Double Indicator Titration of NaOH and Na2CO3 Explained

APPARTUS REQUIRED:

  1. burette
  2. pipette
  3. conical flask
  4. stand with clamp
  5. volumetric flask

CHEMICAL REQUIRED:

  1. sodium hydroxide solution(NAOH)
  2. sodium carbonate
  3. sulphuric acid
  4. phenolphthalein
  5. methyl orange    

THEORY:

 The technique to know the strength of given unknown solution with the help of known standard solution is called as titration.

The mixture of NAOH and NA2CO3 is analyzed by double titration method. The mixture solution is titrated first with phenolphthalein and then with methyl orange.                   

 The reactions involved are as follows:

2NAOH + H2SO4   → NA2SO4 + 2H2O

2NA2CO3 + H2SO4 → 2NAHC03 + NA2SO4

2NAHCO3 + H2SO4 → NA2SO4 + 2H2 +2C02

  The ph at the equivalence point of reaction (i) and (ii) is greater than 7 and phenolphthalein gives their end point but when reaction (iii) begins the evolution of co2 decreases the ph which discharges the pink color of phenolphthalein before the equivalence point so methyl orange is used for this stage .

  The volume of standard and consumed in phenolphthalein corresponds to the whole amount of NAOH and half amount of NA2CO3 present in the mixture.

  Further volume of standard acid consumed in methyl orange corresponds to the remaining half amount of NA2CO3.

           The volume of standard acid consumed in phenolphthalein = X ml

           The volume of standard acid consumed in methyl orange = Y ml

           The volume of acid reacted with NAOH of alkali mixture = (X – Y) ml

           The volume of acid reacted with NA2CO3 of mixture = 2Y ml

           The law of equivalence (chemical) states

             1g equ of acid = 1 g equ of base

            Or 100ml of 1N acid = 100 ml of 1N NAOH

                                               = 100 ml of 1N NA2CO3

                                             = 40 gm of NAOH   

                                             = 50 gm of   NA2CO3

PROCEDURE

  First of all the given apparatus used for titration i.e. burette, pipette, conical flask were cleaned properly with water. Then, the burette was rinsed with standard acid H2SO4 N/10(F= 1.02) and was filled with standard acid. The pipette was rinsed with alkali mixture solution in a conical flask. After that 1 to 2 drops of phenolphthalein indicator was added in conical flask where the pink color appeared as it was titrated till the pink color was discharged. As the experiment was analyzed with the help of double indicator method of titration 1 to 2 drops of methyl orange was added in it where the light yellow color was appeared and again it was titrated then fairly pink color was appeared. Lastly the process was repeated for concurrent reading.

OBSERVATION TABLE:

strength of H2SO4 solution = N/10(f=1.02)

Burette   reading   standard   H2SO4 In phenolphthalein

S. N.Volume    of alkaliinitialfinaldifference Concurrent
1.10 ml01414 
2.10 ml1829.511.5 11.5 ml
3.10 ml3445.511.5 

Burette   reading   standard   H2SO4 In methyl orange

S. N.Volume    of alkaliinitialfinaldifference Concurrent
1.10 ml14184 
2.10 ml29.5344.5 11.5 ml
3.10 ml45.549.54.5 

CALCULATION:

Volume of phenolphthalein of end point (X) = 11.5 ml

Volume of methyl orange of end point (Y) = 4.5 ml

Volume of H2SO4 consumed by NAOH (X – Y) = (11.5 – 4.5) ml

                                                                               = 7ml

Volume of H2SO4 consumed by NA2CO3 = 2Y = 2 × `4.5 = 9 ml

 For NAOH

Volume of NAOH (V1) = 10ml                     strength of NAOH (N1) =?

Volume of H2SO4 (V2) = 7 ml                        strength of H2SO4 (N2) = f/10 = 0.102

Then, N1V1 = N2V2

     Or, N1 = 0.102×7 ml                                                   

                      10 ml

                 =     0.0714

 And gm/liter = 0.0714 × 40

                       = 2.867 gm/liter

For NA2CO3

Volume of NA2CO3 (V1) = 10 ml                   strength of NA2CO3 (N1) =?

Volume of H2SO4 (V2) = 9 ml                        strength of H2SO4 (N2)   = f/10 = 0.102

Then, N1V1 = N2V2

     Or, N1 = 0.102× 9ml                                                   

                      10 ml

                = 0.0918

And gm/liter = 0.0918 × 53

                       = 4.865 gm/liter

RESULT:

From above observation, the amount of NAOH and NA2CO3 was found to be 2.856 gm/liter and 4.867 gm/liter respectively.

CONCLUSION:

 Hence, amount of NAOH and NA2CO3 can be determined by double titration method.

PRECAUTION:

  1. Reading should be noted carefully.
  2. Chemical should be used carefully.
  3. Apparatus should be cleaned before and after use.

More from Chemistry Practical | 1st Year | BSc |T.U.

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