In order to analyze water samples using a spectrophotometer or plate reader, it is necessary to turn the molecules of nitrate into a dye molecule that can be quantified. The first step in turning nitrate (NO3-) into a dye molecule is reducing it to a molecule of nitrite (NO2-). This is done by reacting the NO3- with cadmium.

After the reduction reaction, the NO2- is reacted with two additional reagents. The first reagent, Reagent A, is a solution of sulfanilamide and hydrochloric acid. The second reagent, Reagent B, is a solution of N-(1-naphthyl)-ethylenediamine, called NNED for short. The compounds are mixed with the water sample and produce a purple color. The intensity of the purple color is directly related to the concentration of nitrite in the water sample. We can measure how purple the water turns as absorbance on a spectrophotometer and then convert the absorbance to concentration of nitrate.

To make Reagent A, we will need to make a solution of 10.0 g of sulfanilamide in 1 L of 2.4 molar hydrochloric acid (HCl).
The stock solution of HCl is 12 molar HCl. How many milliliters (mL) of 12 M HCl would you add to produce 0.3 liters (L) of 2.4M HCl? __ mL HCl
After creating 0.3 L of 2.4 molar HCl solution, how many grams of sulfanilamide will be added? g sulfanilamide

Part 2

After reacting the nitrate with cadmium to produce nitrite, the nitrite is then reacting with sulfanilamide and N-(1-naphthyl)-ethylenediamine, to produce a purple dye molecule that can be quantified on a spectrophotometer.
The N-(1-naphthyl)-ethylenediamine, called NNED for convenience, reagent is made by mixing 1 gram of NNED in 1 liter of water. However, we don't always want to make an entire liter of solution because the NNED solution only lasts about 1 month before going bad and turning brown.
How many milligrams of NNED will need to be added to make 0.2 liters of solution?

where Vs: is the HCl volume of the stock solution (12M), [HCl]s: is the HCl concentration of the stock solution, Vd: is the volume of the diluted solution (2.4M) and [HCl]d: is the concentration of the diluted solution.

[tex] V_{s} = \frac{V_{d} [HCl]_{d}}{[HCl]_{s}} = \frac{0.3L \cdot 2.4M}{12M} = 0.06L = 60mL [/tex]

So, we need 60mL of 12M HCl to prepare 0.3L of 2.4M HCl.

Starting with the ratio of 10.0g of sulfanilamide in 1L of 2.4M HCl needed to prepare Reagent A, the mass of sulfanilamide required to produce the same solution in 0.3L of 2.4M will be:

[tex] m_{sulfanilamide} = \frac {10.0g}{1L} \cdot 0.3L = 3.0g [/tex]

Therefore, we need 3.0g of sulfanilamide in 0.3L of 2.4M HCl to prepare Reagent A.

Part 2.

The NNED mass required to make 1.0L of solution is 1.0g, so the mass needed to make 0.2L of the same solution will be:

[tex] m_{NNED} = \frac {1.0g}{1.0L} \cdot 0.2L = 0.2g [/tex]

Hence, we need 0.2g of NNED to prepare 0.2L of the solution.

I hope it helps you!