ₂ SO₄ dissolved in 1L of solution

 
 

TITRATION OF ACIDS AND BASES

Titration: is the process of adding an acid to a base until an indicator shows that the reaction is complete

What is acid base titration?

What is titration? 

This is a process of adding one solution to another until an indicator shows that the reaction is complete

CHOICE OF INDICATORS IN ACID – BASE TITRATION

                                                           
 

In an acid – base titration, the solution of the known concentration is added gradually to a solution of unknown concentration. When the unknown solution is exactly neutralized as shown by the color change of the indicator then at this point the number of moles of the solution with known concentration, for example; acid is equal to the number of moles of solution with unknown concentration, for example; Base. This point is known as End – point of titration.

                            

 MOLAR RATIO
The amount of moles present in the two different solution A and B can be symbolized as n_A and  n_B.
 
But  the moles in standard solution,
                      n = M x V
  Where by,
                   M = Molarity
                   V = Volume taken
Therefore, 
                 n_A = M_A x V_A  
                 n_B  = M_B x V_B  
So that, 
                
This is called molar ratio formula which can be used in titration to determine any unknown, if five items are known. But always n_{A and}   n_B identified from the balanced chemical equation of compound A and B.                                                      

QUESTION

1. 24cm³ of a solution of 0.1M KOH were exactly neutralized by 30cm³ of solution of sulphuric acid

i) mol/dm³       ii) g/dm³

Solution;

 Make the equation;

          

            n_B = 2               n_A = 1

            V_B = 24cm3      V_A = 30cm3

            M_B = 0.1M       M_A =?

                                                                                  

                        M_A = 0.04mol/dm³

ii) Concentration of acid in g/dm³

                                                                                                   

            Mass concentration = molarity x molar mass

                                                = 0.04mol/dm3 x 98g/mol

                                                = 3.92 g/dm³

 
 

ASSIGNMENT:

LAMBAT GENERAL QUESTION 29 ON WARDS

QUESTION

25cm³ of X₂CO₃ contains 10.6g/dm³ required 23.00 cm³   and 0.217M of  HCL for complete neutralization calculating

i)The molarity of X₂CO₃ solution

ii) The molar mass of X₂CO₃

iii) The atomic mass of X

iv)Name of element X

Solution

i) X₂Co₃ + 2HCL      2xCl + H₂O + CO₂

            n_B = 1                                         M_A = 0.217

            V_B = 25cm3                                n_A = 2

            M_B =?                                         V_A = 23cm³

                                                                           n_A = n_B

                           M_A V_A = M_B V_B

                                       

                                     = 0.09982mol/dm³

                                                                       

            0.09982 = 10.6

                  1    =    M

  

            = 106g/mol = 106.2g/mol

iii) X₂ CO₃

            2x + 12 + 48 = 106

            2x + 60=106

2x = 46                       

                        x = 23

                        = 23g

iv)The name of element X is   Sodium (Na)

      PERCENTAGE COMPOSITIONS OF PURE AND IMPURE COMPONENTS
The pure substance may be composed in pure substances without affecting its chemical properties. Through volumetric analysis a percentage composition may be determined.
   – Percentage purity
   – Percentage impurity.
Where by,
              
Density = mass concentration per cm³

EXAMPLES;

3g of impure Na₂CO₃ were made up to 250 cm³ of solution; 25cm³ of this solution required 21.0 cm³ of 0.105M HCl for complete neutralization

i) Write down the balanced equation for the reaction

ii) Calculate the molarity of pure Na₂CO₃

iii) Calculate concentration of pure Na₂CO₃

iv) Calculate % purity of Na₂CO₃

v) Calculate % impurity of Na₂CO₃

Solution i and ii         

              

 n_B = 1              n_A = 2                                

            M_B =?               M_A = 0.105

            V_B = 25cm³       V_A = 21cm³

            n_A = n_B

            M_AV_A = M_BV_B

           

            = 0.0441 M (pure Na₂C0₃)

It’s pure because sand wouldn’t react with the acid

(iii) Concentration = molarity x molar mass

                          = 0.0441 M x 106g/mol      

                        = 4.67g/dm³                                                                                                              

iv) 3.0g=250cm³

       x     = 1000cm³

           

            x = 12 g/dm³  

Percentage purity = 38.9%

           Percentage impurity= 100% – 38.9%

                        = 61.1 %                                                                                      

DILUTION LAW.
Is the law that describe or direct how to change and decrease molar concentration of the solution. The law state that “The product of initial concentration and its volume is equal to the product of final concentration and its volume”

           M₁ V₁   = M₂V₂

           Where, M₁ = initial concentration

                        M₂ = final concentration

                        V₁ = initial volume

                        V₂ = final volume

 
Example.
1. To what volume must 300cm³ of 6.0M NaOH be diluted to give a 0.40M solution?

Solution

            M₁ = 6.0M                   V₁ = 300cm³

            M₂ = 0.04M                 V₂ =?

            M₁V₁ = M₂V₂

                        V₂ = 4500cm³

DETERMINING WATER OF CRYSTALLIZATION.
Water of crystallization is the water that is bound within the crystals of substances. This water can be removed by heating. titration reaction are used to determine the amount of water of crystallization in a substance.

Questions
1. Determine the number of molecules of water of crystallization in the sample (Na₂CO₃ . XH₂O). If R.M.M is 286g/mol.

Solution
        Na₂CO₃ . XH₂O
        2(Na) + C + 3(O) + X(H₂O) = 286
        2(23) + 12 + 3(16) + X(18) = 286
        106  +  X(18) = 286
         18X = 286 – 106
         18X = 180
           X = 180/18
           X = 10
   Therefore, the number of moles of water = 10

2. a) How many grams of concentrated HBr should be used to prepare 500ml of 0.6 M HBr solution?
    b) Find the volume of stored solution.

Solution

 a) Given purity = 48%

            D = 1.5g

            V = 500ml

            M = 0.6M

            n = M x V

            = 0.5l x 0.6mol/l

            = 0.3mol

 

m = 0.3mol x 81g/mol = 24.3g

              

                             x = 50.6g

b) D = 1.5g/ml

   m = 50.6g                                        

   v =?

      

       V = 33.7ml

3. 1.5g of an acid Hx was dissolved in water and its solution made to 250cm³. If 30.2cm3 of this acid solution neutralized 25cm3 of 0.115M KOH, Calculate,
                       i) Molarity of Hx

                       ii) Molecule weight of Hx

                      iii) Name of radical x

Solution

i) V_A = 30.2cm³                       V_B = 25cm³

   M_A =?                                   M_B = 0.115M

   Mass of HX  = 1.5g

            KOH + HX                    KX + H₂O

n_B = 1

n_A = 1   thus M_A V_A = ^MB

                             

                                 = 0.095M                                                                                                       

                   

                   X = 6g/dm³ x 1000

Molar weight = 63g/mol                                                                                                                                                                                                                                           

iii) x must be a radical and the common ones are HCl, HNO₃, H₂SO₄, H₃PO₄, so it’s either HNO₃ or HCl.

            But HNO₃ = 63g/mol

Thus, x is nitrate (v) ion (NO₃^–)