Use the information below to develop the necessary calculations for the rate of

Question

Use the information below to develop the necessary calculations for the rate of reaction from the solutions in part B of the experiment

Calculate the initial molarity of iodide ion in each of the solutions, once your solutions are mixed.

Remember:   The KI stock solution concentration was known: [KI]supplied, and you used a specific volume of the solution, VKI-. However, at the start of the reaction you must account for the fact that the total volume of each solution was 200 mL.  

Calculate the initial molarity of the hydrogen peroxide in each of your solutions, once the solutions have been mixed.

Remember: The H2O2 stock solution concentration was known: [H2O2]supplied, and you used a specific volume of the solution,VH2O2.  However, at the start of the reaction you must account for the fact that the total volume of each solution was 200 mL.

Calculate the number of moles of sodium thiosulfate, Na2S2O3, that was initially present in each solution.

Remember:  The Na2S2O3 stock solution concentration was known: [Na2S2O3]supplied and you used a specific volume of the solution, VNa2S2O3.  This information is used to determine the number of moles.

The following chart will be helpful as a checklist for the preparation of the solutions which will be used in this experiment: Use pipettes to measure the buffer, Na2S203 and the H202 solutions. Use a burette to measure the KI solution and a graduated cylinder for the distilled water. Use the graduated dropper provided to measure the starch. 0.0200 0.1 M Distilled stilledbuffer 0.300 M starch H202 sol'n H2o mL 164.5 162 158 148 128 (mL) KI (mL) (mL)Na,s,O, (mL) Na2S20; (mL) 10 10 10 10 10 10 10 10 20 40 3.5 10 10 10 10 10 Note that the total volume in each case is 200 mL.

Explanation / Answer

1)

stock concentration (C) of KI stock solution = 0.300 M

volume of KI (VKI)= 3.5 ml = 0.0035 L

total volume of solution = 200 ml = 0.2 L

M = n / V

n = M x V

moles of KI = concentration (M) x volume (V)

                   = 0.300 x 0.0035

                   = 1.05 x 10^-3 moles of KI

initial concentration of iodide ion C iodide = moles of KI / total volume

                                                                = 1.05 x 10^-3 / 0.2

                                                                = 5.25 x 10^-3 mol / L

initial concentration of iodide ion C iodide = 5.25 x 10^-3 mol / L

2 )

stock concentration (C) of H2O2 stock solution = 0.100 M

volume of H2O2 (VH2O2) = 10 ml = 0.01 L

total volume of solution = 200 ml = 0.2 L

M = n / V

n = M x V

moles of H2O2 = concentration (M) x volume (V)

                  = 0.1 x 0.01

                  = 1.0 x 10^-3 moles of H2O2

initial concentration of H2O2 C H2O2 = moles of H2O2 / total volume

                                                                = 1.0 x 10^-3 / 0.2

                                                                = 5 x 10^-3 mol / L

initial concentration of H2O2 C H2O2 = 5 x 10^-3 mol / L

3)

stock concentration (C) of Na2S2O3 stock solution = 0.0200 M

volume of Na2S2O3 (VNa2S2O3) = 10 ml = 0.01 L

M = n / V

n = M x V

moles of Na2S2O3 = concentration (M) x volume (V)

                  = 0.0200 x 0.01

                  = 2 x 10^-4 moles of Na2S2O3

initial moles of Na2S2O3 = 2 x 10^-4 moles

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