20 mL of 0.125 M potassium iodide solution is mixed with 10 mL of 0.1 g L–1 sodium thiosulfate (Na2S2O3.5H2O) solution and 0.2 g of solid iodine indicator is added.
To this solution is added 20 mL of 0.025 M peroxodisulfate solution.
After 46 seconds a blue colour appears, indicating that sufficient I2 has been formed from the reaction between KI and peroxodisulfate to consume all the thiosulfate present. The final temperature is 18 °C.
1. The following prelab questions are designed to guide you through calculations and processes that will enable you to use experimental data to i) determine the rate of the reaction, ii) derive the rate law and iii) determine the rate constant, k. You will do this during the laboratory using your own
experimental results.
a) What is the initial concentration of iodide in the reaction mixture?
b) What is the initial concentration of peroxodisulfate in the reaction mixture?
c) What is the number of moles of thiosulfate consumed after 46 seconds of the reaction?
d) What is the number of moles of iodine formed after 46 seconds of the reaction?
e) What is the change in concentration of iodine (I2) formed after 46 seconds of the reaction?
20 mL of 0.125 M potassium iodide solution is mixed with 10 mL of 0.1 g L–1 sodium thiosulfate (Na2S2O3.5H2O) solution and 0.2 g of solid iodine indicator is added. To this solution is added 20 mL of 0.025 M peroxodisulfate solution.
So total volume is 20 mL of 0.125 M KI + 10 mL of 0.1 g/L Na2S2O3.5H2O + 20 mL of 0.025 M Na2S2O8 = 50 mL.
a) What is the initial concentration of iodide in the reaction mixture?
0.125 M KI x (20/50) = ? M
b) What is the initial concentration of peroxodisulfate in the reaction mixture?
0.025 M Na2S2O8 x (20/50) = ? M
c) What is the number of moles of thiosulfate consumed after 46 seconds of the reaction?
All of the Na2S2O3 has been consumed. mols Na2S2O3.5H2O = g/molar mass = 0.1 g/L = 0.1/molar mass Na2S2O3.5H2O = ?
Then mols Na2S2O3.5H2O = M x L = M x 0.010 L = ?
d) What is the number of moles of iodine formed after 46 seconds of the reaction?
mols I2 formed = 1/2 mols KI initially - 1/2 mold Na2S2O3 used. That's 1/2 * moles KI from part a - 1/2 * mols Na2|S2O3 from part c.
e) What is the change in concentration of iodine (I2) formed after 46 seconds of the reaction?
(I2) initially - (I2) when all Na2S2O3 is used.
To answer these questions, we need to understand the given information and apply some concepts of stoichiometry and concentration calculations. Let's go through each question step by step:
a) To determine the initial concentration of iodide (KI) in the reaction mixture, we need to consider the volume and concentration of the potassium iodide solution added.
Given information:
Volume of potassium iodide solution (VI) = 20 mL
Concentration of potassium iodide solution (CI) = 0.125 M
The initial moles of iodide (nI) can be calculated using the formula:
nI = CI * VI (moles)
Let's substitute the values:
nI = 0.125 M * 20 mL = 2.5 mmol (moles)
So, the initial concentration of iodide in the reaction mixture is 2.5 mmol.
b) To determine the initial concentration of peroxodisulfate in the reaction mixture, we need to consider the volume and concentration of the peroxodisulfate solution added.
Given information:
Volume of peroxodisulfate solution (VP) = 20 mL
Concentration of peroxodisulfate solution (CP) = 0.025 M
The initial moles of peroxodisulfate (nP) can be calculated using the formula:
nP = CP * VP (moles)
Let's substitute the values:
nP = 0.025 M * 20 mL = 0.5 mmol (moles)
So, the initial concentration of peroxodisulfate in the reaction mixture is 0.5 mmol.
c) To determine the number of moles of thiosulfate consumed after 46 seconds, we need to understand the stoichiometry of the reaction.
From the given information, we know that thiosulfate (Na2S2O3) reacts with iodine (I2) according to the following balanced equation:
2Na2S2O3 + I2 → Na2S4O6 + 2NaI
This equation shows that 1 mole of iodine reacts with 2 moles of thiosulfate.
Given information:
Number of moles of thiosulfate initially = nP (calculated in part b)
Since the ratio of moles of iodine to moles of thiosulfate is 1:2, the number of moles of thiosulfate consumed is nP/2.
Let's calculate that:
Thiosulfate consumed = nP/2 = 0.5 mmol / 2 = 0.25 mmol (moles)
Therefore, the number of moles of thiosulfate consumed after 46 seconds is 0.25 mmol.
d) To determine the number of moles of iodine formed after 46 seconds, we need to apply stoichiometry again.
From the balanced equation given in part c, we can see that 1 mole of iodine is formed by the reaction between 2 moles of thiosulfate.
Therefore, the number of moles of iodine formed is also equal to nP/2.
Let's calculate that:
Number of moles of iodine formed = nP/2 = 0.5 mmol / 2 = 0.25 mmol (moles)
So, the number of moles of iodine formed after 46 seconds is 0.25 mmol.
e) To determine the change in concentration of iodine formed after 46 seconds, we need to consider the volume of the reaction mixture.
Given information:
Volume of the reaction mixture = (Volume of potassium iodide solution) + (Volume of peroxodisulfate solution) = 20 mL + 20 mL = 40 mL
The change in concentration of iodine can be calculated by dividing the number of moles of iodine formed by the volume of the reaction mixture, using the formula:
Change in concentration = Moles of iodine formed / Volume of the reaction mixture
Let's substitute the values:
Change in concentration = 0.25 mmol / 40 mL = 0.25 mmol / 0.04 L = 6.25 mM (millimoles per liter)
Therefore, the change in concentration of iodine (I2) formed after 46 seconds of the reaction is 6.25 mM.
Note: Remember to convert volumes to liters when performing calculations involving concentrations to ensure consistent units.