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Formation of precipitate

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Strontium Nitrate + Ammonium Hydroxide =

    2. Relevant equations
    Finding net ionic equation

    3. The attempt at a solution
    Balanced Equation:
    Sr(NO3)2(aq) + NH4OH (aq) = Sr(OH)2(aq) + NH4NO3 there is no precipitate correct? I will not need to find the net-ionic equation.
     
  2. jcsd
  3. Jan 8, 2016 #2

    SteamKing

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    Not all hydroxides are water-soluble:

    https://en.wikipedia.org/wiki/Strontium_hydroxide
     
  4. Jan 8, 2016 #3
  5. Jan 8, 2016 #4

    SteamKing

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    Well, it all depends on how much Sr hydroxide you made and how much water there is.

    The solubility figures for Sr hydroxide are given in the Wiki article.

    You can contrast those solubility figures with comparable ones for ammonium nitrate:

    https://en.wikipedia.org/wiki/Ammonium_nitrate
     
  6. Jan 9, 2016 #5

    Borek

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    It is rather easy to estimate.

    pH of a 1 M ammonia solution is around 11.2, so [OH-] = 4.2×10-3

    Different sources list different values for solubility product of Sr(OH)2, but apparently it is around 10-4.

    That means using 1 M ammonia you should not expect precipitate unless concentration of the Sr2+ is around 6 M.

    Using more concentrated ammonia (3 M) will shift the required Sr2+ concentration down to 1.9 M, still quite high.

    No wonder you couldn't see any precipitate. For that you would need a strong base like NaOH or KOH.
     
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