Based on the background information on the experimental reaction, data from Tabl

Question

Based on the background information on the experimental reaction, data from Table 2, and dilution calculations, determine the expected concentration of each species in beakers A and B:

Diluted Concentration in Beaker A (M)

Diluted Concentration in Beaker B (M)

Na2S2O3

TABLE 2: (This is what was given as Table 2)

HCl Volume (mL)

Pre-Lab assignment please Help!!!

Chemicals

Quantity

Hydrochloric acid (HCl), 0.0500 M

100 mL

Hydrogen peroxide (H2O2), 0.5000 M

100 mL

Potassium iodide (KI), 0.0500 M

100 mL

Sodium thiosulfate (Na2S2O3), 0.0100 M

100 mL

Starch indicator

As needed

Water, deionized

As needed

In this experiment, the kinetics of reaction 1 are studied:
2H+ + H2O2 + 2I- 2H2O + I2 reaction 1

How fast reaction 1 produces iodine indicates the rate of the reaction through the following expression.

rate = d[I2] Eq. 1 dt

Reaction 2 consumes this I2 much faster than 1 can make it. So there is no build up of I2 in solution.

2S2O32- + I2 2I- + S4O62- reaction 2

Reaction 2, however, runs out of thiosulfate, S2O32–, very quickly. Once this happens, the I2 from reaction 1 builds up in solution and reacts with the starch indicator to form a blue colored complex. Since all runs are made to have the same concentration of S2O32–, all reactions in the experiment must make the same amount (the same number of moles per liter) of iodine before the color change occurs. As a result, the rate of reaction 1 can be expressed in terms of the concentration of thiosulfate (Eq. 2).

rate = d[I2] = - 1 d[S2O32-] Eq.2 dt 2dt

Equation 2 also accounts for the fact that thiosulfate is being consumed in reaction 2, and that 2 thiosulfate ions react with one I2. Furthermore, since all of the thiosulfate in the reaction is consumed, the change in concentration of thiosulfate is equal to the original concentration of thiosulfate in the reaction. So equation 3 shows an experimentally accessible expression of the rate of reaction 1.

rate = 1 [S2O32-] Eq. 3 2 t

where [S2O3-2] is the initial concentration of thiosulfate in each run, and t is the time it takes for the blue starch/iodine complex to appear.

The rate law for reaction 1 is shown in equation 4:

rate = k [H2O2]x [H+]y [I-]z Eq. 4

where k is the rate constant, [H2O2] represents the molar concentration of hydrogen peroxide and x is the partial order of the peroxide in the reaction. Other reactant concentrations are raised to their own partial orders, respectively.

Explanation / Answer

Beaker A concentration

Run                          H2O2                                                     HCl

1              0.5 M x 10 ml/20 ml = 0.25 M               0.5 M x 10 ml/20 ml = 0.25 ml    

2              0 5 M x 5 ml/15 ml = 0.166 M               0.5 M x 10 ml/15 ml = 0.33 ml

3              0.5 M x 10 ml/15 ml = 0.33 M               0.5 M x 5 ml/15 ml = 0.166 ml

4         0.5 M x 10.99 ml/20.99 ml = 0.26 M       0.5 M x 10 ml/20.99 ml = 0.24 ml

Beaker B concentration

Run                              KI                                                            Na2S2O3

1              0.05 M x 10 ml/30 ml = 0.016 M               0.01 M x 10 ml/30 ml = 0.003 ml    

2              0 05 M x 10 ml/35 ml = 0.014 M               0.01 M x 15 ml/35 ml = 0.0043 ml

3              0.05 M x 10 ml/35 ml = 0.014 M               0.01 M x 15 ml/35 ml = 0.0043 ml

4               0.05 M x 5 ml/30 ml = 0.008 M                0.01 M x 15 ml/30 ml = 0.005 ml

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