The Estimation Of Ethandioic Acid And Sodium Ethandioate In A Mixture Containing Both

Authors Avatar

The Estimation Of Ethandioic Acid And Sodium Ethandioate In A Mixture Containing Both

To determine the percentage of ethandioic acid and sodium ethandioate in the mixture will require two titrations, one in which the reactant will react with both the sodium ethandioate and the ethandioate and a second which will react with only one of them.  

From previous As investigational work we know that Sodium Hydroxide will react with ethandioate acid.  Ethandioic acid reacts with calcium, iron, sodium, magnesium and potassium to form salts called oxalates.  This reaction produces sodium ethandioate.  This reaction will therefore only react with the ethandioic acid and not the sodium ethandioate.  For this titration an indicator is required, phenolphthalein.  Phenolphthalein is a colourless weak acid, which dissociates to form pink anions.  When in alkaline conditions, the equilibrium of the reaction shifts to the right and the concentration of the anions is enough for a pink solution to be observed.  Phenolphthalein works as an indicator in this reaction as it is between a week acid and a strong base to form a solution with an alkaline pH.

From preliminary work we also know that potassium manganate will react with ethandioic acid and sodium ethandioate.  Potassium manganate (VII) is an oxidizing agent and is itself reduced during a reaction.  When using potassium manganate no indicator is required as when the reaction is at the end-point the manganate becomes colourless

We are told that the each of the two components is present by at least 30% by mass.  I have decided to assume that the mixture is 50% by mass of each component.

Join now!

Titration Of Ethandioic Acid With Sodium Hydroxide

2 NaOH     +     (COOH)2                                   Na2C2O4    +    2H2O

From the equation we can work out the required concentration of the Sodium Hydroxide

        Approximate concentration of (COOH)2  = 0.03mol dm3 

        Therefore approximate concentration of NaOH = 2 x 0.03

                                             = 0.06mol dm3

Therefore for this titration I will use Sodium Hydroxide with a concentration of    0.06mol dm3

...

This is a preview of the whole essay