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Topic: Preparation of Salts question  (Read 7972 times)

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Ah Beng

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Preparation of Salts question
« on: September 22, 2005, 05:42:18 AM »

Thanks Borek, for your reply to my last question (posted a while back).

The question that I now need some help with :


Qn : The following passage is about the preparation of lead iodide, and insoluble salt.

"An excess of potassium iodide in solution was shaken with some lead nitrate in a test tube. The lead iodide precipitate was separated from the mixture and then washed several times with water. The lead iodide was dried and placed in a bottle."

a) Suggest a reason why excess potassium iodide is used.


My answer / opinion :

Normally, using the ionic precipitation method to obtain insoluble salts, it should not matter which soluble salt is used in excess, because once the ppt is obtained, simply filter to separate the desired insoluble salt (as the residue) from the excess soluble salt used in excess (as the filtrate).

So why would potassium iodide be desired as the excess, rather than lead nitrate? The only reason I can think of, is that lead salts are considerd more toxic than potassium or iodide salts. Hence, if we were to use excess lead nitrate instead of excess potassium iodide, we would have the difficulty of washing away or dispoosing of the excess toxic lead nitrate, instead of the less toxic potatssium iodide.

Is this correct? Or is there some other explanation?

Thanks in advance for any reply.

Ah Beng

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Re:Preparation of Salts question
« Reply #1 on: September 22, 2005, 09:00:47 AM »
I wonder what the question is about, as it is ambiguous for me.

Is it 'why excess of one of the reagents was added'? Or is it 'why excess of potassium iodide and not lead nitrate was added'?

Best answer should incorporate both problems, just in case. Seems to me you know both answers, so I will not comment further.
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