This is a limit test for chlorides.

1.0g of NaOH sample was dissolved in 5ml of deionized water, acidified with 4ml of concentrated nitric acid (65%) and diluted to 15ml with deionized water. (Test tube A)
10ml of 50ppm Standard NaCl solution was prepared. 5ml of water was added, followed by 1ml of dilute nitric acid. (Test tube B)
1ml of AgNO3 was added to each of the test tubes and compared for opalescence.

Can you explain why we need to add 4ml of concentrated nitric acid and how was the 4ml derived?

Ag ion reacts with strongly basic solution to produce a hydrated Ag2O sometimes simply written as Ag(OH)2. It's a dark brown/black gelatinous ppt when Ag ion is added to a basic solution. So the 4 mL HNO3 is to neutralize the NaOH and make sure the solution is acidic. HNO3 is used because neither HCl nor H2SO4 can be used; HCl for the chloride content and H2SO4 because of solubility concerns. 65% HNO3 is very close to 15 M and 4 mL of that is enough to neutralize all of the NaOH (about 1/40 = 0.025 moles) (4 mL of 15 M HNO3 is about 0.06 mols).

Thank you very much. Another question, why do we need to add 1ml of dilute nitric acid to the 50ppm standard NaCl solution before adding AgNO3?

Same answer. You want the Ag^+ to be an acid solution. AgCl needs to be pptd from an acid solution. HNO3 is the preferred acid to use. Too much can go wrong with Ag^+ from neutral or basic solution.

To understand why 4ml of concentrated nitric acid is added in this chloride limit test, we need to consider the reaction that takes place between chloride ions (Cl-) and silver nitrate (AgNO3). When silver nitrate comes into contact with chloride ions, it forms a white precipitate of silver chloride (AgCl). This precipitate is usually cloudy or opalescent.

In the given test, the purpose is to compare the opalescence formed in two different solutions: Test tube A containing the NaOH sample and Test tube B containing the standard NaCl solution.

Now, let's break down the process:

1. Test tube A:
- 1.0g of NaOH sample is dissolved in 5ml of deionized water.
- Next, 4ml of concentrated nitric acid (65%) is added to the solution.
- Finally, the solution is diluted to a total volume of 15ml with deionized water.

2. Test tube B:
- 10ml of a 50ppm standard NaCl solution is prepared.
- 5ml of water is added to the solution.
- Then, 1ml of dilute nitric acid is added.

By comparing the opalescence between the two test tubes, we can determine the presence or absence of chlorides in the NaOH sample (Test tube A) relative to the known concentration of chlorides in the standard NaCl solution (Test tube B).

Now, to address why 4ml of concentrated nitric acid is added in Test tube A and how it is derived:

Nitric acid is added as it plays two important roles in the test:
1. Acidification: Nitric acid acidifies the solution, ensuring that any potential interfering species or ions are properly dissociated, allowing accurate detection of chloride ions.
2. Avoiding precipitation of other silver salts: Nitric acid helps to prevent the formation of other silver salts (such as silver hydroxide) that might interfere with the formation of the silver chloride precipitate.

The specific volume of 4ml is likely determined through prior experimentation or based on established protocols. It's important to note that the volume of nitric acid is optimized to provide proper acidification without excessive interference or over-acidification. Therefore, in this particular test, 4ml of concentrated nitric acid is deemed appropriate for achieving the desired effects without compromising the accuracy of the chloride limit test.