Question

In a lab, we are required to make a 100ml solution that contain: 54.30mg/L sodium salicylate, 0.020M FeCl3, 0.050 M HCl. I will also need to create a blank solution that contains 0.020 M FeCl3 and 0.050 M HCI. The blank solution is also 100.00 mL. At the end, we are required to get an absorbance at 520 nm. What should I do to prepare the solutions? Please give detailed instructions. Thank you very much!

Answer 1

Step 1: Do the calculations for 100 mL solution !

a. sodium salicylate = 54.30 mg/ L = 5.43 mg in 100 mL

b. 0.050 M HCl.             To prepare 100 mL of 0.05 M HCl

Assuming, molarity of concentrated HCl= 11.65

Using, M1V1 = M2V2     --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 (concentrated HCL)

            M2= molarity of solution 2, V2= volume of solution 2 (0.05 M HCl)

Or, V1 = (M2V2) / M1 = (0.05 M x 100 mL) / 11.65 M = 0.429815 mL

Hence, required volume of concentrated HCl =0.429815 mL = 0.43 mL

c. 0.02 M FeCl₃

0.02 M means 0.02 moles in 1 L

Mass of 0.02 moles of FeCl₃ = moles x molecular mass

                                    = 0.02 moles x 162.20 g mol^-1 = 3.244 g

So, 1 L (= 1000 mL) of 0.02 M FeCl₃ requires 3.224 g.

Amount required for 100 mL = 0.3224 g

Step 2: Preparation of reagent:

Add 5.43 mg sodium salicylate in 100 mL standard volumetric flask.

Dissolve it by adding little amount of distilled water.

Add 0.43 mL concentrated (11.65 M) HCL to the flask, gently swirl to mix.

Add 0.3224 g FeCl₃ to the flask, swirl gently to mix.

Make the final volume upto the mark with distilled water. It is your desired solution.

Preparation of Blank: Prepare the sample blank exactly the same as above solution, except ‘don’t add any sodium salicylate’.