THEORY:
A definite wt. of the given mixture is allowed to react with a known excess volume of standard (strong) NaOH solution. Ammonium chloride reacts with NaOH to give off ammonia gas, but sodium chloride does not. After the reaction is complete, the excess unused alkali is back titrated against a standard acid supplied. Then by knowing the volume of the alkali used in the reaction, the amount of NH4CI in the mixture can be calculated.
REACTION:
NH4CI+ NaOH————>NaCl + NH3 + H₂O
53.5 gm 40 gm
therefore, 40 gm NaOH = 1000 ml (N) NaOH = 53.5 gm NH4CI
APPARATUS :
1) Burette
2) Pipette
3) Conical flask
4) Burette stand
5) Clamps
6) Measuring flask
CHEMICALS:
(1) Phenolphthalein
PROCEDURE :
i) Accurately weigh out about 0.5 – 0.7 gm of the given mixture in a chemical balance. A watch glass may be used for weighing.
ii) Transfer it quantitatively into a clean conical flask with the help of a funnel. Wash down the adhering particles with some distilled water (use a wash bottle)
iii) Add a known volume of fairly strong NaOH into the flask, and heat the flask for some time till the gas (ammonia) evolved changes moist red litmus into blue. When the gas no more affects the red litmus, it shows that the reaction is complete.
iv) Cool the flask, and transfer the residual contents of the flask into a measuring flask of a definite capacity (say 100 ml)
v) Rinse the conical flask a few times with little amounts of distilled water, and transfer all the washings into the measuring flask. Make the volume up to the mark by the addition of distilled water. Shake well to make the diluted residual solution homogeneous.
vi) Titrate 10 ml of this residual solution against the standard acid supplied.
RESULT:
Wt. of the mixture taken = 0.5 gm
Vol. of the strong NaOH taken = 25 ml of 0.95 N NaOH
Solution taken in the burette = standard acid
Residual sol” taken for each titration = 10 ml
Indicator used = methyl orange
End point = from yellow to light pink
Normality of standard acid =N / 10 (f = 1.06) H₂SO4
BURETTE READINGS:
| No. of obs. | Initial reading | Final reading | Vol. of acid (ml) | Mean |
| 1 | 16.5 —-> | (Approx) | ||
| 2 | 16.2 | |||
| 3 | 16.0 | 16.0 | ||
| 4 | 16.0 |
CALCULATION:
Vol. of strong NaOH taken = 25 ml of 0.95 (N) NaOH
= (25 x 0.95) ml of (N) NaOH.
= 23.75 ml of (N) NaOH,
From titration.
10 ml residual sol” = 16 ml of N / 10 (f = 1.06) H₂SO4
= 16 × 1 × 1.06 / 10 ml (N) H₂SO4
= 1.696 ml (N) H₂SO4.
therefore,100 ml residual sol” = 16.96 ml (N) H₂SO4
= 16.96 ml (N) NaOH
= Vol. of NaOH left.
therefore, Vol. of NaOH used by NH4CI = (23.75-16.96) ml (N) NaOH
= 6.79 ml (N) NaOH.
We know,
1000 ml (N) NaOH = 53.5 gm pure NH4Cl
therefore, 6.79 ml (N) NaOH = 53.5 × 6.790 / 1000 gm NH4Cl = 0.3633 gm NH4Cl.
therefore, % of pure NH4Cl in the mixture = 0.3633 × 100 / 0.5 = 72.66%.