Question

A medication B is available as a concentrated paste (of unknown concentration – see later). 0.1406 g of the paste was made up to 100 mL with water. This “original” solution was diluted serially twice further, each time a hundred-fold (i.e. 1 mL to 100 mL). The final dilute solution was analyzed and found to contain 36.68 ng of B per mL. What was the concentration of the paste in terms of %B by weight of paste?

Answer 1

Ans. The paste is diluted as follow-

I. 0.1406 g paste is diluted to 100.0 mL. Label it as solution 1.

II. Solution 1 is diluted 100-fold to produce another solution, label it as solution 2.

Current dilution (from solution 1 to solution 2) = 1 : 100

III. Solution 2 is again diluted 100-fold to produce another solution, label it as solution 3.

Current dilution (from solution 2 to solution 3) = 1 : 100

Total dilution required to make solution 3 (from solution 1 to solution 3)

= Current dilution x previous dilution

= Current dilution x dilution factor for solution 1 to 2

                                                = (1 : 100) x (1 : 100)

                                                = 1 : 10000

Now,

            Final [B] in solution 3 = [B] in solution 1 / dilution factor from solution 1 to 3

            Or, 36.68 ng / mL = [B] in solution 1 / (1 : 10000)

            Or, [B] in solution 1 = (36.68 ng / mL) x 10000 = 3.668 x 10⁵ ng/ mL

            Or, [B] in solution 1 = (3.668 x 10⁵ x 10^-9 g)/ mL                    ; [1 g = 10⁹ ng]

            Hence, [B] in solution 1 = 3.668 x 10^-4 g/ mL

Now,

Total amount of B in solution 1 = [B] in solution 1 x Volume of solution 1

                                                = (3.668 x 10^-4 g/ mL) x 100.0 mL

                                                = 3.668 x 10^-2 g

                                                = 0.03668 g

Therefore, total amount of B in solution 1 = 0.03668 g.

Since solution 1 is prepared from 0.1406 g of paste, the total amount of B in solution 1 is equal to total amount of B in 0.1406 g paste.

Therefore, total amount of B in 0.1406 g paste = 0.03668 g

Now,

            % B (w/w) in paste = (Mass of B / Mass of paste) x 100

                                                = (0.03668 g / 0.1406 g) x 100

                                                = 26.09 %