QUANTITATVE ANALYSIS [NEUTRALIZATION TITRATION]

Name of teacher:           

Week:                                                                   Date:                                                     Time:

Period:                                 Duration: 1 HR 20 MIN.                                                  Average age of learners: 16YEARS

Subject:                               CHEMISTRY                                                                         Class: SS THREE

Topic:                                    QUANTITATIVE ANALYSIS

Sub topic:

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: beaker, beam balance

Entry behavior: The students have been taught density.

Objective: At the end of the lesson the students should be able to:

1.       State the apparatus commonly used in volumetric analysis.

2.       Titrate, Tabulate and record titration value.

3.       Solve problems involving molar concentration, mass concentration, and molar mass.

CONTENT

ANALYSIS VOLUMETRIC

The apparatus commonly used in volumetric analysis are: the volumetric flask; burette; pipette; conical flask; retort stand; funnel and a white tile.

The apparatus that are used in the preparation of a standard solution are: weighing bottle; measuring cylinder; spatula; stirring rod; Bunsen burner; tripod stand and wire gaze.

Recording in Titration

Titration work could be recorded thus:

1.       State the size of the pipette used in cm3

2.       Name the indicator used.

3.       Record your titrations in tabular form as shown below:

Burette Reading (cm3)

Rough or Trial(cm3)

1st Titre (cm3)

2nd Titre (cm3)

3rd Titre (cm3)

Final

 

 

 

 

Initial

 

 

 

 

Volume of acid used

 

 

 

 

Find the average volume of acid used from any two or more titre values that do not differ by more than 0.20cm3 (Rough titre value may be used in averaging as long as the difference of any two titre values is within 0.20cm3).

Titration is a technique for determining the volumes of two solutions which a chemical reaction is complete.

MOLE

This is the amount of substances which contains as many elementary entities as there are atoms in 12.00g of the  isotopes. The elementary entities may be atoms, molecules, ions, electron etc.

Mole =

Example

1.       Convert 5.3g Na2CO3 to mole.

Solution

 mass of Na2CO3 = 5.3g;                 Molar mass Na2CO3 = 2Na + C + 3O = (2x23) + 12 + (3x16) = 106gmol-1.

Mole =  =  = 0.05mol

2.       Convert 20g NaOH to mole.

Solution

Mass = 20g; molar mass of NaOH = Na + O + H = 23+16+1= 40gmol-1

Mole =  =  = 0.5mol

Standard Solution: This is a solution of known concentration or is a solution which contain definite amount of solute in a definite amount of solution.

Molar Solution: it is a solution which contains one mole of solute in 1dm3 solution.

Concentration

(a)    Moldm-3 – it is the amount of solute in mole dissolved in 1dm3 solution. C=  ; where C- concentration in moldm-3 or molar concentration, n= Amount of mole, V= volume in dm3.

(b)   Gram dm-3 – it is the amount of solute in grams dissolve in 1 dm3 of solution.

End Point- is the point at which the chemical reaction is complete during titration. The end point is detected with the help of indicator.

 Example

3.       Calculate the concentration in moldm-3 of 5g of Ca(OH)2 dissolved in 500cm3 solution.

Solution

Mass= 5g; Ca(OH)2 = Ca + 2O + 2H = 40 + (2x16) + (2x1) = 74gmol-1

Amount  =  =  = 0.0676mol

Amount = Molar conc. (moldm-3) x volume (dm3)

0.0676mol = Molar conc. x 500/1000 ; Molar conc. = 0.0676/ 0.5 = 0.1352 moldm-3.

    4.   2NaOH + H2SO4                                 Na2SO4 + 2H2O

If in the above reaction, 20cm3 of the acid solution requires 30cm3 of 0.12M NaOH solution. Find the concentration of the acid in moldm-3.

Solution

       2NaOH + H2SO4                                 Na2SO4 + 2H2O

        2mol       1mol                                             1mol    2mol

Na= 1; Nb =2; CA = ? ; CB = 0.12M; VA = 20cm3; VB = 30cm3

  =  ; CA =  =  = 0.09M

5.   Solution B is prepared by dissolving 6.5g of impured Na2CO3 per dm3. Solution A contains 1.1g H2SO4 per 250cm3 portion of B using methyl Orange as an indicator and stop titration when the end point is reached.

Tabulate your burette reading and find the average volume of acid used from your table to calculate:

(a)    Conc. of B in moldm-3     (b) Conc. of B in gdm-3   (c ) % purity of B (d) volume of CO2 released in cm3.

