Lesson Notes By Weeks and Term - Senior Secondary 3
Qualitative and quantitative analysis 1
Qualitative and quantitative analysis 1
TERM - 2nd TERM
WEEK TWO
Class: Senior Secondary School 3
Age: 17 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Chemistry
Topic: Qualitative and quantitative analysis 1
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, indicator extracts from flowers, Bomb calorimeter, acids and bases.
INSTRUCTIONAL PROCEDURES
PERIOD 1-2
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PRESENTATION |
TEACHER’S ACTIVITY |
STUDENT’S ACTIVITY |
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STEP 1 INTRODUCTION |
The teacher explains the percentage purity of substances and also water of crystallization. |
Students pay attention |
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STEP 2 EXPLANATION |
Teacher guide students to calculate percentage purity of substances. |
Students pay attention and participate |
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STEP 3 DEMONSTRATIO N |
Teacher guide students to calculate the water of crystallization in a solution.. |
Students pay attention and participate |
|
STEP 4 NOTE TAKING |
The teacher writes a summarized note on the board |
The students copy the note in their books |
NOTE
QUANTITATIVE AND QUALITATIVE ANALYSIS 1
Redox titration cont'd.
In redox titration, the solutions commonly used are the oxidizing agents and reducing agents. Examples of oxidizing agents include KMnO4, I2, Na2C2O4, and Fe2+.
In acid base titration, the neutralization reaction are represented by molecular equation. e.g.
HCl(aq) + KOH(aq) 
KCl(aq) + H2O(l)
But in redox titrations only the net ionic equation is represented e.g.
2Mn-4 + 5C2O²-4 + 16H+ 2Mn²+ + 10CO2 + 8H2O
When an oxidizing agent (OA) is titrated against a reducing agent (RA) the concentrations of their solutions can be determined as in acid base titrations.
Example 1
A is a solution on KMNO4, B is 0.02 mol dm³ of ethanedioc acid and solution C is 1.0 mol dm³ of the concentrated tetraoxosulphate (VI) acid. Titrate A solution against 25 cm³ of solution B which has been acidified with 5 cm³ of solution C. Average volume of 16.5 cm³ of A was required for complete oxidation of solution B. From the results obtained calculate,
(a) concentration of solution A in mol dm³
(b) concentration of solution A in g dm³
Equation of reaction
2Mn-4 + 5C2O²-4 + 16H+ 2Mn²+ +10CO2 + 8H2O
(K = 39, Mn = 55, O = 16)
Solution
(a) CoA x VoA. + nCA
CRA x VRA. nRA
CoA x 16.5 = 2
0.02 x 25 5
CoA = 0.02 x 25 x 2
16.5 x 5. = 0.012112 mol dm³
= 0.0121 mol dm³
(b) Conc in g dm³ of B
Molar mass of KMnO4
= 39 + 55 + 64 + 158 g mol
Mass conc. of A = molar conc. x molar mass
= 0.0121 x 158
= 1.91 g dm³
NOTE: In this redox titration, add KMNO4 from the burette until the solution is permanently pink, that is the end point.
KMNO4 is useful oxidizing agent. However, it has the disadvantage that it is sometimes too strong.for example, it will oxidize chloride ion to chlorine
2Cl- Cl2 + 2e-
And it must not be used in the titrations, if chloride is present. In this case, a weaker oxidizing agent must be used and potassium heptaoxodichromate (VI) is a suitable one
Cr2O²-7 + 14H + 6e- 2Cr³+ + 7H2O
The chromate ion is yellow and chromium(III) salt is green, so that there is not the sharp end point as with KMNO4. In this case an indicator must be used and a common one is diphenylamine. The reduced form is colourless, which the oxidized form is dark blue.
Percentage purity
Volumetric analysis is often used to determine the purity of substances (usually expressed as percentage purity). It is especially useful in determining the amount of dissolved impurities in water from various sources.
Example 2. 12 g of a mixture of anhydrous sodium trioxicarbonate (IV), Na2CO3, and sodium chloride, NaCl were made up to 1 dm³ of aqueous solution. 25 cm³ of this mixture were neutralized 0.2M hydrochloric acid, HCl.
Calculate the mass of the sodium chloride impurity. Find also the percentage purity of trioxicarbonate IV sample. (Na2CO3 = 106, HCl = 36.5)
Equation of reaction
Na2CO3 (aq) + 2HCl(aq) 2NaCl(aq) + H2O (l) + CO2 (g)
mole ratio 1 : 2
CAVA. = NA. mole ratio of acid
CBVB. NB base
= 0.2mol dm-¹ x 20 cm³ CB. = 2
CB x 25 1
CB = 0.2 x 20
25 x 2 = 0.08 mol dm³
i.e. molar mass of Na2CO3 = 0.08 mol dm3
Mass concentration of Na2CO3 = 0.08 x 106 = 8.48g dm³
p = C x M = 0.08 mol dm³ 106 g/mol
= 8.48 g dm³
Therefore, mass of NaCl impurity = 12 - 8.48 = 3.52g
Percentage purity of the Na2CO3 sample
= 8.48. x 100%
12
= 70.7%
Molar mass, water of crystallization and solubility of the substance
Volumetric analysis can be used to calculate
(a) the molar mass of the metal acid, or base;
(b) the number if molecules of water of crystallization of a hydrate,
(c) the solubility of a substance in water.
Example 3. Some crystals of washing soda were exposed to the atmosphere for efflorescence to take place. 60g of this partly effloresced washing soda, Na2CO3.yH2O, were dissolved in the 500cm³ of water. 25cm3 of this trioxicarbonate IV solution is required 32.10 cm³ of 0.097 mol dm³- hydrochloric acid, HCl, for complete neutralization. Calculate y. Hence, write the formula of the effloresced washing soda.( Na = 23, H = 1, C = 12, Cl = 35.5, O = 16)
Equation of reaction
Na2CO3.yH2O(s) + 2HCI(aq) 2NaCl(aq) + H2O (l) + CO2(g) + yH2O(l)
mole ratio 1 : 2
CAVA = 2
CBVB. 1
0.097 x 32.1. = 2
CB = 0.097 x 32.1
2 x 25
= 0.0623 mol dm³
Therefore, molar concentration of Na2CO3.yH2O = 0.0623 mol dm³
Mass concentration of Na2CO3.yH2O = 6.02 x 1000
500
= 12.04 g dm³
Molar mass of Na2CO3.yH2O = (106 + 18y) g
Molar mass = mass concentration
molar concentration
= M = C
C
106 + 18y = 12.04
0.0623
18y = 87.3
y = 4.9
Therefore, the formula of the effloresced washing soda is Na2CO3.yH2O
EVALUATION: A solution trioxonitrate(V) acid contained 0.67g in 100cm³. 31.0 cm³ of this solution neutralized 25cm³of a sodium trioxicarbonate IV solution. Calculate the concentration of the trioxicarbonate IV solution. (HNO3 = 63, Na2CO3 = 106)
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively