Given the situation: Burrette 10mL of sodium thiosulfate solution into a clean 2

Question

Given the situation: Burrette 10mL of sodium thiosulfate solution into a clean 250 mL Erlenmeyer flask. Using 50mL pipette, next add 50mL of buffered iodide solution containing the starch indicator. In a clean 150 mL beaker, burette 10mL of stock H2O2 and 20mL distilled water.

a) Immdediately after mixing, what are the initial concentrations of H2O2 and S2O32- after the four stock solutions have been combined in the above situation? Assume that the stock H2O2 and Na2S2O3 solutions are 0.210 and 0.0200 mol/L, respectively, and the volumes specified in the procedure were buretted exactly.

b) A blue colour appears exactly 3 min 5s after the reagents are combined. What concentrations ofH2O2 and S2O32- have reacted at the instant this happens?

c) Calculate the initial rate of reaction for this hypothetical case.

Explanation / Answer

Total volume of the solution after the four stock solutions are mixed = 10mL + 50mL+ 10mL +20mL = 90mL

Calculations for H2O2:

M1V1 = M2V2

M1 = 0.210 mol/L, V1 = 10mL, V2 = 90mL, M2 = ?

M2 = M1V1/M2 = 0.210 mol/L * 10mL/90mL = 0.023 mol/L

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Calculation for Na2S2O3

M1V1 = M2V2

M1 = 0.020 mol/L, V1 = 10mL, V2 = 90mL, M2 = ?

M2 = M1V1/M2 = 0.020 mol/L * 10mL/90mL = 0.0022 mol/L

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(b)

In the first step of the reaction I2 will be liberate which is then reacts with Na2S2O3. As long as there are thiosulphate ion present in the solution, i2 will react with it. After all the thiosulphate is consumes, it reacts with starch to give a blue color. The balanced equation between the reaction of I2 and S2O3^2- is as follows:

I2 + 2Na2S2O3 ----> 2NaI + Na2S4O6

According to the above reaction , 1 mol I2 reacts with 2 moles of Na2S2O3.

Moles of thiosulphate present = 0.0022 *0.09 L =1.98*10^-4 moles

Moles of I2 reacted = 1.98*10^-4 moles/2 = 9.9*10^-5 mol

The balanced reaction of h2O2 and na2S2O3 is as follows:

2I^- + 2H+ + H2O2 -----> I2 + 2H2O

Amount of I2 formed in the above reaction :

moles of I2 formed = moles of H2O2 reacted = 9.9*10^-5 mol

Moles of S2O3^2- reacted in this instant = 1.98*10^-4 mol

moles of H2O2 recated = 9.9*10^-5 mol

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time required = 3 min 5s = 3*60 sec + 5 sec = 185 sec

Initial rate = - [final conc of Na2S2O3- initial conc of Na2S2o3]/Final time-inital time = 1.07 *10^-6 mol/sec

               = - [0-1.98*10^-4]/185sec-0 =

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