Reaction Of NaI with nitrate salt..Please Help!

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[Thung Su Ling]

I have a question here. In rate of reaction question, we usually put potassium iodide to react with a salt Z nitrate to produce precipitate potassium iodide. However, there is a question here. Why can't we use this method for sodium iodide? Is it because it doesn't produce precipitate?

[Lim Tze Khai]

..put potassium iodide to react with a salt Z nitrate to produce precipitate potassium iodide.

Please check is there any mistake in the precipitate form??

[Ting Tin Chung]

I have a question here. In rate of reaction question, we usually put potassium iodide to react with a salt Z nitrate to produce precipitate potassium iodide. However, there is a question here. Why can't we use this method for sodium iodide? Is it because it doesn't produce precipitate?

Hi, just want to know any mistake for this sentence : we usually put potassium iodide to react with a salt Z nitrate to produce precipitate potassium iodide.

As i know, potassium iodide is not a precipitate, because potassium / sodium salt are all soluble salts..... correct me if i'm wrong...
TQ

[Thung Su Ling]

Sorry...mistake..the question goes like that....
The formula of an iodide of an element Z can be determined by using the following method : 5 cm3 of 0.5 moldm-3 potassium iodide is poured into each seven test tubes labeled I, II, III,..... to VII. 0.5 cm3 of 0.5 moldm-3 of a solution of Z(NO3)2 is added to test tube I, 1.0 cm3 to test tube II, and so on until 3.5 cm3 to test tube VII.
Now, the question is Explain why the formula of sodium iodide cannot be determined by using the above experiment.

[Ting Tin Chung]

Sorry...mistake..the question goes like that....
The formula of an iodide of an element Z can be determined by using the following method : 5 cm3 of 0.5 moldm-3 potassium iodide is poured into each seven test tubes labeled I, II, III,..... to VII. 0.5 cm3 of 0.5 moldm-3 of a solution of Z(NO3)2 is added to test tube I, 1.0 cm3 to test tube II, and so on until 3.5 cm3 to test tube VII.
Now, the question is Explain why the formula of sodium iodide cannot be determined by using the above experiment.

This one actually is the precipitation reaction / double decomposition reaction. We are using this method to determine the formula for insoluble salt like lead(II) iodide. Sodium iodide is a soluble salt. We need to measure the height of the precipitate form to ensure what is the volume needed to react completely. If no precipitate form, how can we determine the volume of the reactants use? Without the volume of the reactants, definitely we cant determine the formula of the NaI..... correct me if I'm wrong 

[Liew Hui Lee]

Yes, I agree with Mr. Ting.  That is actually double decomposition reaction or precipitation to prepare an insoluble salt from aqueous solutions of two soluble salts, X and Y.  In which X provides us with the cation of the insoluble salt, while Y provides the anion of the insoluble salt.  In this reaction, we use lead(II) nitrate ( coz only lead(II) nitrate soluble in water ) and any soluble iodide salt, normally we use sodium iodide or potassium iodide. 
This is only my opinion, maybe the others chemistry teacher have other sharring...

[Thung Su Ling]

This means that if the student's answer is no reaction would occur is acceptable??

[Liew Hui Lee]

I think in this reaction, Z(NO₃)₂ is actually the lead(II) nitrate.  The method can not be used to determine the formation of sodium iodide because sodium iodide is not an insoluble salt, so it is not suitable to be prepared by double decomposition reaction...

[Tay Guan Piang]

Sorry...mistake..the question goes like that....
The formula of an iodide of an element Z can be determined by using the following method : 5 cm3 of 0.5 moldm-3 potassium iodide is poured into each seven test tubes labeled I, II, III,..... to VII. 0.5 cm3 of 0.5 moldm-3 of a solution of Z(NO3)2 is added to test tube I, 1.0 cm3 to test tube II, and so on until 3.5 cm3 to test tube VII.
Now, the question is Explain why the formula of sodium iodide cannot be determined by using the above experiment.

For your information, we did learn another method when studying preparation of insoluble salt, which is continuous variation method. Sodium iodide is a kind of soluble salt and continuous variation method can't be used to prepare this salt.
A good method to prepare this salt will be the reaction of acid and alkali as below.

HI + NaOH ----> NaI + H₂O

Do correct me if i'm wrong.

[Sim Thiam Huat]

The question says explains why ....
    So if my student answers ' no reaction would occur" . i will give him zero mark. the answer has to be an explanation like the one given by mr. ting & mr, Liew

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