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Titration question help needed

So I dont get d ii . My working out is below. Can somebody explain to me where I have went wrong? I ended up with the number of moles of I2 remaining as 1.02 x 10^-4 but it's wrong. MS below as well
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Original post by Kalabamboo
So I dont get d ii . My working out is below. Can somebody explain to me where I have went wrong? I ended up with the number of moles of I2 remaining as 1.02 x 10^-4 but it's wrong.


(i) 25ml of 0.005M = 1.25 x 10-4 mol iodate

hence moles of iodine produced = 3 x 1.25 x 10-4 mol = 3.75 x 10-4 mol iodine

Iodine reacts with E300 in a 1:1 ratio

Unreacted iodine = mol thiosulfate/2 = (0.0204 x 0.005)/2 mol = 5.1 x 10-5
Reply 2
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by charco
(i) 25ml of 0.005M = 1.25 x 10-4 mol iodate

hence moles of iodine produced = 3 x 1.25 x 10-4 mol = 3.75 x 10-4 mol iodine

Iodine reacts with E300 in a 1:1 ratio

Unreacted iodine = mol thiosulfate/2 = (0.0204 x 0.005)/2 mol = 5.1 x 10-5


Thanks a lot. But could you please let me know where I have went wrong in my working and what I can do to fix it below? Sorry if I make you repeat anything. I have a feeling I have to do something to the top equation but I don't know what. Also, the top equation is not in the question. It's just in my working out.
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Original post by Kalabamboo
Thanks a lot. But could you please let me know where I have went wrong in my working and what I can do to fix it below? Sorry if I make you repeat anything. I have a feeling I have to do something to the top equation but I don't know what. Also, the top equation is not in the question. It's just in my working out.
45.PNG

you made a mistake where you wrote "therefore, there are 2 x 1.02 x 10^-4 moles of S203^2-"

There are 1.02 x 10^-4 moles of thiosulphate ions as you mentioned.

But there are half as many iodine molecules as there are of thiosulphate ions. ( I2: S203^2- = 1:2)
What you should do then is 1.02 x 10^-4 / 2 = 5.1 x 10^-5 mol I2 [mol]
What you did is 1.02 x 10^-4 x 2 = 2.04 x 10^-4 mol I2. This is where you made the mistake.

This means there are 5.1 x 10^-5 mol of I2 unreacted with E300.
(edited 6 years ago)
Reply 4
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by dip0
you made a mistake where you wrote "therefore, there are 2 x 1.02 x 10^-4 moles of S203^2-"

There are 1.02 x 10^-4 moles of thiosulphate ions as you mentioned.

But there are half as many iodine molecules as there are of thiosulphate ions. ( I2: S203^2- = 1:2)
What you should do then is 1.02 x 10^-4 / 2 = 5.1 x 10^-5 mol I2 [mol]
What you did is 1.02 x 10^-4 x 2 = 2.04 x 10^-4 mol I2. This is where you made the mistake.

This means there are 5.1 x 10^-5 mol of I2 unreacted with E300.


Thanks a lot! Here is my correction: Could you check it please? Also, the first equation is not from the question. I've just used it to help in my working out. Does that make sense? Have I balanced it correctly? I just made the thiosulfate ion have the same no. of moles in both equations for it to work.
49.PNG
Original post by Kalabamboo
Thanks a lot! Here is my correction: Could you check it please? Also, the first equation is not from the question. I've just used it to help in my working out. Does that make sense? Have I balanced it correctly? I just made the thiosulfate ion have the same no. of moles in both equations for it to work.
49.PNG


correct
Reply 6
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by dip0
correct


Brilliant!! Thanks a lot for helping me with another question!! :biggrin::smile:
Original post by Kalabamboo
Brilliant!! Thanks a lot for helping me with another question!! :biggrin::smile:


np

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