Question

Design a protocol with a list of glassware needed and calculations to prepare a 100-mL homogeneous solution of HCl/FeCl3 with 0.025 M of FeCl3 and 0.10 M of HCL.

The following reagants are available:

3 M HCl and Solid FeCl3 . 6 H2O

Answer 1

Ans. Part 1: Solution to be prepared 0.025 M FeCl₃

Moles of FeCl3 in 100 mL of 0.025 M solution = Molarity x volume in L

                                    = 0.025 M x 0.100 L = 0.0025 moles

Source of FeCl3 is FeCl3.6H2O. 1 mol FeCl3.6H2O gives 1 mol FeCl3. So, moles of FeCl3.6H2O require to prepare the desired solution = 0.0025 moles

Mass of FeCl3.6H2O required = moles x molar mass

                                                = 0.0025 moles x 270.29 g mol^-1

                                                = 0.675725 g

Part 2: Preparation of 0.10 M HCl, 250 mL    

Using, M1V1 = M2V2     , where

M1= molarity of initial solution 1, V1= volume of initial solution 1      ;(standard stock solution)

M2= molarity of final solution 2, V2= volume of final solution 2        ;(aliquot 1)

            3M x V1 = 0.10 M x 250 mL

            Or, V1 = 8.33 mL

Transfer 8.30 mL (you can use micropipette for taking 8.333 mL) of 3M HCl into a 250-mL standard volumetric flask using graduated glass pipette. Make the final volume upto the mark with distilled water. It is 250 mL solution of 0.1 M HCl.

Part 3: Preparation of desired solution

Weigh and transfer 0.676 g FeCl3.6H2O into a 100-mL standard volumetric flask. Add a small volume of 0.1M HCl. Gently swirl the flask to dissolve the crystals. Make the final volume upto the mark with 0.1M HCl. It is 0.025 M FeCl₃ solution.

Glassware required:

10-mL graduated glass pipette

            100-mL standard volumetric flask

            250-mL standard volumetric flask