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Author: Subject: Golden rain experiment ruined . Did not recrystallise out on cooling
vibbzlab
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[*] posted on 29-12-2019 at 09:34
Golden rain experiment ruined . Did not recrystallise out on cooling


I was trying to make lead iodide precipitate by heating the precipitate formed by mixing potassium iodide with lead nitrate. On heating the ppt dissolved but didn't recrystallised out on cooling.
P .S I had heated it for quite a long time.
Did it decompose or any chemical change occured? Please explain





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[*] posted on 29-12-2019 at 12:34


You'll have to be more specific than that if you want help. What procedure you followed, how much of each substance you used, ...
Lead iodide is slightly soluble in cold water (0.044 g/100 ml according to wiki) so maybe all of it stayed dissolved.
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[*] posted on 29-12-2019 at 13:33


Did it properly cool down? If yes you maybe have a supersaturated solution. I would try scraping the beaker with a glass rod to scratch the surface in order to form a crystalisation germ. Or, if you have any, throw a small crystal of PbI2 in so it can aid in crystalisation.
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[*] posted on 29-12-2019 at 13:44


have you tried evaporating some of the water? as mentioned above, maybe it's still dissolved in solution. you can also try freezing a small sample of it in the freezer just for a test.

I think reducing the water volume is the best way to go, because either way - something would precipitate.
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[*] posted on 29-12-2019 at 16:55


I tried to crystallise out by evaporating and keeping in refrigerator. But still nothing happened.i remember the color was odd . It had a light brown tinge. Very light brown. I wonder what happened




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[*] posted on 29-12-2019 at 20:06


Any purple 'smoke' leaving on heating? Maybe you drove off and lost all of your iodine.
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[*] posted on 30-12-2019 at 16:30


No there wasn't any smoke




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[*] posted on 31-12-2019 at 03:01


did you use store bought lead nitrate? or did you make it yourself?
did you use tap water? all glassware?
anything unusual about the procedure you followed?





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[*] posted on 31-12-2019 at 07:44


How did you measure out the reactants? In other words; what was the molar ratio of the reactants? Lead iodide is freely soluble in excess iodide ions so its very important you get the ratio right. An excess of lead nitrate is better. Have you tried adding more lead nitrate solution?

When asking questions like this we need quantitative data ie how many grams of lead nitrate and KI did you actually use?
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vibbzlab
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[*] posted on 31-12-2019 at 17:33


I made lead nitrate crystals by dissolving nitric acid with lead acetate.
It was dissolved in distilled water .
And I think I took excess of iodides. Did not try adding lead nitrate into it .I will try it today





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[*] posted on 31-12-2019 at 18:05


Do you mean that you dissolved lead acetate in nitric acid to produce a solution of lead nitrate?
Why do you think that you used excess iodideS? Did you use more than one iodide salt? What quantities of lead acetate and nitric acid did you use (mols, weights, concentrations, volumes)?

Boffis asked you about the quantities of reagents that you used. Your reply ~12 hours later did not answer his question. I am respectfully asking you the same question.

p.s. Happy New Year
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[*] posted on 1-1-2020 at 07:26


Why did you use lead acetate and then add nitric acid? This just adds another layer of unnessecary complexity! I am sure lead acetate will work perfectly well on its own. But what I was getting at was did you actually weigh out the reagents at some point so we can calculate what the molar ratios are? If not only people with ESP or a functioning crystal ball will be able to answer.

The reaction is simple: Pb(NO3)2 + 2 KI = PbI2 + 2 KNO3
But when there is an excess of iodide: PbI2 + 2 KI = K2PbI4 or similar complex salt.

The addition of acetate ions also complicates the issue because lead forms soluble complexes with high concentrations of acetate ions (but not with free acetic acid as far as I am aware) to the extent that lead sulphate is soluble in strong ammonium acetate solution. When lead acetate is used with potassium iodide in stoichiometric proporations the acetate ion concentration isn't enough to prevent precipitation completely, though it may reduce the yield somewhat. However, without some actual weights or molar quantities it is impossible to say what has happened.
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[*] posted on 1-1-2020 at 08:56


That wasn't done in stoichiometric amounts. Sorry about that. But I wanted to know the reason . Like this is new information for me excess iodides would not cause precipitation




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[*] posted on 2-1-2020 at 07:16


I'll be the third person to ask: What amounts of each chemical did you use?

You said it wasn't stoichiometric; that's fine, but we still need to know the amounts you did use. Or did you just throw things together without measuring anything?
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[*] posted on 2-1-2020 at 08:11


This experiment should not be difficult at all if decent chemicals are used. Even if you throw things together (in roughly equal amounts to the eye) you will get a precipitate and on heating you get nice golden glittering crystals. Only if you use a massive excess amount of iodide you may not get a precipitate, or only a transient precipitate, due to formation of a complex ion PbI4(2-).

I suspect that the "lead nitrate", used in this experiment, is not OK.

If you have lead acetate, then do the expeirment with that instead of lead nitrate. It works equally well. Dissolve a small spatula of lead nitrate in 10 ml of water, to which a few drops of dilute acetic acid (8% cleaning vinegar without additives is perfectly fine) were added. In another 10 ml of water dissolve a similarly sized spatula of potassium iodide. Mix the liquids. You get your precipitate, it is a very fine bright yellow precipitate. Heat the liquid. Most (or maybe even all) of the precipitate dissolves and the particles clump together, making the part of the precipitate which does not dissolve more compact. Let cool down and you get nice glittering golden yellow crystals. The amount of crystals looks big in the water, but on evaporating the liquid it becomes clear that in reality the amount is very small. The thin platelets of PbI2 take up a lot of space, with a lot of water between the platelets.




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[*] posted on 2-1-2020 at 16:07


But the thing is I first got that beautiful yellow ppt of lead nitrate. Then I just heated it. On heating it up, the whole ppt dissolved and solution had a light brown. I am pretty sure I over heated it. I kept the flask containing about 50ml fluid with a lot of ppt directly on high flame for about 15min. It kept boiling.




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[*] posted on 3-1-2020 at 00:12


PbI2 in solution cannot be overheated. It simply dissolves (to some extent) in boiling water and does not react. I think that you only had a very little amount of the precipitate, which locally formed, and that you dissolved all on heating. The brown color may be because you oxidized a lot of iodide to iodine with excess HNO3. Even dilute HNO3 is capable of oxidizing iodide at boiling hot temperatures. The iodine may have caused the brown color. Iodine may even be further oxidized to iodate/iodic acid by HNO3. If that occurred, then you don't get any precipitate on cooling down, HIO3 does not react with lead(II) at low pH.

So, my suggestion is to retry the experiment without any HNO3. If you solution of lead acetate is not entirely clear, you can add a few drops of dilute acetic acid, as I wrote before.

[Edited on 3-1-20 by woelen]




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vibbzlab
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[*] posted on 4-1-2020 at 08:27


I had retried it and I got the ppt. I used it to create some crystal balls as well for the Christmas
https://youtu.be/LzQXkSR60vg here you go





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