Lesson Notes By Weeks and Term - Senior Secondary 2

TERM׃ 1^ST 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

1. Define acid base reaction.
2. Discuss the types of qualitative analysis.

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