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-
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?
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1 answer