Sodium Determination

Question

Determine the amount of sodium chloride in potato chips by the Mohr method. Potato chips are boiled with water and filtered. 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 62.1 mL of AgNO₃ titrant to fully reach the equivalence point of the reaction.

You then use the same silver nitrate solution to analyze the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 48.5 mL of titrant is added.

If the sample of chips used to make the filtrate weighed 88.5 g, how much NaCl is present in one 155 g serving of chips?

Answer

First off, we need to determine the concentration of our silver nitrate solution. Before we begin computation we have to figure out the reaction. Searching Mohr method online gives results if you were unsure of what the method really involved to give you an idea of the reaction. The idea behind the Mohr method is that silver nitrate is soluble, silver chloride is not. Adding a solution of silver nitrate to a solution containing chloride anions precipitates out silver chloride which can then be weighed to determine how much chloride was present. The reaction for the precipitation is:

AgNO_3(aq)+ NaCl_(aq) ---> AgCl_(s) + NaNO_3(aq)

Note that any chloride anions, not just those that came from the sodium would cause a precipitate, however for simplicities sake this is usually not assumed to be an issue. So, now we have a reaction, time to go back to the numbers provided in the problem. You know how much KCl was titrated, let's convert that to the number of mols first by dividing by the number of grams in a mol.

0.500 g KCl / 74.55 (g/mol) = 0.006707 mol KCl

1 mol KCl gives 1 mol Cl^-therefore you have the same number of mols of chloride floating around for the reaction above. Now, because you know it took 62.1 ml of silver nitrate solution to react with this amount of chloride, you know there was the same number of mols of silver nitrate in that 62.1 ml of solution as there were mols potassium chloride. To help us out here, calculate the molarity (mol/L) of your silver nitrate solution by dividing the number of mols of silver nitrate (again, equal to the number of mols of potassium nitrate) by the volume of solution used in liters.

M = 0.006707 mol AgNO₃ / 0.0621 L = 0.1080 mol/L

Finally, you know from above that it took 48.5 ml of this silver nitrate solution to precipitate all of the chloride. Figure out how many mols of silver nitrate there are in 48.5 ml of the solution by multiplying the molarity (M) by the volume of solution used in liters.

0.1080 (mol / L) x 0.0482 L = 0.005206 mol

This is the mols of silver chloride produced which is equivalent to the number of mols of sodium chloride. Multiply by the molecular weight of sodium chloride to get the number of grams of sodium chloride that were originally in the sample.

0.005206 ~~mol NaCl~~ x 58.44 (g/~~mol~~) = 0.3042 grams NaCl

Almost there, this is the number of grams of sodium chloride in your 88.5 g sample that was used to prepare the solution you are analyzing, what you want is the number of grams of sodium in a 155 g serving. The easiest way to go from here to there is simply obtain a ratio of the of the sample to the serving by dividing the 155g serving by 88.5 g sample and then multiplying the answer by the number of grams of sodium chloride obtained from the sample.

155 g / 88.5 g = 1.7514

0.3042 g NaCl x 1.7514 = 0.533 g NaCl in 155 g Chips

Things of note here. There is of course more than one way to come up with a solution to this problem, I chose this way because it seems the most intuitive and easy to follow but there are other faster ways that give the same answer. Secondly, there are a number of numerical manipulations. Normally one is supposed to carry through answers on a calculator and round only for the final answer, but in this case due to the medium of communication I rounded at each step which can lead to an answer being off by a few percent. Finally, the 'rusty brown' precipitate mentioned could be due to contamination from the silver chromate used to determine the end point of the titration which could signal that the titration went too far which could lead to erroneous results. However I assumed in this case, due to there being no additional information that this was not the case.