Someone help me with my titration molarity calculation please!?
I have titrated 1.000 mL of Ca(IO3)2 with 2.197 ml of 0.022 M thiosulfate.
First I found the moles of thiosulfate present with
0.002197 L x 0.022 M = 4.8388E-5.
The mole ratio of S2O3^-2 and IO3- should be 6:1 (Please tell me if that is correct or not!)
So 4.8388E-5/6 = 8.0648E-6 moles.
I found the Molarity with this equation:
M = 8.0648E-6 moles/.001 L
= .00806 M IO3-
Can anyone please check if this is correct? I want to know before I continue and make a tremendous amount of mistakes if it's wrong.
First I object to saying you titrated Ca(IO3)2 with S2O3^-. You didn't do that; you added I^- to IO3^- and titrated the liberated I2 with S2O3^-
Here are the equations.
Ca(IO3)2 ==> Ca^2+ + 2IO3^-
IO3^- + 5I^- ==> 3I2 (which isn't complete but the redox part is balanced).
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
So 1Ca(IO3)2 = 2IO3^- = 6I2 = 12S2O3^2-
I think your first calculation is wrong.
0.002197 x 0.022 = 4.833E-5 mols S2O3^2-.
mols IO3^- = 4.833E-5 x (1 mol Ca(IO3)2/12 mols S2O3^2-) = ?
Then M = mols Ca(IO3)2/L Ca(IO3)2 = ?
Check that ratio. Just follow the multipliers fromj one equation to the next and you get 1:12.
Your calculations are correct up until the mole ratio.
The correct mole ratio between thiosulfate (S2O3^-2) and IO3- is actually 3:1. This is because in the balanced chemical equation for the reaction, 2 moles of thiosulfate react with 1 mole of IO3-.
So when you divide 4.8388E-5 moles by 3, you get 1.61296E-5 moles of IO3-.
Now, to calculate the molarity of IO3-, you divide the moles of IO3- by the volume in liters (0.001 L):
Molarity (M) = 1.61296E-5 moles / 0.001 L
= 0.0161296 M IO3-
Therefore, the molarity of Ca(IO3)2 is 0.0161296 M.
Please double-check your calculations with these corrections to ensure accuracy.