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Balancing acid equation-net ionic equation

  1. Mar 22, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    I was asked to take NaCl then add HNO3, then add AgNO3, and to write a net ionic equation... then add NH3 to the same mixture and write another net ionic equation. I have been trying between work and everything else to try to do this one. it wasn't well understood in class and now I haven't found any resources to figure it out. I need whatever tips possible to guide me in the correct direction. At least for now I don't know what to do. Someone showed me an answer of Ag(aq)+ + 2H+ + Cl- + NO3----->> AgCl(s) + NO2(g) + H2O. I have tried 3 or four times, but I just have ZERO IDEA how that answer arrives... I know Ag+ binds with Cl- and as far as I know, because of the nitric acid H2O and a salt will be in the product. I am Having a hard time seeing how the NO2 gets in the picture. Thank You in advance for any help.
     
  2. jcsd
  3. Mar 22, 2016 #2

    Borek

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    And that's all you need for the first step.

    There is no gaseous product.

    Ammonia is known to complex metals.
     
  4. Mar 22, 2016 #3
    OK. Thank You. How does the Nitric Acid on the reagent side fit into the 'mix' (no pun intended). I really am trying here. Where am I missing the formation of H2O and NO2? In the attached picture, on the second line 20160322_185929 (1).jpg , to the right, is where I am getting stuck. (Is it correct to say that 'because of the acid (Nitric in this case), a salt and H2O are OBLIGATORY products'?)
     
    Last edited: Mar 22, 2016
  5. Mar 23, 2016 #4

    Borek

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    Net ionic reaction means - whatever was not changed during reaction is removed from the equation. These removed molecules/ions are called "spectators".

    Did the Na+ change? Did the NO3- change during the reaction?
     
  6. Mar 23, 2016 #5
    Thank you again. Na+ for sure doesn't change. It is removed. How do the 2 NO3's become 1 NO2-? (Also, how do you input the elevated + and - signs in the text? [this question isn't as important to me as the first])
     
  7. Mar 23, 2016 #6

    Borek

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    It doesn't, there is no reduction taking place. Whoever told you about NO2 being a product was wrong, that's what I told you in my very first post in the thread.

    There are buttons for formatting the post, above the edit window (at least if you are using a normal browser).
     
  8. Mar 23, 2016 #7
    OK . I am following you. Right now, in the picture, it looks like almost all of them, the Na+, both NO3's, and the H+ as well, do not change in the product side. I just do not think that is correct.
     

    Attached Files:

  9. Mar 23, 2016 #8

    Borek

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    That's perfectly correct, most of the ions present are spectators. Can you write the net ionic equation for the first reaction now?
     
  10. Mar 23, 2016 #9
    Na+ + Cl¯ + Ag+ + NO3¯ - - - >> AgCl?
     
  11. Mar 23, 2016 #10

    Borek

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    As written, it is not balanced, as Na+ and NO3- are present only on the left side of the equation.

    However, do they change during the reaction, or are they spectators? And if they are spectators, do you need to include them at all?
     
  12. Mar 25, 2016 #11
    Ag+(aq) + Cl- (aq) ----->>>> AgCl(s) ?? (Since, as you mentioned, NO2 is not part of the product, this is all i can think of right now). I very much appreciate your patience with me. I see you are in Poland and you seem to have a very good understanding of English too.
     
  13. Mar 25, 2016 #12

    Borek

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    Yes, that's perfectly correct.

    The second one will be a bit more tough, as actually there are two reactions taking place. Each needs a separate equation.
     
  14. Mar 26, 2016 #13
    So it will become AgCl(s) + NH3(aq)+ - - - - > ? :wideeyed:
     
  15. Mar 27, 2016 #14

    Borek

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    No such thing as NH3+(aq).

    One reaction will be more or less what you wrote - dissolving precipitate in ammonia.

    The other - sample was initially acidified with nitric acid, wasn't it? What do you know about ammonia and acids?
     
  16. Apr 3, 2016 #15
    Thank You again for helping me. I have since gotten that answer. I am learning enthalpy now, and it is going well. I just wanted to convey good wishes to you for your time.
     
  17. May 3, 2016 #16
    I scored a %98! Thank you again for your time and help! I have new questions but first I wanted to share the good news :smile:
     
  18. May 3, 2016 #17
    Here is the test. test.jpg
     
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