Please show work. Thanks Preparation of Iron(III) Chloride Solution. Prepare 250

Question

Please show work. Thanks
Preparation of Iron(III) Chloride Solution.

Prepare 250ml of a solution that is 0.02M in iron (III) chloride, 0.025 M in hydrochloric acid and 0.025 M in potassium chloride. Please show work. Thanks
Preparation of Iron(III) Chloride Solution.

Prepare 250ml of a solution that is 0.02M in iron (III) chloride, 0.025 M in hydrochloric acid and 0.025 M in potassium chloride.
Preparation of Iron(III) Chloride Solution.

Prepare 250ml of a solution that is 0.02M in iron (III) chloride, 0.025 M in hydrochloric acid and 0.025 M in potassium chloride.
Prepare 250ml of a solution that is 0.02M in iron (III) chloride, 0.025 M in hydrochloric acid and 0.025 M in potassium chloride.

Explanation / Answer

Ans. Volume of solution = 220.0 mL = 0.250 L

#1. Mass of FeCl3 required = (Vol. of soln. in litters X Molarity) x Molar mass

                                    = (0.250 L x 0.02 M) x (91.2997 g/ mol)

                                    = (0.250 L x 0.02 mol/ L) x (91.2997 g/ mol)         ; [1 M = 1 mol/L]

                                    = (0.005 mol) x (91.2997 g/ mol)

                                    = 0.456 g

#2. Mass of KCl required = (Vol. of soln. in litters X Molarity) x Molar mass

                                    = (0.250 L x 0.02 M) x (74.551 g/ mol)

                                    = 0.373 g     

#3. The molarity of concentrated HCl available in lab = 11.65 M.

            C1V1 = C2V2             - equation 1

C1= Concentration, and V1= volume of initial solution 1     ; [HCl] in Conc. HCl

C2= Concentration, and V2 = Volume of final solution 2      ;[HCl] in final solution

            V1 = ?                         , C1 = 11.65 M

            V2 = 250.0 mL          , C2 = 0.025 M

Putting the values in equation 1-

            V1 = (C2V2)/ C1 = (0.025 M x 250.0 mL ) / 11.65 M = 0.536 mL

Thus, required volume of HCl to get a final [HCl] = 0.025M in final solution = 0.536 mL

Preparing of Solution:

Step I. To a small volume of distilled water in 250.0 mL standard volumetric flask, add 0.456 g FeCl3 and swirl to dissolve.

Step II. To it add 0.373 g KCl and mix well.

Step III. Add 0.536 mL concentrated (11.65 M) HCl.

Step IV. Make the final volume upto mark with distilled water. Mix well.

This is the desired solution.   

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