Question

I am quantitating lead in a solution using anodic stripping Voltammetry. I am asked to use the Reagents as follows: 0.1 M KNO3 /50 mM HNO3 containing an unkown concentration of lead. For the purposes of preparing the method of standard additions, I am to assume the concentration in the container to be 1ppm. 0.1 M KNO3/50 mM HNO3 (take only what needed) 500 ppm Hg ^2+ in 0.1 M KNO3/50 mM HNO3 1000 ppm Pb and the preparation of standards and unknowns: 1. you will need to prepare a solution for your ASV analysis that contains the unkown Pb (us 2 mL of the stock Pb unkown) spiked with Hg ^2+ at a concentration of 50 ppm, diluted to a final volume of 5 mL in 0.1 M KNO3/50 mM HNO3. You should know precisely the diltuion of your Pb unkown as you will have to back calculate to your original concentration. 2. You will need to prepare a standard concentration of Pb for your additions. There is a Pb standard of 1000 ppm to be used for serial dilutions. Determine a standard concentration from your knowledge of standard additions and the fact that your unknown is near 1 ppm Pb. Check the standard concentration you determine before continuing with preparation of your solution. I don't understand how we are to back calculate to original concentration and how to prepare the standards and unkowns. PLEASE HELP!

Answer 1

The unknown solution will e provided to you whose concentration you need to find using ASV. In order to prepare the solution of ASV you have to take a given volume of the unknown solution ( and since you are asked to take 2mL and dilute that further, you need to know accurately how muchu you have diluted)

*NOTE: rememer on dilution your concentration will change so whatever value of concentration you will obtain after the experiment, you have to "back calculate" it to the concentraton that you had efore diluting your unknown solution. For that you can use the formula C1 = C2 * V2 / V1 where you would know the values of volumes before and after dilution and concentration after dilution and determine the concentration of the original unknown.

In order to prepare standard lead solution of 1000ppm (1000ppm is equivalent to 1000 mg/l or 1g/l of lead), if you use a salt of lead like lead nitrate then you need to calculate it as per

(331.3 g/mol * 1000 mg Pb) / 207.2 g/mol = 1598.9 mg lead nitrate.

Molar mass of Pb = 207.2 g/mol and molar mass of lead nitrate = 331.2 g/mol

dissolve the given amount with 1l water in a volumetric flask to prepare the standard solution