Question

A solution to be tested by AAS is prepared by performing two dilutions in series: First. 0.10 mL of the starting solution is diluted to 10.0 mL. Second, 1.0 mL of the solution resulting from the first dilution is diluted to 100.0 mL. This solution is then tested by AAS and found to contain 3.4 ppm Cd^2 + ion. What was the Cd^2 + ion concentration (in ppm and Molarity) in the starting solution?

Answer 1

Dilution law: M1V1 = M2V2

M1 and V1 are concentration and volume before dilution.

M2 and V2 are concentration and volume after dilution.

M1 x 0.10 mL = 3.4 ppm x 100.0 mL

M1 = ( 3.4 ppm x 100.0 mL)/ 0.10 mL = 3400 ppm

Concentration of Cd²⁺ in starting solution is 3400 ppm.

ppm = 1 micro gram/1 L

3400 ppm = 3400x10^-6 g/1L

Atomic mass of Cd is 112 g/mol

So, (3400x10^-6 g)/(112 g/mol)/1L = 3.036x10^-5 mol/L = 3.036x10^-5 M