Na2CO3 + H2SO4                 Na2CO3 + H2O + CO2

Solution

Volume of pipette used = 25.0cm3; Indicator used = Methyl Orange

Burette Reading (cm3)

Rough or Trial(cm3)

1st Titre (cm3)

2nd Titre (cm3)

3rd Titre (cm3)

Final

23.40

23.00

23.10

23.00

Initial

0.00

0.00

0.00

0.00

Volume of acid used

23.40

23.00

23.10

23.00

Average volume of acid used =   =  = 23.33cm3

Mass = 1.1g; molar mass of H2SO4 = 2H + S + 4O= (2x1) + 32 + (4x16) = 98gmol-1.

Amount  =  =  = 0.0112mol

Amount = Molar conc. (moldm-3) x volume (dm3)

CA = 0.0112/ 0.25 = 0.0448 moldm-3

Na= 1; Nb =1; CB = ? ; CA = 0.0448M; VA = 23.33cm3; VB = 25.0cm3

  =  ; CB =  =  = 0.042M

(b)   Find concentration of B in gdm-3

Molar mass of Na2CO3 – 2Na + C + 3O = (2x23) + 12 + (3x16) = 106gmol-1

Mass concentration of B = molar concentration of B x Molar mass = 0.042 x 106 = 4.452g/dm3.

 ( c ) Find the % purity of B

6.50g of impure B contains 4.45g pure Na2CO3

100g of impure B will contains 4.45/6.50 x 100 = 68.49%

% purity of B = 68.49%

(d) Find the volume of CO2 released in cm3 from pure B

Na2CO3 + H2SO4                 Na2CO3 + H2O + CO2

106g of Na2CO3 is required to released 22400cm3 of CO2 at Stp

4.45g of Na2CO3 is required to release 22400/106 x 4.45g = 940.4cm3 of CO2.

6. Solution B contained 7.76g of hydrated washing soda crystal in 500cm3, 25cm3 of solution B neutralized 29.0cm3 of 0.092M Hydrochloric acid solution A. Calculate :

(i) the mass concentration of solution B in moldm-3.  (ii) the mass of X.

Solution

The equation for the reaction is : Na2CO3 + 2HCl               2NaCl + H2O  + CO2

na = 2; nb = 1; CA= 0.092M; CB= ?; VA = 29.0cm3; VB = 25cm3.

Mass concentration in gdm-3 of sol. B = 0.053 X 106 = 5.62gdm-3.

(ii) To find the value of X

If 500cm3 contains 7.76g of Na2CO3. XH2O

1000cm3 will contains 7.76/500 x 1000 = 15.52gdm-3

Using the formula,

 =

 = ;       = ;    106 x 15.52 = 5.62 (106 + 18x)

1645.12 = 592.72 + 101.16x;  101.16x = 1645.12 – 592.72;  101.16x  = 1049.4; x = 1049.4/ 101.16; x = 10.

Hence the value of X = 10. The formula is Na2CO3. 10H2O.

Note – To calculate the % of H2O of Crystallization =   X 100

PRESENTATION

Step I: The teacher explains the apparatus used in titration.

Step II: The teacher explains how to record the reading of titration.

Step III: The students chorus the apparatus needed for titration.

Step IV: The teacher leads the students in solving problem on volumetric analysis.

Step V: The teacher allows the students to ask questions.

EVALUATION

The teacher asks the following questions:

1.       State the apparatus commonly used in volumetric analysis.

2.       Explain how to record the reading of titration

ASSIGNMENT

In an experiment, 20.0cm3 portion of 0.065moldm-3 NaOH were titrated against dilute HCl. The table shows the result of the titration

Burette Reading

Rough

1st Titre

2nd Titre

Final reading (cm3)

23.50

46.60

47.40

Initial reading (cm3)

0.00

23.50

24.00

Volume of acid used (cm3)

23.50

23.10

23.40

 

i.      Name suitable indicator for the titration. Give a reason for your answer.

ii.     Give the colour of  the indicator :

I.                    In base     II. At the end point

iii.    What type of reaction is demonstrated by the experiment..

bi.   Write a balance equation for the reaction.

bii. Determine the average volume of acid used.

c.     Calculate (i) concentration of the acid in moldm-3.

cii.   Concentration of the acid in gdm-3.

ciii. Mass of HCl in 20cm3 of solution 

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