10g of AgI is dissolved in 100cm3 of water and Na2SO4(s) is added to the solution with the ppt.After about 10mins.25cm3 of the solution is taken and added Cl2 and heated.The remains Cl2 is removed and a small amount of KI(s) is added and then the solution is titrated with 0.02 M NaS2O3 under starch as the indicator.Find the mass of the ppt of the initial solution.
I know the solution is titrated with Na2S2O3 to find the moles of I2 created.
But I want to know the reasons for
1)adding Cl2 and then heating
2)adding Na2SO4 ( guess it is done so that the ppt Ag2SO4 will appear)
3)and again why KI solid is added (guess it is done to find the I- moles in the solution.But why not simply titrate the solution with Na2S2O3 and find the I- moles.or is it because it is I2 we usually titrate with.
Or because I2+ I- ---> I3(-) )
4) By calculating the moles of I2 with the results of the titrations and then calculating the moles of I- by the stoichiometry of the reaction I've mentioned in my 3rd question , do we simply get the free I- ion moles in the solution, so that we can find the remaining mass of the ppt AgI?
Can I know why dont we simply find the remaining I- after keeping the solution for some time,after adding AgI.
Why do we have to wait 10mins after adding Na2SO4? Is it because dissolving Ag2SO4 in water takes time?
I don't know the purpose of the procedure in your post. I need to know the purpose of the process but I can answer some of them. My best guess is that you are determining the Ksp for AgI.
Adding Cl2 and heating probably is to oxidize some of the I^- in AgI to I2.
KI is added to dissolve the I2 formed from the Cl2 oxidation by forming I3^- because I2 itself is not too soluble. Titrating I2 directly is difficult because of the limited solubility of I2 in aqueous solution. Adding KI takes care of that by dissolving all of the I2 forming I3^-,
As for Ag2SO4 I'm lost without knowing the procedure because Ag2SO4 is far more soluble than AgI. Look at the Ksp for AgI vs Ksp Ag2SO4 (but caution: Ksp predicts solutility ONLY for the same kind of salt; i.e., AgI vs AgCl vs AgBr or but not AgI vs Ag2SO4.).
This is a given to us at an exam and they haven't told a reason for the process but have given those details(I've mentioned before MY questions ) for us to answer the question.
I was curious about procedure and the reasons.
My best guess about the addition of Ag2SO4 is that Ag2SO4 is not all that soluble (not all that insoluble either) but it provies a small amount of added Ag^+ to act as a common ion and reduce the solubility of AgI even more than it already is.
Well...If this is about finding the ksp of AgI,why do we reduce of the solubility of AgI by adding Ag+ to act as a common ion? So it is more like finding the solubility of AgI in a solution with the same cation
That didn't make sense to me either.
1) Adding Cl2 and then heating is done to oxidize the iodide ions (I-) to iodine (I2). The reaction between chlorine (Cl2) and iodide ions is as follows:
2Cl2 + 2I- → I2 + 2Cl-
By adding Cl2 and heating, the iodide ions are converted to iodine, which will make it easier to titrate with sodium thiosulfate (Na2S2O3) later on.
2) Adding Na2SO4 is done to precipitate silver sulfate (Ag2SO4). When sodium sulfate (Na2SO4) is added to the solution, it reacts with silver ions (Ag+) present from the dissolution of AgI. The reaction is as follows:
Ag+ + SO42- → Ag2SO4(s)
The formation of the silver sulfate precipitate helps isolate the silver ions from other interfering species in the solution.
3) KI solid is added to convert any remaining iodine (I2) into iodide ions (I-). This step ensures that all the iodine present in the solution is converted to iodide ions before titration with Na2S2O3. The reaction between iodine and potassium iodide is as follows:
I2 + 2KI → 2K+ + 2I-
By converting the remaining iodine to iodide ions, it ensures that the moles of iodine formed can be accurately determined during titration.
4) By calculating the moles of I2 with the results of the titrations, you can determine the moles of iodine produced. Then, using the stoichiometry of the reaction you mentioned (I2 + I- → I3-), you can calculate the moles of iodide ions present in the solution. This allows you to determine the moles of iodide ions not used in the reaction and subsequently the moles of AgI that were present initially. By knowing the moles of AgI and its molar mass, you can calculate the mass of the precipitate formed.