Question

In the preparation of unknown samples, explain why acid (e.g. HNO₃) should be added to dissolve sample II, but not sample I. This is a flame atomic absorption spectroscopy experiment.

The preparation of the samples are as followed:

Sample I: (Preparation of unknow sample for solubility product determination of CaCO₃) Use about 100mg of the unknow solid sample. Place the unknown solid sample into a 100mL beaker and ass about 50mL of (boiled then cooled to room temp.) boiled distilled water. This suspension should be stirred often, and allowed to equilibrate for 30 minutes. With the solid sample still in the solution, filter the suspension. Measure the pH of this sample.

Sample II: (Preparation of unknown sample for total metal content determination) Weigh out 125mg of your unknown sample. Put in 10mL of distilled water. Add a minimum amount of acid (1M HNO₃) to dissolve the unknow solid. Transfer the solution to a 100mL volumetric flask and then dilute ut to the mark with 2% HNO_3.

Answer 1

In the preparartion of samples for analysis by flame absorption spectroscopy,

Sample 1 : We dissolve the solid in the absence of any acid or HNO3. This is because in presence of acid H+ ions would react with the dissolved CO3^2- from CaCO3 as,

CaCO3(s) <==> Ca2+(aq) + CO3^2-(aq)

CO3^2-(aq) + 2H+(aq) <==> H2CO3(aq)

This would change the solubility of CaCO3. More CaCO3 would dissolve in solution and exact measurement of soubility product constant is not possible.

Sample II : In here, HNO3 (acid) is added to the solution with metal in it. The metal does not react with the added acid and it hepls the sample to dissolve in solution which can then easily be analysed for the amount of metal in solution.