Gravimetric Analysis

[Shubhashish Banerjee]

Hey guys! I am new to this ATAR Notes stuff so please pardon my noob questions/mistakes. I had a question on gravimetric analysis that I solved but I am not sure if the method I used was correct (I feel like I just did random stuff and the answer happened to be right).

Q. The iodide ions in a solution containing 0.300g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed?

Thanks!

[brightsky]

Find M(NaI).
Use n=m/M to find n(NaI).
This will equal n(AgI).
Find M(AgI).
Use m=nM to find req mass.

[Bad Student]

Hey Shubh

Firstly, you need to write out the full balanced equation.



Secondly you need to find the amount of iodide ions.



As you can see, the amount of silver iodide is equal to the amount of iodide ions.

Therefore,

[Stick]

Hey guys! I am new to this ATAR Notes stuff so please pardon my noob questions/mistakes. I had a question on gravimetric analysis that I solved but I am not sure if the method I used was correct (I feel like I just did random stuff and the answer happened to be right).

Q. The iodide ions in a solution containing 0.300g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed?

Thanks!

It's all in the ratios. It's nice here because n(NaI)=n(AgI) but sometimes the ratio isn't always 1:1. Good luck.

[teletubbies_95]

@Shubh: Hey!!

There is no such thing as "noob " question! It's great that your asking! Next time, can you post on the Chem 3/4 question thread , so its more organised ! yeh?

Thank you very much for asking questions and keep them coming!

@Badstudent: if Na(solid) wouldnt that be a precipitate ?

[Bad Student]

@Badstudent: if Na(solid) wouldnt that be a precipitate ?

OMG you're right. This is why you shouldn't take advice from a bad student.

Should it be instead?

[teletubbies_95]

Haha. Dont say your a bad student! Confidence !
Yeh I think so , but i usually put like a soluble anion with the Ag ( ie. PO4 3- ) , so it becomes aqueous in the equation . But i guess , Na+ would be right? I'll let the seniors answer that. Unsure?

[Bad Student]

So basically there's a spectator ion that the question has left out for the sake of simplicity, is that right?

[teletubbies_95]

Haha. Yep. I would just include the soluble anion so it seems like im doing everything right.
But i could be wrong ?

[Shubhashish Banerjee]

Thanks for all the help guys (I'll remember to post on the Chemistry 3/4 thread next time!). That is what I did but my only question was that how do you know that n(NaI)=n(AgI). That was the part I guessed and why I felt unsure about the question :S. Like when @Badstudent said that NaI reacts with Ag how can you know that? Or does that not matter? Thanks again guys.

[Bad Student]

Thanks for all the help guys (I'll remember to post on the Chemistry 3/4 thread next time!). That is what I did but my only question was that how do you know that n(NaI)=n(AgI). That was the part I guessed and why I felt unsure about the question :S. Like when @Badstudent said that NaI reacts with Ag how can you know that? Or does that not matter? Thanks again guys.

From the equation, we see that for every sodium iodide that reacts, one silver iodide is produced. This shows that the ratio of NaI to AgI is 1:1.

[Shubhashish Banerjee]

That's my question. How did you know to react sodium iodide with silver? Sorry if I am being stupid it's just that I like to get my understanding perfectly right. I really appreciate the help.

[Jaswinder]

 The iodide ions in a solution containing 0.300g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed?

So from the question you get something like this

NaI+some element = AgI+ Na, hence you can find out what the "some element" is going to be ie silver

Also for it to be any other ratio suppose 2:1 you would need 2NaI. For that they would need to state something like Ag(II)(2+) then you would need 2 Iodine and the whole equation would change.

[Shubhashish Banerjee]

Oh cool that makes sense! I was just thinking that it didn't necessarily have to be Ag, it could have been AgNO3 but I guess that makes the same molar ratio. Thank you everyone for your help .

[Shubhashish Banerjee]

Hey @teletubbies_95. You said to post on the Chemistry 3/4 Thread right? But everyone seems to be talking mostly about just general Chemistry stuff and I feel like I would be interrupting them. Same thing with Specialist Maths. They have a question thread but I don't know if I should just randomly put in a question or like wait for the last person to finish their query (which could take ages). What should I do? Oh and @Badstudent how did you do write out your equations like that. I feel so stupid right now haha. Thanks guys!