Week 9 – SS3 Second Term Chemistry Notes

WEEK NINE
TOPIC: VOLUMETRIC ANALYSIS
CONTENT

• Calculation Based on Percentage Purity and Impurity of substances.
• Percentage/amount of water of crystallization,
• Molar mass of the acidic base
• Solubility of substances
• Volume of gases
• Mole ratio of acid to base

 Volumetric Analysis
Volumetric analysis involves acid base titration.

 Mole Ratio
Mole ratio is the ratio of the reacting species. This determines the ratio of the acid that would react with the base.
Examples are
1.     H₂SO₄ + 2NaOH         Na₂SO₄ + 2H₂O

 CaVa = ½
CbVb
2.    2HCl + Na₂CO₃ 2NaCl +H₂O + CO₂
    CaVa = 2
    CbVb 1

 EVALUATION

1. What is volumetric analysis
2. Give the ratio of the reaction species in the following chemical reactions

   a. CaCO₃ + 2 HCl      CaCl₂ + H₂O + CO₂
   b. KHCO₃ + 2HCl     KCl + H₂O + CO₂

 Calculation Involving Titration
1. Mole Ratio
A is a solution of an acid hydrogen chloride .B is a solution of sodium trioxocarbonate(iv) containing 0.05 mole per dm³ solution A was titrated against 25cm³ of solution B, using methyl orange as indicator during the process, the following data were obtained.

 Burette reading (cm³)          Rough         1^st         2^nd         3^rd
Final burette reading (cm³)    24.65        48.95        24.30        24.30
Initial burette reading (cm³)    0.00        24.65        0.00        0.00
Volume of acid used (cm³)    24.65        24.30        24.30        24.30.

1. Calculate the average titre value
2. Calculate the concentration of the acid in moldm³.
3. Calculate the concentration of the acid in g/dm³.

The equation of the reaction
NaCO₃ + 2HCl      2NaCl +H₂O + CO₂
Solution
1. Average titre value = 24.30 + 24.30 + 24.30
    3
         = 24.30cm³

 2. Concentration of A in moldm³
from    

CaVa = Na
CbVb Nb
        Ca x 24.30 = 2
0.05 x 25     1
    Ca = 0.05 x 25 x 2
24.30
    Ca = 0.103moldm³.
        OR
From no of mole = Conc. In moldm-3 X vol/dm³
No of moles = 0.05 x 25
1000
equation of the reaction.
Na₂CO₃ + 2HCl     2NaCl + H₂O + CO₂

1. : 2

   1 mole of Na₂CO₃ react with 2 moles of HCl
   :. 0.00125 mole of Na₂CO₃ will require 0.00123 x 2 of HCl
   :. No of mole of A = 0.0025 mole
   From conc of A in moldm-3 = No of mole
        Volume in dm3
       = 0.0025 × 10000
        24.30
       1000
       
         0.0025 x 1000
           24.30.
       = 0.103moldm³
   3. Concentration of A in g/dm³
   
    From:- conc in g/dm3 = conc in moldm^-3 x molar mass

    Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.
   :. Conc in g/dm3 = 0.103 x 36.5
   = 3.76g/dm³

    PERCENTAGE PURITY AND IMPURITY
   During the titration process of an impure acid or base is titrated only the pure part of either acid or base react with the base or acid. Therefore the percentage (%) purity or impurity can be calculated.
   % purity = Conc in g/dm³ of pure solution X 100
   Conc in g/dm³ of impure solution 1
   % impurity = conc of impure – conc of pure X 100
   conc in g/dm³ of impure 1
   Mass of pure substance = Conc of pure in moldm^-3 x Molar Mass
   Mass of impurity = Conc of impure – pure

    Example
   A is a solution of 020mole of HCl per dm³. B is a solution of an impure sodium trioxocarbonate(iv) containing 3.0g per 250cm³.
   a. Calculate the
   (i)   percentage purity of A
   (ii) percentage impurity of A
   Va = 20.40cm³ Vb = 25.00cm³

    The equation of reaction
   Na₂CO₃ + 2HCl     2NaCl + H₂O + CO₂
   (Na = 23 C= 12 O = 16 H = 1, Cl = 35.5)
   Solution

   CaVa = na
   CbVb nb

    
   0.20 x 20.40 = 2
   25 x cb 1

    Cb = 0.20 x 20.40 x 1
   25 x 2
   Cb = 0.0823 moldm³
   Conc in g/dm³ of pure
   From
   Conc in g/dm³ = Moldm³ x molar mass
   Molar mass of Na₂CO³ = 2(23) + 12 + 3 (16)
   Molar mass of Na₂CO₃ = 106g/mol
   :. Conc in g/dm³ of pure = 0.082 x 106
   = 8.692 g/dm³

