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Estimation of oxalic acid - Volumetric Analysis | Chemistry Practical Laboratory Experiment | Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail |

Chapter: 12th Chemistry : Practicals

Estimation of oxalic acid

To estimate the amount of oxalic acid dissolved in 1250 ml of the given unknown solution volumetrically. For this you are given with a standard solution of HCl solution of normality 0.1010 N and sodium hydroxide solution as link solution.

Estimation of oxalic acid

 

Aim :

To estimate the amount of oxalic acid dissolved in 1250 ml of the given unknown solution volumetrically. For this you are given with a standard solution of HCl solution of normality 0.1010 N and sodium hydroxide solution as link solution.

 

Principle:

Neutralization of Sodium hydroxide by HCl is given below. To indicate the end point, phenolphthalein is used as an indicator.

NaOH + HCl NaCl + H2O

Neutralization of Sodium hydroxide by oxalic acid is given below. To indicate the end point, phenolphthalein is used as an indicator.


 

Short procedure:


 

Procedure :

Titration–I

(standard HCl )Vs (link NaOH)

Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up to the zero mark. Exactly 20 ml of NaOH is pipetted out into the clean, washed conical flask. To This solution 2 to 3 drops of phenolphthalein indicator is added and titrated against HCl solution from the burette. HCl is added drop wise till the pink colour disappears completely. Burette reading is noted and the same procedure is repeated to get concordant values.

Titration –I

(standard HCl )Vs (link NaOH)


 

Calculation :

Volume of  NaOH(link) solution       V1      =       20 ml

Normality NaOH(link) solution          N1    =       ? N

Volume of standard HCl solution     V2      =       ml

Normality of standard HCl solution N2      =       0.1010 N

 

According to normality equation:

V1 x N1  = V2  x N2

N1  = ? × 0.1010 / 20  =


Normality NaOH (link) solution N1 = ------------X---------N

Titration–II

(Unknown oxalic acid ) Vs (Link NaOH)

Burette is washed with water, rinsed with oxalic acid solution and filled with same oxalic acid solution up to the zero mark. Exactly 20 ml of NaOH solution is pipetted out into the clean, washed conical flask. To This solution 2 to 3 drops of phenolphthalein indicator is added and titrated against oxalic acid solution from the burette. oxalic acid is added drop wise till the pink colour disappears completely. Burette reading is noted and the same procedure is repeated to get concordant values.

Titration –II

(Link NaOH )Vs (Unknown oxalic acid solution)


 

Calculation :

Volume of Unknown oxalic acid solution V1      =       ml

Normality of Unknown oxalic acid solution             N1 =       ? N

Volume of  NaOH solution     V2      =       20 ml

Normality NaOH solution       N2      =       N

According to normality equation:

V1 x N1 = V2  x N2

N1 = V2  x N2 / V1


Normality of Unknown oxalic acid solution N1   = _____Y__________N

 

Weight calculation:

The amount of oxalic acid dissolved in 1 lit of the solution = (Normality) x (equivalent weight)

The amount of oxalic acid dissolved in 1250 ml of the solution = Normality x equivalentweight x 1250 / 1000


 

Report :

The amount of oxalic acid dissolved in 1250 ml of the solution = g

Tags : Volumetric Analysis | Chemistry Practical Laboratory Experiment Volumetric Analysis | Chemistry Practical Laboratory Experiment
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail


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