AIM- Find the concentration in gm per litre of unknown Oxalic acid solution given in bottle B. You are provided with standard crystalline Oxalic acid solution whose 12.68 gm of oxalic acid is disolved in 1 lit in bottle A. Intermediate solution is sodium Hydroxide and Phenolphthalin is used as an indicator.
Theory-It is acidic-base titration. It is a double titration. With phenolphthalin indicator it gives pink colour in base adn becomes colourless in acidic medium.
Chemical Reaction -
H2C2O4 + 2 NaOH Na2C2O4 + 2H2O
Apparaturs and Matericals required -
Burette, pipette, conical flask, funnel, burette stand, NaOH solution, Oxalic acid solution, indicator Phenolphthalin , distiled water etc.
Method-
a) Titration of std. Oxalic acid solution with NaOH solution
Fill burette with NaOH intermidiate solution and Pipette out 20 ml known Oxalic acid solution in conical flosk add 2-3 drops of phenolphthalin indicator. Do the titration as discussed in Ex. No. 01 and find the Concordant Reading.
b) Titration of Unknown Oxalic acid solution with intermediate NaOH solution
Fill burette with intermediate NaOH solution and pipette out 20 ml of unknown oxalic acid. Add 2-3 drops of phenolphthalin indicator. Repeat the process of titration and find the concordant reading.
Obseration Table 1
a) Titration between known Oxalic acid and intermediate solution.
Obseration Table 2
a) Titration between unknown Oxalic acid and intermediate solution.
Calculation -
a) Normality of std. sodium carbonate
Normality =
= g/l
b) Normality of std. Hydrochloric acid solution
N1V1 = N2V2
N1 = Normality of std.oxalic acid solution.= N
N2 = Molarity of Intermediate HCl solution. = ?
V1 = Volume of std. Oxalic acid solution. = 20 ml
V2 = Volume of Intermediate HCl solution. = x ml
N2 = =N
c) Normality of unknown Sodium Carbonate solution.
N3V3 = N4V4
N3 = Normality of unknown Na2CO3 solution.= ?
N4 = Normality of Intermediate HCl solution. = y N
V3 = Volume of unknown Na2CO3 solution.= 20 ml
V4 = Volume of Intermediate HCl solution. =y ml
N3 = =N
=N
Result- Normality of unknown sodium carbonate is N.
Theory-It is acidic-base titration. It is a double titration. With phenolphthalin indicator it gives pink colour in base adn becomes colourless in acidic medium.
Chemical Reaction -
H2C2O4 + 2 NaOH Na2C2O4 + 2H2O
Apparaturs and Matericals required -
Burette, pipette, conical flask, funnel, burette stand, NaOH solution, Oxalic acid solution, indicator Phenolphthalin , distiled water etc.
Method-
a) Titration of std. Oxalic acid solution with NaOH solution
Fill burette with NaOH intermidiate solution and Pipette out 20 ml known Oxalic acid solution in conical flosk add 2-3 drops of phenolphthalin indicator. Do the titration as discussed in Ex. No. 01 and find the Concordant Reading.
b) Titration of Unknown Oxalic acid solution with intermediate NaOH solution
Fill burette with intermediate NaOH solution and pipette out 20 ml of unknown oxalic acid. Add 2-3 drops of phenolphthalin indicator. Repeat the process of titration and find the concordant reading.
Obseration Table 1
a) Titration between known Oxalic acid and intermediate solution.
Obseration Table 2
a) Titration between unknown Oxalic acid and intermediate solution.
Calculation -
a) Normality of std. sodium carbonate
Normality =
= g/l
b) Normality of std. Hydrochloric acid solution
N1V1 = N2V2
N1 = Normality of std.oxalic acid solution.= N
N2 = Molarity of Intermediate HCl solution. = ?
V1 = Volume of std. Oxalic acid solution. = 20 ml
V2 = Volume of Intermediate HCl solution. = x ml
N2 = =N
c) Normality of unknown Sodium Carbonate solution.
N3V3 = N4V4
N3 = Normality of unknown Na2CO3 solution.= ?
N4 = Normality of Intermediate HCl solution. = y N
V3 = Volume of unknown Na2CO3 solution.= 20 ml
V4 = Volume of Intermediate HCl solution. =y ml
N3 = =N
=N
Result- Normality of unknown sodium carbonate is N.
