Class 11 Record Experiment 4

 EXPERIMENT 4 OBSERVATIONS AND CALCULATIONS:- (write in LHS OF THE RECORD)

AIM a):- PREPARATION OF STD SODIUM CARBONATE SOLUTION – 

 Molar mass of solid Na2CO3 is 106  g/mol.

Weight of watch glass                                                                    = 10g

Weight of watch glass + Na2CO3 salt                                           = 12.65 g

Weight of Na2CO3 dissolved in 250ml of distilled water         = 2.65 g

Molarity of Na2CO3 solution prepared                                        = (2.65 × 4 )/106 = M/10

AIM b) and c) :-

Volume of Na2CO3 solution           (V1)                                     = 20ml 

Molarity of  Na2CO3 solution         (M1)                                    = (M/10)

SL.NO.	VOLUME OF sodium carbonate  SOLUTION    (V1)	INITIAL  READING OF BURETTE	FINAL READING  OF BURETTE	VOLUME OF  HCl USED
1.	20 ml	..........ml	..........ml	..........ml
2.	20 ml	..........ml	..........ml	..........ml
3.	20 ml	..........ml	..........ml	..........ml

Standard Na2CO3 solution (M/10) against unknown HCl solution using methyl orange indicator                               END POINT : GOLDEN YELLOW TO  PALE RED PINK COLOUR

 PRINCIPLE OF VOLUMETRIC ANALYSIS    Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + H2O (l) + CO2 (g)

From  equation:-  (M1V1) / n1 = (M2V2) / n2

n1  = stoichiometric coefficient of reactant 1 that is sodium carbonate  = 1

n2  = stoichiometric coefficient of reactant 2 that is HCl          = 2

Volume of HCl solution      = (V2) = ...........ml

Molarity of HCl solution = (M2) = (M1V1n2)/(n1V2)    =  ---M 

Molecular mass of HCl solution        = 36.5 g/mol

Strength of HCl =   M2X 36.5 = Molarity X molar mass =............... × 36.5 = ................g/l

CONCLUSIONS (LHS)   a) Mass of Na2CO3 required to produce 0.1M 250ml = 2.65g

b) Molarity of the HCl solution = ……………M 

c)The strength of the given hydrochloric acid solution is _______ g/L.

EXPT NO : 4              AIM (write in RHS OF THE RECORD)                                        DATE : 

To determine the molarity and strength of a given solution of hydrochloric acid solution by titrating it against a standard solution of 0.1M sodium carbonate solution.

CHEMICALS AND APPARATUS REQUIRED:-

M/10 sodium carbonate solution, HCl solution, methyl orange indicator, Burette stand, burette, pipette, conical flask, white tile, wash bottle.

THEORY – (write in RHS)

Sodium carbonate is Na2CO3.  The molar mass of solid sodium carbonate is 106 g/mol. 

The acid base titration follows the chemical reaction 

 Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + H2O (l) + CO2 (g)

Since hydrochloric acid is not a primary standard, a standard solution of sodium carbonate is prepared and used for standardisation of hydrochloric acid.

In acid base titration at the end point the amount of acid becomes chemically equivalent to the amount of base present. 

PROCEDURE (write in RHS)

(a) Preparation of 0.1M Standard Na2CO3 Solution

2.650 g of Na2CO3 is carefully weighed into a 250 ml standard flask and the dissolved in distilled water and made up to the mark.

(b) Titration of std Na2CO3 and unknown HCl Solution

1. Rinse the burette with the HCl solution and fill and note the initial reading.

2. Rinse the pipette with Na2CO3. Pipette out 20ml of the Na2CO3 solution into a clean conical flask. Add 1-2 drops of methyl orange indicator into it, the solution turns golden yellow and place it just below the nozzle of the burette over a white glazed tile.

3. Now run HCl solution slowly and dropwise into the conical flask till a very faint permanent light red pink colour is obtained. Repeat the titration for concordant values (similar values)

Burette - solution unknown HCl solution                                    

Pipette –  standard Na2CO3 (0.1M )

Indicator = Methyl orange   

End point = from golden yellow colour to Light red pink colour

Results (write in RHS)   

 a) Mass of Na2CO3 required to produce 0.1M 250ml = 2.65g

b) Molarity of the HCl solution = ……………M 

c)The strength of the given hydrochloric acid solution is _______ g/L.