Determination of ammonia in the adduct. Method 1.11

1. Standardisation of HCl_(aq)

1.3129g Na₂CO_{3 anh.} (AR grade, dried at 150°C for 5 hours) was dissolved in distilled water and made up to a final volume of 250 cm³.

This gave a solution of concentration 0.0495 mol dm^-3. Sodium carbonate was used as it a primary standard.

8.7 cm³ of conc. hydrochloric acid was pipetted out and diluted to a final volume of 1 dm. This gave a concentration of 0.1 mol dm^-3.

A 10 cm³ aliquot of the standard carbonate was pipetted into a 100 cm³ conical flask. Three drops of indicator (phenolphthalein) was added. The solution turned pink. The acid was poured into a 50 cm³ burette. The solutions were titrated in triplicate and the end - point determined by the solution going clear. A pH probe and meter were also used to ensure the end - point was found accurately.

Solution number	Start titre (cm3)	End titre (cm3)	Titre (cm3)
1	0.2	2.7	2.5
2	3	5.6	2.6
3	6	8.5	2.5

The three titres gave an average titre of 2.53 cm³.

2. Concentration of ammonia evolved.

This test was carried out three times. The concentration of NI₃ was determined using the mass of I₂ used and assuming a 100% conversion of I₂ to NI₃.

Sample number	Mass of I2 used	No. moles I2	No moles NI3 produced
1	0.3017	1.1887 x 10-3	2.5478 x 10-4
2	0.3651	1.4385 x 10-3	3.0832 x 10-4
3	0.2874	1.1323 x 10-3	2.4270 x 10-4

10 cm³ of the standardised HCl_aq was used for each test. At the end of each test, the acid was titrated against the standard sodium carbonate_(aq) and the concentration of ammonia determined.

Sample no.	Start titre (cm3)	End titre (cm3)	Titre (cm3)
1	0.52	3.09	2.57
2	3.2	5.78	2.58
3	6	8.58	2.58

The average titre was 2.58 cm³. This meant that the concentration of the acid after the experiment was 0.0255 mol dm^-3. The difference being 0.0733 mol dm^-3 in HCl.

Taking the literature¹ giving the formulae of the adduct as NI₃.NH₃ or NI₃.3NH₃, a relationship of one mole of NI₃ to either one or three moles of NH₃ was expected.

If the amount of NI₃ is known, then the amount of the NH₃ can be calculated thus

If there is one mole of NI₃, there will be a corresponding x number of moles of NH₃. Therefore, if there is y moles of NI₃, there will be xy moles of NH₃.

If the amount of NI₃ is 5 x 10^-4 moles and the ratio is 1 mole NI₃ to 3 moles NH₃, then 3 ´ (5 x 10^-4) = true number of moles NH₃ which will be in the adduct.

From my results, 7.22 x 10^-4 moles NH₃ was evolved from the adduct. If the number of moles NH₃ evolved is divided by the average amount of NI₃ the ratio is found