12. To determine the strength of a given sodium chloride solution. We are provided with M/20 oxalic acid.

Aim:- To determine the strength of a given sodium chloride solution. We are provided with M/20 oxalic acid.

Theory:- Here, the sodium hydroxide solution is taken in burette and a known volume (20.0 ml) of oxalic acid is taken in titration flask. The titration is carried out using phenolphthalein as indicator.

Chemical equation:-

COOH + 2NaOH → COONa +2H₂O

     |                                |

COOH                        COONa

Indicator:- Phenolphthalein

End point:- colourless to pink (alkaline in burette)

Procedure:- 

1. Take a burette and wash it with water.
2. Rinse and then fill the burette with the given sodium hydroxide solution. Clamp it vertically in a burette stand.
3. Rinse the pipette with the given oxalic acid solution.
4. Pipette out 20 ml of oxalic acid solution in a washed titration flask.
5. Add 1-2 drops of phenolphthalein indicator into it and place it just below the nozzle of the burette over a white glazed tile.
6. Note down the lower meniscus of the solution in the burette and record it as the initial burette reading.
7. Now run sodium hydroxide solution and drop wise into the flask till a very faint permanent pink colour is obtained. Read the lower meniscus of the solution again and record it as the final burette solution.
8. Repeat the procedure until three concordant reading are obtained.

Observation:- Molarity of the given oxalic acid solution = M/20 volume of oxalic acid solution taken for each titration = 20 m/v.

S.No.	Initial reading of the burette	Final reading of the burette	Volume of the sodium hydroxide solution
1.	0	3.2	3.2 ml
2.	0	3.4	3.4 ml
3	0	3.2	3.2 ml

Calculation:- According to the equation, one mole of oxalic acid react with two mole of sodium hydroxide.

Therefore,

V_{oxalic acid}                         No. of moles of    x M oxalix acid              oxalic acid in the                                                balanced equation
 -----------------------        =   -------------------------------

V_NaOH  x  M_NaOH           No. of moles of                                                    NaOH in the                                                      balanced equation.

Molarity of the given oxalic acid solution = M/20

Therefore,

20.0 x 1/20             1

 --------------------- = -----

3.2 x M_NaOH         2

→ M_NaOH  = 2/3.2

→ M_NaOH = 20/32

→ M_NaOH = 5/8

Strength of NaOH = molarity x molar mass

                                 = 5/8 x 40

                                 = 25 g/l

Result:- The strength of the given sodium hydroxide solution is 
25g/l.