I am trying to find moles of IO3 from Cu(IO3)2 10 ml and standardized sodium thiosulfate (Na2S2O3.5H20) .28 M.
My calc is:
0.28 Moles Na2S2O3.5H20 * (2 moles of S2O3/1 mole Na2S2O3.5H20)* (2 moles of (IO3)/1 mole Cu(IO3)2= (1.12/.01 L)
but that is why to high for (M) what am I doing wrong?
What's the .28 M by Na2S2O3.5H2O? Is that molarity of the thiosulfate? Where's the equation. You can't do anything without an equation. You aren't titrating the IO3 directly are you? I presume you may have added KI to the solution, generated I2, and titrating the I2 with the thiosulfate. Clarify what the experiment is and we can help. As it stands now, however, the post is too disjointed to know what is going on.
I apologize for the disjointedness,
I pipetted 5 ml of KIO3, 25 ml of H20, 10ml of Acetic acid, and 1.3g of KI
I then Titrated the sodium thiosulfate did 3 trials.
Trial one gave me 28.22 ml, I then calculated (M) by taking KI 1.3 g/MW(166.00)then dividing that number by (28.22/1000 ml) to get (M)of Na2S2O3
Then i did 3 more trials consisting of
10 ml of Cu(IO3)2, 25 ml of H20, 10ml of Acetic acid, and 1.3g of KI
I now need to find (M) of IO3 so i can find the Ksp of the reaction
I know the Ksp equation is
S=[Cu2+]=(1/2)[IO3-]
Ksp=4S^3
Frankly, it makes (almost) no difference how much KI you took, as long as it is enough to react with the KIO3. You don't list a mass KIO3 but I suspect you weighed KIO3 as a primary standard. Then you added the KI to generate I2, then titrated the I2 with the thiosulfate. First you must write the equation for KIO3 and KI.
IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O
Then you titrate the liberated I2 with S2O3^-2.
I2 + 2S2O3 ==> 2I^- + S4O6^-2
Now you can do some calculation.
moles KIO3 = grams/molar mass. You don't have the mass KIO3 in your post but you must have it in your notes.
Then moles KIO3 x 3 = moles I2.
moles I2 = 2 moles Na2S2O3; therefore, 1 mole KIO3 = 6 moles Na2S2O3 or
1/6 mole KIO3 = 1 mole Na2S3O3.
So you take the grams of KIO3, divide by 6 to give you moles Na2S2O3, then use
M = moles/L. Plug in moles Ma2S2O3 from above and divide by L Na2S2O3 (which is 28.22 mL or 0.02822 L). That will give you M Na2S2O3.
Now that I know you are trying to calculate Ksp for Cu(IO3)2, what is the 0.28 M? Is that the molarity of the Na2S2O3 solution?
Unfortunately, my KIO3 was given to me in mL from my professor so i have no clue what the grams of KIO3. I tried to figure out the moles of Na2S2O3 solution but looking over it i believe that 0.28 M is incorrect.
The info I have is:
2.6g of Na2S2O3 was added to roughly 1 mL of distilled water Titrated this solution into a beaker filled with
5 ml of KIO3, 25 ml of H20, 10ml of Acetic acid, and 1.3g of KI
so I figured I could find Molarity of Na2S2O3 solution by taking (grams of KI/MW of KI)/(mL of Na2S2O3/1000)
I know I have to find the (M) of the Na2S2O3 in the standard, then find the (M) of IO3- to find the Ksp but Im not finding (M)Na2S2O3 and I have no clue how to find (M) of IO3-
To find the moles of IO3 from Cu(IO3)2 and standardized sodium thiosulfate (Na2S2O3.5H2O), you need to use the following steps:
1. Convert the volume of Cu(IO3)2 to liters. Given that you have 10 mL, divide by 1000 to convert to liters: 10 mL / 1000 = 0.01 L.
2. Calculate the moles of Na2S2O3.5H2O using its molarity and volume. You have a concentration of 0.28 M and a volume of 0.01 L. Use the equation: moles = concentration × volume = 0.28 M × 0.01 L = 0.0028 moles Na2S2O3.5H2O.
3. Determine the stoichiometry between Na2S2O3.5H2O and IO3 in the reaction. From the balanced equation, it appears that for every 2 moles of S2O3, there are 2 moles of IO3.
4. Now use the stoichiometry to find the moles of IO3. Multiply the moles of Na2S2O3.5H2O (0.0028 moles) by the ratio of moles of IO3 to moles of Na2S2O3.5H2O, which is 2 moles of IO3 / 2 moles of S2O3: moles of IO3 = 0.0028 moles Na2S2O3.5H2O × (2 moles IO3 / 2 moles S2O3) = 0.0028 moles IO3.
Based on these calculations, the number of moles of IO3 from Cu(IO3)2 and standardized sodium thiosulfate is 0.0028 moles. The error in your calculation may have been due to a discrepancy in the conversion from mL to L; always ensure you have the correct volume in liters to accurately calculate moles.