Experiment 3 PRE-LAB Questions (show instructor before starting; include in lab

Question

Experiment 3 PRE-LAB Questions (show instructor before starting; include in lab report) The hydrogen peroxide solution that you are using in this experiment is labeled as a 3% solution. mass/volume. However, in order to complete the calculations, the concentration must be in molarity. Calculate the molarity of a 3% mass volume l 1202 solution (Part I. II, and IV) and a 1.5% mass/volume H2O2 solution (Part III) and record these values in the table below. 1. Complete the following table. NOTE: 3% mass/volume H202 solution means 3g H2O2 100ml of solution and you need to convert it to moles of H202/ L of solution. volume H02[H:O] in Molarity Volume KI before mixing Part KI] before mixing 0.50 M 0.25 M 0.50 M 0.50 M Trial) (mL) 4 of 3% 4013% 40f 1.5% 4of3% mL) 44 M 2. Reading the table for Trial 1, what would be the new concentration of [KIl after mixing 4 ml of the H202 with 1 ml of the [KI]? What is the concentration of [I-1 in the same trial? (HINT Usethedilution equation MM-MM) ,50·k-M&L; k1 K+1 3. Considering trial III fill in the blanks. I will be mixing ml of tMH:0; with ml of 5 M KI which will ive me a total volume of 5 ml. What is the new concentration of H:O, after mixing from trial I1? 4. You will need to do this for each Trial to complete the '[H O:l after mixing' and I-1 after mixing columns in the data table for the Postlab/Data analysis section.

Explanation / Answer

Trial 1:

M1V1 = M2V2

4 x 0.88 = 5 x MH2O2

MH2O2 = 0.704 M

And 1 x 0.5 = 5 x MI-

MI- = 0.1 M

Trial 2:

M1V1 = M2V2

4 x 0.88 = 5 x MH2O2

MH2O2 = 0.704 M

And 1 x 0.25 = 5 x MI-

MI- = 0.05 M

Trial 3:

M1V1 = M2V2

4 x 0.44 = 5 x MH2O2

MH2O2 = 0.352 M

And 1 x 0.5 = 5 x MI-

MI- = 0.1 M

Trial 4:

M1V1 = M2V2

4 x 0.88 = 5 x MH2O2

MH2O2 = 0.704 M

And 1 x 0.5 = 5 x MI-

MI- = 0.1 M

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