Question

As a food chemist for a major potato chip company, you are responsible for determining the salt content of new potato chip products for the packaging label. The potato chips are seasoned with table salt, NaCl. You weigh out a handful of the chips, boil them in water to extract the salt, and then filter the boiled chips to remove the soggy chip pieces. You can analyze the chip filtrate for Cl- concentration using the Mohr method.

First, you prepare a solution of silver nitrate, AgNO3, and titrate it against 0.500g of KCl using the Mohr method. You find that it takes 63.5mL of AgNO3 titrant to reach the equivalence point of the reaction.

You then use the same silver nitrate solution to make the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 46.1mL of titrant is added.

If the sample of chips used to make the filtrate weighed 88g, how much sodium chloride is present in one serving (155g) of chips?

Answer 1

Ans:

In the above experiment is done in two step titrations. First step is the determination of the silver nitrate solution’s strength in g/mL and in the second step finding out the quantity of sodium chloride in the chips per serving. The balanced reactions are happening in1:1 mol ratio as givne below:

Step-1:            AgNO3 + KCl → AgCl (ppt) + KNO3

Mol. Wt.         169.87    + 74.55

It is given that, in total 0.500 g of KCl is reacted / titrated with the unknown strength silver nitrate solution.

We know from the above balaned reaction that 74.55 g of KCl reacts completely with 169.87 g of AgNO3

Hence, the amount of AgNO3 reacted with 0.500 g of KCl will be,

= (169.87 / 74.55) * 0500 = 1.139 g

The equivalence point is reached by consuming 63.5 mL of the silver nitrate.

Therefore, 63.5 mL of the silver nitrate solution will have 1.139 g of it.

1 mL of the above silver nitrate solution contains = 1.139/63.5 = 0.0179 g/mL of silver nitrate

Step-2:            AgNO3 + NaCl → AgCl (ppt)+ KNO3

Mol. Wt.          169.87 + 58.44

The extract in water having sodium chloride, from chips requires 46.1 mL of the above silver nitrate.

Hence, the silver nitrate quantity required by the chips extract = 46.1 mL * 0.0179 g/mL = 0.8252 g of AgNO3

From the above balanced reaction, we know that 169.87 g of silver nitrate reacts completely with 58.44 g of the sodium chloride.

Hence, 0.8252 g of AgNO3 of silver nitrate will react with

= (58.44 g/ 169.87 g) * 0.8252 g = 0.2839 g of sodium chloride

This amount of sodium chloride is extracted from the 88 g of chips.

Hence, one serving of 155 g of chips will contain

= (0.2839 g / 88 g) * 155 g

= 0.500 g of sodium chloride