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    I understand that the Kw constant refers to the equilibrium:

    H20 + H20 --> Oh- + H30+

    is written as Kw = [OH-][H30+] and that at 25 degrees the constant is equal to 1 * 10^-14. I also know that for the equilibrium OH- and H30+ are formed in equal quantities so you can deduce the concentration of H30+ ions and therefore why at 25 degrees pure water has a Ph of 7.

    Firstly, am i right in saying that at 25 degrees a ph of 7 for all aqueous solutions would be neutral?

    And secondly this is how i see how an acid dissolved in water produces a lower ph but the same value for Kw could i just have some confirmation that this is actually what happens thanks.

    I thought that the acid will dissolve and react with water molecules to form H30+ so the concentration of this species will increase. In response the equilibrium will shift to the left to use up both OH- and H30+. What is left is obviously an acidic solution because there is more H30+ than Oh- but does the Ph decreases because the concentration of H30+ actually never reaches its initial level and if you went on to multiply the concentrations for H30+ and Oh- it would give a value of 1*10^-14.

    Thanks for your help
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    (Original post by 111davey1)
    Firstly, am i right in saying that at 25 degrees a ph of 7 for all aqueous solutions would be neutral?
    Yes.

    And secondly
    H2O <-> H+ + OH-

    If you add H+, the equilibrium shifts to the left, OH- ions are used up reacting with the added H+

    If you added 1 mol of H+ ions to 1 dm3 of water, [H+] = 1 mol dm-3. Since Kw is constant, to make Kw = [H+] [OH-], all you need to do is lower [OH-] to 1x10-14
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    (Original post by Pigster)
    Yes.



    H2O <-> H+ + OH-

    If you add H+, the equilibrium shifts to the left, OH- ions are used up reacting with the added H+

    If you added 1 mol of H+ ions to 1 dm3 of water, [H+] = 1 mol dm-3. Since Kw is constant, to make Kw = [H+] [OH-], all you need to do is lower [OH-] to 1x10-14
    Thanks,
    so when you dissolve an acid in water am i wrong to say that the acid will dissociate with the H+ reacting with water to form H30+. the concentration of this species increases so equilibrium shifts to the left using up both the added H30+ and existing Oh- ions in the solution. What you are left with is a lower than initial level of OH- and a higher level of H30+(but lower than immediately after the acid dissolves) in the solution hence an acidic solution and if you multiply both concentrations you will get 1times 10^-14.

    Or is it that the H30+ is in great excess which means that eventhough it reacts with the OH- ions its concentration stays pretty much constant Meaning that in practice we only really see the concentration of OH- lower.

    Thanks
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    H+ is basically a quick way of writing (the more correct) H3O+. They mean the same thing.

    If you add 1 mol of HCl to 1 dm3 of water (assuming 100% dissociation) you would form 1 mol of H+ (on top of the 1x10-7 mol of H+ present from the dissociation of water). There is 1x10-7 mol of OH- present. Assuming all of that could react, the amount of H+ left would be 1 mol. Even if we ignore the H+ from the water and only worry about the OH-, the amount of H+ would be 1-1x10-7 = 0.9999999 mol i.e. 1 mol.

    Your final sentence pretty much sums it up.
 
 
 
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