    Conc of impure Na₂XO₃
   250 cm³ dissolve 3.0g of Na₂CO₃
   1 cm3 dissolves 3.0 X 1000
   250
   = 12.0g/dm³
   1. :. % purity = Conc of pure X 1000
   Conc of impure 1
   = 8. 69 X 100
   12 1
   = 72.4%

    % impurity = Conc of impure – pure X 100
   Conc of impure 1
   % impurity = 12 – 8.6g X 100
   12         1
   = 27.6%

    PERCENTAGE AMOUNT OF WATER OF CRYSTALLIZATION
   Water of crystallization in the wager given off when an hydrated salt is heated or exposed to the atmosphere
   Hydrated salt does not contain water
   Amount of water of crystallization is calculated as follows:

   Conc of anhydrous     = moalr mass of anhydrous
   Conc of hydrated         molar mass of hydrated

    Percentage Water of Crystallization is calculated as follows:
   % water of crystallization    = Hydrated – Anhydrous X 100
   Hydrated 1

    Example
   Solution A is a solution of hydrogen chloride acid containing 0.095 moldm₃ of solution.
   B is a solution of hydrated salt Na₂CO₃. XH₂O containing 3.94g which was made up to 250cm₃ of solution with distilled water
   Va = 29.00cm³, Vb = 25.00cm^3.
   Calculate the
   I. value of X
   II. percentage of water of crystallization.

    
    Equation of the reaction
   Na₂CO₃.XH₂O + 2HCl         2NaCl + H₂O + H₂O + CO₂
   Solution
   i. Value of x
   From

   CaVa = Na     CaVa = 2
   CbVb Nb CbVb 1

    
   0.095 x 29 = 2
   Cb x 25 1
   Cb = 0.095 x 29 x 1
   25 x 2.
   Cb = 0.0550moldm³
   Conc in g/dm3 of Na₂CO₃ = moldm^-3 x m.m
   Molar mass of Na₂CO₃ = 2 (23) + 12 + 3(16) = 106 g/mol
   Conc in g/dm³ = 0.055 x 106 = 5.83 g/dm³
   Conc in g/dm³ of hydrated:

   Mass X 1000
   Volume 1

    Conc in g/dm³ = 3.94 x 1000
   250
   = 15.8g/dm³

   Conc of anhydrous = molar mass of anhydrous

   Conc of hydrated     molar mass of hydrated.

    
   5.83 = 106

   15.76 106 x 18
   (106 x 18x) 5.83 = 106 x 15.76
   106 + 18x = 106 x 15.76
   5.83
   106 + 18x = 286. 55
   18x = 286.55 – 106
   18x = 180.55
   x = 180.55
   18.
   x = 10
   The salt is Na₂CO₃.10H₂O

    READING ASSIGNMENT
   Practical Chemistry by Makanjuola pages 1-15.
   New School Chemistry by Osei Yaw Ababio pages 165 – 183
   Practical Chemistry for Schools and Colleges pages 100 – 170

    GENERAL EVALUATION
   

   1. What is volumetric analysis
   2. Name five apparatus used in volumeric analysis.
   3. Define the following terms; a. Indicator b. Buffers c. pH scale

    WEEKEND ASSIGNMENT
   

   1. C + water give colourless solution (a) c is a soluble salt (b) c is partially dissolve in water (c) c is a filterate (d) c is a residue
   2. ____ is the apparatus use to convert vapor into liquid during distillation. (a) conical flask (b) distillation column (c) lie-big condenser (d) round bottom flask
   3. X which fumes in most air can be suitably stored (a) under paraffin or naphtha (b) In a white bottle (c) inside a corked conical flask (d) inside a burette.
   4. The observation in bubbling SO₂ into acidified KMnO₄ solution is (a) The solution turns to green (b) the solution becomes decolourized (c) no visible reaction (d) the solution turns steam
   5. The two substances that can give both H₂ and ZnSO₄ when added to H₂SO₄ are: (a) Magnesium and Zinc (b) Magnesium and CuO (c) Sodium and NaOH (d) iron and copper

    THEORY
   

   1. State what would observe on
      1. mixing Zinc dust with CuSO₄ solution
      2. adding concentrated HNO₃ to freshly prepared FeSO₄ solution
   2. A salt sample was suspected to be either Na₂CO₃ or NaHCO₃. A student who was required to identify it, tested a portion for solubility in water and for effects on litmus paper.
      1. What was the observation in each case?
      2. State the reason why the student’s procedure was unsuitable.
      3. Describe briefly how you would have identified the salt.