Question

A second calibration standard solution of an iron(III) salicylate complex was prepared in two steps. First, 30.8 mL of a 0.185 M stock solution was diluted with solvent to make 100.0 mL of the first calibration standard solution and, secondly, 78.5 mL of that first calibration solution was diluted to 100 mL to make the second calibration standard. What is the molar concentration of the second calibration standard solution

Answer 1

0.0447 M.

Step-by-step explanation:

We'll begin by calculating the molarity of the diluted solution of the first calibration standard solution. This can be obtained as follow:

Molarity of stock solution (M1) = 0.185 M

Volume of stock solution (V1) = 30.8 mL

Volume of diluted solution (V2) = 100 mL

Molarity of diluted solution (M2) =?

M1V1 = M2V2

0.185 × 30.8 = M2 × 100

5.698 = M2 × 100

Divide both side by 100

M2 = 5.698 / 100

M2 = 0.05698 M

Therefore, the molarity of the diluted solution of the first calibration standard solution is 0.05698 M

Finally, we shall determine the molarity of the second calibration standard solution as follow:

Molarity of first calibration standard solution (M1) = 0.05698 M

Volume of first calibration standard solution (V1) = 78.5 mL

Volume of second calibration standard solution (V1) = 100 mL

Molarity of second calibration standard solution (M2) =?

M1V1 = M2V2

0.05698 × 78.5 = M2 × 100

4.47293 = M2 × 100

Divide both side by 100

M2 = 4.47293 / 100

M2 = 0.0447 M

Therefore, the molarity of the second calibration standard solution is 0.0447 M.