Class 11 Record Experiment 2

 EXPERIMENT 2  OBSERVATIONS AND CALCULATIONS:- (write in LHS)

AIM a):- PREPARATION OF STANDARD SOLUTION OF crystalline oxalic acid  H2C2O42H2O (Molar mass of solid oxalic acid is 126 g/mol.)

Weight of watch glass                                                                    = 10g

Weight of watch glass + oxalic acid  salt                                    = 13.15 g

Weight of oxalic acid  dissolved in 250ml of distilled water = 3.15 g

Molarity of oxalic acid  solution prepared                                 = (3.15 × 4)/126     =M/10

AIM b) and c) :-

Volume of oxalic acid solution           (V1)                                     = 20ml 

Molarity of oxalic acid  solution         (M1)                                     = (M/10)

SL.NO.	VOLUME OF oxalic acid   SOLUTION    (V1)	INITIAL  READING OF BURETTE	FINAL READING  OF BURETTE	VOLUME OF  NaOH USED
1.	20 ml	..........ml	..........ml	..........ml
2.	20 ml	..........ml	..........ml	..........ml
3.	20 ml	..........ml	..........ml	..........ml

Standard Oxalic salt solution (M/10)  against  unknown NaOH solution using phenolphthalein indicator END POINT : APPEARANCE OF PERMANENT PALE PINK COLOUR

 PRINCIPLE OF VOLUMETRIC ANALYSIS  The equation is  H2C2O4 + 2NaOH → Na2C2O4  + 2H2O

From  equation:- 

n1  = stoichiometric coefficient of reactant 1 oxalic acid = 1

n2  = stoichiometric coefficient of reactant 2 NaOH          = 2

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

Therefore  Molarity of NaOH  solution                               = (M2)             =  (M1V1n2)/(n1V2)

  =  -----------M 

Molecular mass of NaOH solution        = 40g/mol

Strength of sodium hydroxide =   M2X 40   =Molarity X molar mass 

                            =............... × 40                                 = ................g/l

EXPT NO :2                VOLUMETRIC ANALYSIS (write in RHS OF THE RECORD )           DATE :                                                                     

AIM (write in RHS)

To determine the molarity and strength of a given solution of sodium hydroxide solution by titrating it against a standard solution of oxalic acid.

(M1V1) / n1 = (m2V2) / n2

CHEMICALS AND APPARATUS REQUIRED:-

M/10 oxalic acid solution,NaOH solution, phenolphthalein indicator, Burette stand, burette, pipette, conical flask, white tile, wash bottle.

THEORY – (write in RHS)

Crystalline oxalic acid is H2C2O42H2O.  The molar mass of solid oxalic acid  is 126 g/mol. 

The acid base titration follow the chemical reaction 

H2C2O4 .2H2O + 2NaOH → Na2C2O4  + 4H2O

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

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 Oxalic Acid Solution

3.15 g of crystalline oxalic acid is carefully weighed into a 250 ml standard flask and the dissolved in distilled water and made upto the mark.

(b) Titration of Sodium Hydroxide and Oxalic Acid Solution

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

2. Rinse the pipette with oxalic acid . Pipette out 20ml of the oxalic acid solution into a clean conical flask. Add 1-2 drops of phenolphthalein indicator into it and place it just below the nozzle of the burette over a white glazed tile.

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

Burette – standard oxalic acid solution

Pipette – unknown NaOH solution

Indicator = Phenolphthalein

End point = Light pink colour

RESULTS (write in RHS)     

i)  Mass of oxalic acid  dissolved in 250ml of distilled water  to make 0.1M solution = 3.15 g

ii) Molarity of the NaOH solution = ……………M 

iii)The strength of the given sodium hydroxide solution is _______ g/L.