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Topic: Regarding forming of precipitates  (Read 308 times)

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Offline TheNicklePickle

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Regarding forming of precipitates
« on: August 05, 2019, 04:07:28 AM »
I am an Australian student currently studying the new NSW Chemistry syllabus, and am up to the final module, "Applying Chemical Ideas".

I had a general question behind the formation of precipitates. Let me paint a picture. Say you had a system where there was one silver ion, a chromate ion, and a chlorine ion. Assume equal temperatures, and are all dissolved in water. What anion does the silver cation react with to form a precipitate? Am I correct in assuming it is chlorine? Is it because of electronegativity? If not, what?

Thank you.

Offline AWK

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Re: Regarding forming of precipitates
« Reply #1 on: August 05, 2019, 04:28:14 AM »
Calculate from solubility products and compare solubilities them
AWK

Offline TheNicklePickle

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Re: Regarding forming of precipitates
« Reply #2 on: August 05, 2019, 04:35:24 AM »
Calculate from solubility products and compare solubilities them
I probably am just misunderstanding what you're saying, but I am not sure what you mean.

The solubility products of silver chloride is higher than silver chromate, making it more soluble than silver chromate. Yet in Mohr's method of precipitation titration, if you add chromate ions to a solution with silver ions and chlorine ions, why won't the silver react with the chromate until the chlorine is used up?

Offline mjc123

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Re: Regarding forming of precipitates
« Reply #3 on: August 05, 2019, 05:51:24 AM »
Consider the concentrations of chloride and chromate in a sample for titration, according to the numbers used e.g. in this reference:
https://pdfs.semanticscholar.org/90e7/ac0f90bf50c0cbf0811242dc59c436008e2a.pdf
Calculate, from the solubility products, the minimum concentration of Ag+ needed to precipitate AgCl and Ag2Cr2O7. Which will precipitate first?
Although the number for the Ksp of silver chromate is smaller than that of silver nitrate, bear in mind
(i) the initial concentration of chromate is much lower than that of chloride
(ii) the Ksp of silver chromate involves the square of the silver concentration. (What are the units of Ksp? Are the two numbers directly comparable?)

Offline Enthalpy

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Re: Regarding forming of precipitates
« Reply #4 on: August 05, 2019, 06:28:37 AM »
[...] Is it because of electronegativity? If not, what?

Nope. Solubility is complicated, it doesn't relate with other simple attributes, and is hard to predict.

Whether a pair of ions is soluble, or rather how much soluble it is, depends on how easily the ions are solvated and on how nicely they fit together in a crystal. You can already infer that solubility is a matter of ion pairs. This explains why mixing two salts can let a new one precipitate.

How easily an ion is solvated depends on the solvent very much, and on how many charges the ion bears, how big it is (including polyatomic ions that may spread the charge), and half a zillion more reasons more or less unknown.

How favourable a crystal is depends on the relative radii of the ions, how they can be ordered, what charge each one carries, and so on and so forth.

Any prediction, or even computation, hence relies on experimental tables essentially. Sorry for the bad news. What theories can bring is limited (but useful): not much more than extrapolate the solubility from one temperature to a not too different one.

Offline AlirezaDehghani

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Re: Regarding forming of precipitates
« Reply #5 on: August 07, 2019, 10:57:02 AM »
silver ion would participate with chlorine ion first because of their relative solubility constants.
actually participation doesn't depend on electronegativity and it best can be judged by the HSAB theory.
And I have to say the question you asked is somehow related to the Mohr titration. I suggest you to read an article about it.

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