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Alkali metal oxide reaction with water.

  1. Oct 14, 2013 #1
    I am confused as to why the following happens:
    [tex]K_2O + H_2O\rightarrow 2KOH^-[/tex]
    Does the potassium hydroxide no longer react with the water? Why exactly is this a base? How does the potassium end up bonding with hydroxide?
     
    Last edited: Oct 14, 2013
  2. jcsd
  3. Oct 14, 2013 #2
    Is this what happens?
    [tex]K_2O + H_2O \rightarrow 2K^+ + 2OH^-\rightarrow 2KOH[/tex]
    I think I have a better understanding now. The potassium oxide is the base and produces hydroxide ions in water!
     
    Last edited: Oct 14, 2013
  4. Oct 14, 2013 #3

    Zondrina

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    The potassium oxide will be dissolved with the water in a solution. A solution is NOT a reaction because dissolving is a physical change. The polarity of the water molecules will cause the ions to dissociate and form an aqueous solution:

    ##K_2O_{(s)} → 2K^{+1}_{(aq)} + O^{-2}_{(aq)}##
     
  5. Oct 14, 2013 #4
    I see and then I assume the oxygen will react with the water to create hydroxide ions, which will then produce potassium hydroxide.
     
  6. Oct 14, 2013 #5

    Zondrina

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    ##K_2O_{(s)}## is neither an acid nor a base. It's a basic anhydride (which basically means its a metal oxide).

    When metal oxides react with water they produce bases like so:
    ##K_2O_{(S)} + H_2O_{(L)} → 2KOH_{(S)}##
     
  7. Oct 14, 2013 #6

    Zondrina

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    Sorry for the double, I just had something to add. As for why it's a base, it ionizes almost completely in water. If you dissolve ##KOH_{(S)}## in more water, the ions will dissociate to release an ##OH_{(aq)}## ion:

    ##KOH_{(S)} + H_2O_{(L)} → KOH^{+1}_{2_{(aq)}} + OH^{-1}_{(aq)}##
     
  8. Oct 15, 2013 #7

    Borek

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    O2- in the solution? No way, it reacts with water too fast to exist in this form.

    Alkali metal oxides are deliquescent - they are strongly hygroscopic and will react even with traces of water in the air, producing bases. They absorb water producing bases even as solids, and they are converted to bases long before they absorb enough water to start to dissolve.

    If anything, process IMHO goes like

    K2O(s) + H2O(l) → KOH(s) → KOH(aq)

    (not balanced, but it is not intended to be the reaction equation, rather a scheme - and it assumes excess liquid water is present all the time).
     
  9. Oct 15, 2013 #8

    Zondrina

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    I figured the oxygen atom would be attaching itself very quickly to a water molecule so I wasn't sure whether to write it or not. I see now that a base will be formed long before the ##K_2O## even dissolves and it will be the base that mixes with water, not the ##K_2O##. So it would be fair to say ##K_2O## does not dissolve in water to form a solution.

    I'm curious now myself though. I know that ##KOH## is going to ionize almost completely when dissolved in water because it's such a strong base. Though I got confused when you wrote: KOH(s) → KOH(aq).

    EDIT: Never mind, further research into deliquescence yielded my answer for me. So ##KOH## is a deliquescent salt, which means it has a very high affinity for moisture. So the base does not dissolve, but it rather soaks up the water and forms a solution without dissociating. I believe the reaction you wrote is fine then:

    ##K_2O_{(S)} + H_2O_{(L)} → 2KOH_{(S)} → 2KOH_{(aq)}##
     
    Last edited: Oct 15, 2013
  10. Oct 15, 2013 #9
    Now I'm lost, how does the potassium hydroxide produce more potassium hydroxide when in water ??

    EDIT: Oh ok, I the equation makes more sense now. But how exactly does the potassium hydroxide soak up water? I suppose I should do some reading about deliquescence.

    EDIT: I see ! The potassium oxide reacts with the water so it "absorbs" it. It then becomes a solution if there is excess water.
     
    Last edited: Oct 15, 2013
  11. Aug 8, 2016 #10
    I read that LiOH is the only alkali metal hydroxides that is not deliquescent. I was wondering why is this is so?
     
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