TERM׃ 1ST TERM
WEEK TEN
Class: Senior Secondary School 2
Age: 16 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Chemistry
Topic: ACID BASE REACTION 2
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, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 1-2
PRESENTATION |
TEACHER’S ACTIVITY |
STUDENT’S ACTIVITY |
STEP 1 INTRODUCTION |
The teacher discusses acid base reaction
|
Students pay attention |
STEP 2 EXPLANATION |
Teacher explains the types of qualitative analysis. |
Students pay attention and participates |
STEP 3 NOTE TAKING |
The teacher writes a summarized note on the board |
The students copy the note in their books |
NOTE
ACID BASE REACTION 2
Titration as discussed earlier is a method used in volumetric analysis. In this method, a solution from a graduated vessel is added to a known volume of a second solution until the chemical reaction between the two is just completed. This is shown by a colour change in the resulting solution or in an added indicator.
In any titration, a standard solution, i.e. one with a concentration which is accurately known, must be used to react with unknown concentration. The reacting volumes of the solution are then used to calculate the unknown concentration of the solution.
NOTE: No suitable indicator for titration of weak acid and weak base. Any indicator could be used for strong acid and base.
At this level, volumetric analysis usually includes titration of
- acid against base or trioxicarbonate(IV),
- oxidizing agent against reducing agent,
- one substance against another substance, giving a precipitate.
Acid-base or acid-trioxocarbonate(IV) titrations are usually neutralization reactions. We use indicators to determine the end point of this type of titrations.
Molar concentration
Molar concentration is the concentration of a solution in moldm^-3. In titrations, the concentration of the reacting solution give an accurate picture of the quantitative behavior of the reacting particles. The following equations are useful in calculations involving concentration.
Amount of a substance = Number of particles
6.02 x 1023
Amount of a substance = Mass of a substance
molar mass
= N/M = mol-1
Amount of a substance = Mass of a substance
Mass of 1 mole of a substance
= m/M = gmol-1
Standard solution is a solution of known concentration. The amount of solute in a given volume of solvent must be known. For example, 1.06g of Na2CO3, dissolved in 1dm3 of water gives a standard solution.
Molar solution
A molar solution of a compound is one which contains bone mole or the molar mass of the compound in one dm3 of the solution.
The molar mass of sodium and potassium hydroxide are 40g and 56g respectively. Therefore, a molar solution of sodium hydroxide contains 1 mole or 40g of the hydroxide in 1dm3 of the solution, while a molar solution of the potassium hydroxide contains 1 mole or 56g of the hydroxide in 1dm3 of the solution.
Relationship between the molar concentration and mass concentration
Example 1. Calculate the volume of 0.25 moldm-3 solution of tetraoxosulphate(vi) acid, H2SO4, that will contain a mass of 4.5g of raw acid. (H= 1, S = 32, O = 16)
Solution
Molarity of the given H2SO4 = 0.25M
Molar mass of H2SO4 = 98gmol-1
Mass concentration(gmol-1) = molar concentration x molar mass
e = C x M = 0.25 x 98gmol-1
= 24.5gdm-3
This means 24.5g of H2SO4 are combined in 1000cm3
4.5g H2SO4 are contained in 4.5 x 1000
24.5
= 184cm3
I.e 184cm3 of 0.25moldm-3 solution of H2SO4 contains 4.5g of the raw acid.
Using formula method.
Conc. In mol dm-3= mass concentration x 1000/vol
molar mass
Molar mass of H2SO4 = 98gmol^-¹
0.25moldm-³ = 4.5/98 x 1000/volume
= 4.5 x 1000/98 x 0.25
= 183.7cm³
= 184cm³
Example 2. Calculate the volume of hydrogen chloride gas at STP that would yield 1.2 dm³ of 0.15moldm-3 aqueous hydrogen chloride solution.(Molar volume of all the gases at STP = 22.4dm³)
Solution
Concentration of HCl solution. = 0.15 moldm-3
1 dm³ of the HCl solution contains 0.25 mol of HCl.
Therefore, 1.2 dm³ of the solution contains 1.2 x 0.15
1
= 0.18mol of HCl.
1 mole of HCl gas at STP occupies 22.4dm³
Thus, 0.18 mole at STP occupies 0.18 x 22.4
= 4.03dm³
Hence, 4.03 of HCl gas at STP would be required to yield 1.2 dm³ of 0.15moldm-3 aqueous solution.
EVALUATION: - Differentiate between molar concentration and mass concentration
- A solution of sodium chloride (NaCl) with a molar concentration of 0.5mol/L. Calculate the mass concentration of the solution in grams per liter (g/L).
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively