THEORY:
Reaction HCL + NaOH ———–> NaCl + H₂O
Ionically, H ++ OH- ————–>H₂O
Here, the bench alkali NaOH is much stronger than the standard acid supplied. Hence the bench NaOH should be diluted to the required extent such that the diluted alkali and the acid neutralize in equal volumes. Then, by knowing the strength of the diluted alkali, we can calculate the strength of the original bench alkali NaOH.
APPARATUS :
1) Burette
2) Pipette
3) Conical flask
4) Burette stand
5) Clamps
6) Measuring flask
CHEMICALS:
Phenolphthalein indicator
PROCEDURE :
i) Wash the whole apparatus with water.
ii) Rinse the burette with the standard acid two times, and then fill it up with the same acid.
iii) Rinse the graduated pipette with the bench NaOH once, and pipette out 1 ml bench NaOH into the clean conical flask, add about 10 ml water and 1-2 drops of phenolphthalein indicator.
iv) Titrate the alkali by adding acid from the burette rapidly say 1 ml at a time till the pink color just disappears.
N.B. you will see that 1 ml of the bench NaOH can neutralize a large volume of the acid [Caution:- do not suck bench NaOH by mouth]
So this is rough titration.
Rough titration reading:
Suppose, 1 ml of bench NaOH = 20.4 ml acid
V₁ N₁ V₂ N₂
N₁ = 20.4N₂
Which means, the bench NaOH is 20.4 times stronger than the acid. Hence, the alkali should be diluted 20.4 times.
Dilution :
Suppose, a 250 ml measuring flask is provided for dilution.
Then, the volume of bench NaOH required for dilution up to 250 ml =250 / 20.4 = 12.25 ml.
v) For convenience, take 12 ml of bench NaOH from the graduated pipette into the 250 ml measuring flask, and make up to the volume by addition of water. Shake well to make it homogeneous.
Vi) Now, titrate 10 ml of this diluted alkali against the standard acid in the burette as usual.
RESULTS:
Volume of bench NaOH taken for dilution = 12 ml.
Solution taken in the burette = standard acid.
Diluted alkali taken for each titration = 10 ml
Indicator used = phenolphthalein
End point = from pink to colorless
Normality of standard acid =N / 10 (f = 1.05)
BURETTE READING:
| No. of obs. | Initial reading | Final reading | Vol. of acid (ml) | Mean |
| 1 | = | 10.0 —–> | (approx) | |
| 2 | 9.9 | |||
| 3 | 9.8 | 9.8 | ||
| 4 | 9.8 |
CALCULATIONS :
Acid diluted alkali
V₁ = 9.8 ml V₂ = 10 ml
N₁ =n / 10 (f = 1.05) N₂ = ?
From the normality equation.
N₂ = 9.8 xNx1.05 / 10 x 10 = N / 10 (f= 1.029)
Since, 12 ml bench NaOH are diluted to 250 ml.
12 ml of bench NaOH = 250 ml of diluted alkali
Applying the normality equation again,
Normality of bench NaOH = 250 N 1.029 / 12 × 10 = 2.1437N.