Question

A dietary supplement capsule weighing 4.84 g was ground into a fine powder. Two portions of the solid, both weighing 0.137 g, were dissolved in dilute acid and transferred to 50 ml volumetric flasks. To one of these, 5 ml of 40 ppm Mn2+ was added, and then both flasks were diluted to the mark with distilled water. When aspirated into the flame of an atomic absorption spectrometer set at a manganese absorption wavelength, the absorbances of the unknown and unknown plus standard were 0.374 and 0.641, respectively. Calculate the percentage of Mn in the capsule.

Answer 1

Ans. Solution 1: Absorbance = 0.374

Solution 2: Abs = 0.641

Increase in absorbance due to adding standard = 0.641 -0.374 = 0.267

Both the solutions are prepared identically except to the ‘standard addition’ in solution 2.

# About Solution 2   

5 mL of 40 ppm Mn⁺² is added to it. Final volume made upto 50.0 mL.

Calculate the [Mn²⁺] increased due to addition of the standard.

Using C1V1 = C2V2 - equation 1

C1= Concentration, and V1= volume of initial solution 1         ; i.e. std solution

C2= Concentration, and V2 = Volume of final solution 2         ; 50.0 mL dilution

Putting the values in above equation-

            40 ppm x 5 mL = C2 x 50 mL

            Or, C2 = (40 ppm x 5 mL) / 50 mL = 4 ppm.

That is, due to addition of standard Mn²⁺, there is increase of 4ppm Mn²⁺ in solution 2 with respect to solution 1 in which standard was not added.

# So an absorbance (increase) of 0.267 correspond to a [Mn²⁺] of 4 ppm.

Now,

            [Mn²⁺] in solution 1 = Abs of solution 1 x (4 ppm / 0.267 abs)

                                                = 0.374 x (4 ppm / 0.267)

                                                = 5.60 ppm                          ; [1 ppm = 1 mg/ L]

                                                = 5.60 mg/ L

                                                = 0.0056 g/ L                      

# Calculating % Mn²⁺ in capsule

Total amount of Mn²⁺ in solution 1 = [Mn²⁺] x Volume in liters

                                                            = (0.0056 g/L) x 0.050 L       ; [1 mL = 10^-3 L]

                                                            = 0.00028 g

Note that this solution is prepared from 0.137 g of the sample.

Now, % Mn²⁺ = Mass of Mn²⁺ / Mass of sample) x 100

                        = (0.00028 g / 0.137 g) x 100

                        = 0.204 %