Problem with determining the mass value of a soil sample for gravimetric calculations [closed]

Question

I'm having some trouble with how to determine the representative mass value of a prepared soil sample for calculating its $\ce{Cl-}$ concentration in ppm (mg/kg).

Recently I carried out a gravimetric analysis for determining the concentration of $\ce{Cl-}$ in a soil sample by the addition of $\ce{AgNO3}$. The sample was prepared from 10 g of soil, which were filtered with 50 mL of distilled water. Then, an aliquot of 15 mL was taken from the filtrate. The aliquot was centrifuged in order to separate the solid phase from the liquid phase. The liquid was transfered to a volumetric flask and diluted to 50 ml with distilled water. This final volume was the one used for the analysis. The gravimetric processes were developed normally and at the end it turned out that the produced mass of the AgCl precipitate was 0.003 g and by stoichiometry I calculated its content of $\ce{Cl-}$ (which resulted to be 0.00074204 g).

Since I want the $\ce{Cl-}$ concentration in terms of mg/kg, I expressed the mass of the aformentioned ion in mg but my conflict comes when trying to establish the mass of the used soil sample (in kg) because of the dilutions. As far as I'm concerned, the preparation of the sample could be mathematically arranged as follows: $$\frac{\pu{10 g}\,\text{soil}(\frac{\pu{1 kg}}{1000\,g})}{\pu{50 ml}\,\text{filtrate}}\cdot\frac{\pu{15 ml}\,\text{aliquot}}{\pu{50 ml}\,\text{dilution}}=\pu{6e-5 kg ml-1}$$

but as you can see, the dimensional analysis doesn't result in just kg. I don't really know where the problem is, but I intuit that either I'm missing something or adding more of something, or I'm just having an erronous mathematical concept about the sample preparation.

It will help me a lot if you can help me to identify my mistake.

Answer 1

First I would say that your measurement is not reliable enough. You say that you obtain 3 mg AgCl. 3 mg AgCl is nearly nothing. If you are not able to be more precise, it means that your amount of AgCl is between 2.5 mg and 3.5 mg. The uncertainty is about ± 17%. That is a huge uncertainty.

Anyway your 3 mg AgCl contains 0.074 mg Cl which were in the 50 mL flask. As this solution was made from 15 mL of the first solution, the entire initial solution contains 50/15 times 0.074 mg Cl. This is 0.248 mg Cl. As this amount of Chloride is coming from 10 g soil, it means the the proportion of Cl in the soil is 0.248 mg/0.01 kg soil = 24.8 mg/kg. As the uncertainty on the initial weight (3 mg) is high, you may admit that the content of Cl in the soil is 25 mg/kg. With the uncertainty, it is 25 ± 5 mg/kg.

If I were at your place, I would determine the chloride content with a volumetric method. The result of such a titration would be much more reliable. A precision of ±1% is easily obtained.