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    Name:  rfg.png
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Size:  90.0 KB I have a few questions about this question.

    1. Adding silver nitrate will form 3 white precipitate right? Since silver also reacts with the sulphate ions and this also forms a white precipitate.

    2.So wouldn't 2 white precipitates and 1 chloride precipitates form and ALSO one cream precipitate from AgBr ?

    The answer does not talk about sulfate ions reacting with silver nitrate but I thought it will form a white precipitate, no?

    Thanks
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    (Original post by coconut64)
    Name:  rfg.png
Views: 244
Size:  90.0 KB I have a few questions about this question.

    1. Adding silver nitrate will form 3 white precipitate right? Since silver also reacts with the sulphate ions and this also forms a white precipitate.

    2.So wouldn't 2 white precipitates and 1 chloride precipitates form and ALSO one cream precipitate from AgBr ?

    The answer does not talk about sulfate ions reacting with silver nitrate but I thought it will form a white precipitate, no?

    Thanks
    Yes, that is correct.

    The idea is that you choose both the reagents to add and the order in which you do the tests to narrow down and finally identify each solution.
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    (Original post by charco)
    Yes, that is correct.

    The idea is that you choose both the reagents to add and the order in which you do the tests to narrow down and finally identify each solution.
    The mark scheme mentions that the way this is marked is by talking about how the halide ions and the sulfate ions can be identified separately which doesn't make sense. Name:  ikh.png
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Size:  90.8 KB For this example, it says that adding silver nitrate only forms 2 precipitates which are the bromide and chloride but this doesn't make sense though ... The sulfate ions would have reacted as well as you have agreed..

    Is this answer right because this apparently scored full marks...

    Thanks
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    You are right, in so far as what YOU say you should do won't help.

    You are extending the procedure given, so you need to do some things to the test tubes given as well.

    1. You know which two are sulfates, and you know which two are halides. Let's start by distinguishing which sulfate is which.

    What are the differences in reactivity between the NH4+ and the Na+ ions based on the reagents you're given?
    You need to remember your analytical tests. Identification of the ammonium cation is done by adding dilute alkali, and testing the vapour above the solution with moist litmus paper (or indicator). Adding alkali affects the reaction; OH- + NH4+ -----> H2O + NH3. Ammonia vapour is alkaline, and will turn the universal indicator blue. Only one of the sulfates will respond to this test, obviously, so you now know which is which.

    2. The other two

    You have solutions of barium chloride with sodium chloride and potassium bromide.
    Silver nitrate reacts with halides, this you know and forms precipitates as you say. So adding silver nitrate to these two will form precipitates of the chloride from the barium, and either a chloride or bromide from the two tubes. You will not be able to tell based on the colours really, the cream will be undetectable, so another test is needed. Silver chloride dissolves in dilute ammonia solution. Silver bromide does not. Add dilute ammonia to these test tubes; the AgCl precipitate will dissolve, the AgBr precipitate will not. Hence, the KBr solution is the one whose precipitate with AgNO3 will not dissolve in dilute ammonia solution.

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    (Original post by coconut64)
    The mark scheme mentions that the way this is marked is by talking about how the halide ions and the sulfate ions can be identified separately which doesn't make sense. Name:  ikh.png
Views: 66
Size:  90.8 KB For this example, it says that adding silver nitrate only forms 2 precipitates which are the bromide and chloride but this doesn't make sense though ... The sulfate ions would have reacted as well as you have agreed..

    Is this answer right because this apparently scored full marks...

    Thanks
    It doesn't explicitly say the silver nitrate will form 2 precipitates, just that the halides will form precipitates with silver nitrate.

    The precipitates will be formed with the sulphates, but if you mark which tube you got the sample from, you can crossreference the silver nitrate test results with the sodium hydroxide and barium chloride test results and see which results are false positives- e.g if you know the solution bubbled with NaOH but also formed a white precipitate with silver nitrate, it must be ammonium sulphate, despite the halide positive. Or, if you got a white precipitate with both barium chloride and silver nitrate, it must be a sulphate as all sulphates produce a white precipitate with silver nitrate, but not all halides produce a white precipitate with barium chloride.
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    (Original post by Infraspecies)
    You are right, in so far as what YOU say you should do won't help.

    You are extending the procedure given, so you need to do some things to the test tubes given as well.

    1. You know which two are sulfates, and you know which two are halides. Let's start by distinguishing which sulfate is which.

    What are the differences in reactivity between the NH4+ and the Na+ ions based on the reagents you're given?
    You need to remember your analytical tests. Identification of the ammonium cation is done by adding dilute alkali, and testing the vapour above the solution with moist litmus paper (or indicator). Adding alkali affects the reaction; OH- + NH4+ -----> H2O + NH3. Ammonia vapour is alkaline, and will turn the universal indicator blue. Only one of the sulfates will respond to this test, obviously, so you now know which is which.

    2. The other two

    You have solutions of barium chloride with sodium chloride and potassium bromide.
    Silver nitrate reacts with halides, this you know and forms precipitates as you say. So adding silver nitrate to these two will form precipitates of the chloride from the barium, and either a chloride or bromide from the two tubes. You will not be able to tell based on the colours really, the cream will be undetectable, so another test is needed. Silver chloride dissolves in dilute ammonia solution. Silver bromide does not. Add dilute ammonia to these test tubes; the AgCl precipitate will dissolve, the AgBr precipitate will not. Hence, the KBr solution is the one whose precipitate with AgNO3 will not dissolve in dilute ammonia solution.

    Happy?
    I sort of understand what you mean. You are saying that the first step should be identifying which are the sulfate ions. But only one out of the 4 will turn the indicator paper blue. So You are left with 3 other unknowns. At this point when you add silver nitrate, surely this will form 2 white precipitates and 1 cream precipitate. Am I on the right line? Thanks
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    (Original post by coconut64)
    I sort of understand what you mean. You are saying that the first step should be identifying which are the sulfate ions. But only one out of the 4 will turn the indicator paper blue. So You are left with 3 other unknowns. At this point when you add silver nitrate, surely this will form 2 white precipitates and 1 cream precipitate. Am I on the right line? Thanks
    You only need to do the sodium hydroxide test on the two ones that form a precipitate with barium chloride on its own, because you know the two other ones aren't sulfates.

    And therefore you know which sulfate is which.

    Which means you can leave it aside and just do the silver nitrate test on the other two test tubes.
 
 
 
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