Table of Contents
Exp. 1
AIM- Calculate the molarity of unknown sodium carbonate given in bottle B. You are provided with M/20 standard solution of Sodium carbonate solutin in Bottle A. By Titration method using methyl orange as an indicator and intermediate solution is HCl.
Theory-a) this is
neutralisation titration.
b) This is
double titration.
c) Methyl
orange gives orange colour in basic and pink colour in acidic medium.
Chemical Reaction -
Na2CO3 + 2HCl
2NaCl + H2O + CO2
Apparaturs and
Matericals required -
Burette,
pipette, conical flask, funnel, burette stand, Na2CO3 solution, HCl solution, indicator methyl
orange, distiled water etc.
Method-
a)
Titration of std. Na2CO3 solution with HCl
solution
Wash
Burette with water followed by washing with HCl solution. Then fill burette
with HCl solution by placing it in stand.
New
pitpette out 20 ml of Std. solution in conical flask. Add 2-3 drops of
indicator in the flask. Allow the solution of burette to add in the flask drop
by drop with continous stinning.
At the
equvatence point, pink colour appears. Note the burette reading to calculate
the molority of solution repeat the process to get concordant reading.
b)
Titration of Unknown solution with intermediate solution
Take
intermediate solution in burette, and pipette out 20 ml in conical flask. Add
2-3 drops of indicator.
Now add
solution from burette in the flask drop by drop, fill pink colour comes at end
point.
Obseration Table 1
Obseration Table 2
Calculation
-
a) Molarity of Given Na2CO3 =
M/20
b) a1M1V1 = a2M2V2
a1 = 2
a2 = 1
M1 = M/20 Molarity of std.
Na2CO3 solution.
M2 = Molarity of
Intermediate HCl solution.
V1 = Volume of std. Na2CO3 solution.
V2 = Volume of Intermediate
HCl solution.
M2 = =2
M
c) Molarity of unknown Sodium Carbonate
solution.
2M3V3 = M4V4
M3 = Molarity of unknown Na2CO3 solution.= ?
M4 = Molarity of
Intermediate HCl solution. =2
M
V3 = Volume of unknown Na2CO3 solution.= 20 ml
V4 = Volume of Intermediate
HCl solution. =
M3 =
=
M
=
M
Result- Molarity of unknown sodium carbonate is
................... mol/Lit.
Exp. 2
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.