Answer:
a) The number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b) The number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c) rate = 2.5 x 10-6M/L/s
Explanation:
a)
The reactions taking place in this experiment are represented by the ionic equations,
S₂O₈ 2- + 2I- ----------> 2SO₄2- + I₂ --------------------(1)
2 S₂O₃2- + I₂ -----------> S₄O₆ +2 I- -------------------(2)
The persulphate ions react with the iodide ions to produce free iodine which is in turn reduced by the thiosulphate ions to produce iodide ions again. This reaction proceeds till all the thiosulphate ions are used up. Therefore the rate of the reaction will be the rate at which iodine is formed and used up.
When there is free iodine the reaction mixture, the solution gives a dark blue coloration. This happens when all the thiosulphate ions are used up.
The volume of sodium thiosulphate ( Na₂S₂O3) solution added to the reaction vessel = 10ml
Molarity of sodium thiosulphate ( Na₂S₂O3) solution = 0.01M
Number of moles of ( Na2S2O3) = 0.01ml x 10M /1000ml = 0.0001M (molarity x volume in L)
The number of moles of thiosulphate reacted will be equal to the number of moles taken since the reaction proceeds till all the thiosulphate is consumed.
Hence, the number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b)
To calculate S₂O₈ 2-
The permanent blue color is produced once all the thiosulphate ions are used up and persulphate reacts with iodide ions to produce iodine, so, the number of moles of persulphate ions will be equal to the number of moles of iodine formed
According to the stoichiometry of equation 1.
1 mole of S₂O₈ 2-produces 1 mole of iodine.
According to the stoichiometry of equation 2,
1mole of iodine produced consumes 2 moles of S₂O₃2-
The number of moles of S₂O₃2- taken = 0.0001M.
2 moles of S₂O₃2- is equivalent to 1 mole I₂
therefore
0.0001 mole of S₂O₃2- = 0.0001/2 = 0.00005M of I₂
Since stoichimetrically,
1 mole of S₂O₈2- is equivalent to 1 mole I2, the number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c)
The initial reaction rate is given by
rate =change in concentration of persulphate ion [S₂O₈2-] / time
rate = change in Concentraion of I₂ / t
since initial concentration of I₂ = 0.
rate = Concentraion of I₂/ t
The concentration of I₂ = number of moles of iodine / total volume of solution in L
= 0.00005M/ 0.1L = 0.0005M/L (Volume of the solution = 100ml = 0.1L)
rate = Concentraion of I₂ / t
= 0.0005 /200s
rate = 2.5 x 10-6M/L/s
