20 mL of 0.125 M potassium iodide solution is mixed with 10 mL of 0.1 g L–1 sodium

thiosulfate 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.

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?

a. 0.125M x 20/50 = ?

b. 0.025 x 20/50 = ?
c. mass = 0.1g/L x 0.010 L = ?
Then mols = grams/molar mass
d. mols I2 = M I2 x L I2 = ?
e. I2 + 2S2O3^2- ==> 2I^- + S4O6^2-
Therefore, delta mols I2 = 1/2 mols S2O3^2-

To solve this problem, we'll use the principles of stoichiometry and the given information.

a) What is the initial concentration of iodide in the reaction mixture?

We have 20 mL of 0.125 M potassium iodide (KI) solution. To find the initial concentration of iodide, we need to convert mL to L:

20 mL = 20/1000 L = 0.02 L

Using the formula: concentration (M) = moles (mol) / volume (L)

Concentration of iodide = moles of iodide / volume of solution
0.125 M = moles of iodide / 0.02 L

Rearranging the formula, moles of iodide = concentration x volume
moles of iodide = 0.125 M x 0.02 L = 0.0025 mol

Therefore, the initial concentration of iodide in the reaction mixture is 0.0025 M.

b) What is the initial concentration of peroxodisulfate in the reaction mixture?

We have 20 mL of 0.025 M peroxodisulfate (peroxodisulfate) solution. Similar to the previous calculation, we need to convert mL to L.

20 mL = 20/1000 L = 0.02 L

Concentration of peroxodisulfate = moles of peroxodisulfate / volume of solution
0.025 M = moles of peroxodisulfate / 0.02 L

Rearranging the formula, moles of peroxodisulfate = concentration x volume
moles of peroxodisulfate = 0.025 M x 0.02 L = 0.0005 mol

Therefore, the initial concentration of peroxodisulfate in the reaction mixture is 0.0005 M.

c) What is the number of moles of thiosulfate consumed after 46 seconds of the reaction?

The question states that the thiosulfate was completely consumed. This means that all the moles of thiosulfate reacted with iodine.

Since we don't have the reaction equation, we can assume that the reaction between thiosulfate (S2O3^-2) and iodine (I2) produces tetrathionate (S4O6^-2) and iodide (I^-).

The stoichiometry of the reaction can be represented as:

2 S2O3^-2 + I2 → S4O6^-2 + 2 I^-

From the stoichiometry, we can see that 2 moles of thiosulfate react with 1 mole of iodine.

Since we know the initial concentration of iodide (0.0025 M), the moles of thiosulfate consumed will be twice that of iodide.

moles of thiosulfate consumed = 2 x moles of iodide
moles of thiosulfate consumed = 2 x 0.0025 mol = 0.005 mol

Therefore, the number of moles of thiosulfate consumed after 46 seconds of the reaction is 0.005 mol.

d) What is the number of moles of iodine formed after 46 seconds of the reaction?

From the stoichiometry given above, we can see that 2 moles of thiosulfate react with 1 mole of iodine. Therefore, the number of moles of iodine formed will be half that of the moles of thiosulfate consumed.

moles of iodine formed = moles of thiosulfate consumed / 2
moles of iodine formed = 0.005 mol / 2 = 0.0025 mol

Therefore, the number of moles of iodine formed after 46 seconds of the reaction is 0.0025 mol.

e) What is the change in concentration of iodine (I2) formed after 46 seconds of the reaction?

The change in concentration of iodine can be calculated by comparing the initial concentration of iodide (0.0025 M) to the number of moles of iodine formed (0.0025 mol).

Change in concentration = moles of iodine formed / volume of solution
Change in concentration = 0.0025 mol / 0.02 L

Rearranging the formula, we get the change in concentration as:

Change in concentration = 0.125 M

Therefore, the change in concentration of iodine formed after 46 seconds of the reaction is 0.125 M.

To answer these questions, we can use the principles of stoichiometry and the given information about the reaction mixture.

a) To find the initial concentration of iodide, we need to calculate the moles of iodide present in the 20 mL of 0.125 M potassium iodide solution. To do this, we can use the formula:

moles of iodide = volume of solution (in L) * concentration of iodide (in mol/L)

Given that the volume is 20 mL (or 0.02 L) and the concentration is 0.125 M, we can calculate:

moles of iodide = 0.02 L * 0.125 mol/L
moles of iodide = 0.0025 mol

Therefore, the initial concentration of iodide in the reaction mixture is 0.0025 mol.

b) Similarly, to find the initial concentration of peroxodisulfate, we can use the same formula:

moles of peroxodisulfate = volume of solution (in L) * concentration of peroxodisulfate (in mol/L)

Given that the volume is 20 mL (or 0.02 L) and the concentration is 0.025 M, we can calculate:

moles of peroxodisulfate = 0.02 L * 0.025 mol/L
moles of peroxodisulfate = 0.0005 mol

Therefore, the initial concentration of peroxodisulfate in the reaction mixture is 0.0005 mol.

c) To find the number of moles of thiosulfate consumed after 46 seconds, we need to know the stoichiometric ratio between thiosulfate and iodine. According to the balanced equation, the ratio is 2:1:

2 Na2S2O3 + I2 → Na2S4O6 + 2 NaI

Since the stoichiometric ratio is 2:1, the number of moles of thiosulfate consumed will be half the number of moles of iodine formed. Therefore, the number of moles of thiosulfate consumed after 46 seconds is:

moles of thiosulfate consumed = 0.0025 mol (moles of iodine formed) / 2
moles of thiosulfate consumed = 0.00125 mol

d) Similarly, the number of moles of iodine formed after 46 seconds is equal to the number of moles of thiosulfate consumed (as per the stoichiometric ratio):

moles of iodine formed = 0.00125 mol

e) To find the change in concentration of iodine formed after 46 seconds, we need to consider the initial and final volumes of the reaction mixture. The final volume is the sum of the initial volumes of the solutions added (20 mL KI + 10 mL thiosulfate + 20 mL peroxodisulfate). Therefore, the final volume is 50 mL (or 0.05 L).

Now we can calculate the change in concentration of iodine using the formula:

change in concentration = (moles of iodine formed) / (final volume)

change in concentration = 0.00125 mol / 0.05 L
change in concentration = 0.025 M

Therefore, the change in concentration of iodine after 46 seconds of the reaction is 0.025 M.