I don't get his about water dissociation?
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I don't get his about water dissociation?

[From: ] [author: ] [Date: 14-03-06] [Hit: ]
It is a mistake to think on the basis that water is neutral when the pH is 7.00 - In fact you have seen that this is not true .What is correct to say is that : Water is neutral when [H+] = [OH-] . This is always true of water .It so happens that at 25.0°C ,......
So water can dissociate according to the reaction:
H2O ⇌ H+ + OH-

The water is then heated so that is temperature rose by 5°C. How is it possible that the water remains neutral when its pH is decreasing?? I don't get this - it makes no sense to me!

Please explain this. Any help is appreciated. Thanks!

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Changing temperature changes the equilibrium constant, i.e. you get a different proportion of protons and hydroxide anions compared to water. However, pH is only a function of [H+], which will obviously change if you do shift the equilibrium.

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It is a mistake to think on the basis that "water is neutral when the pH is 7.00" - In fact you have seen that this is not true .
What is correct to say is that : Water is neutral when [H+] = [OH-] . This is always true of water .
It so happens that at 25.0°C , the [H+] = [OH-] = 1*10^-7M
With this H+ concentration the pH = 7.00 and the pOH = 7.00. Under these conditions , it can be said that neutral water has pH = 7.00 at 25°C ( and only at 25°C)
The dissociation of water
H2O ↔ H+ + OH- is endothermic.
If you add heat energy to the water , the temperature increases and so does the dissociation of the water - That means that the [H+] and [OH-] is now greater than 1.00*10^-7.
At 50°C the Kw for water has increased to 5.476 x 10-14 This makes [H+] and [OH-] = 2.34*10^-7M
The pH is then calculated
pH = -log [H+]
pH = -log ( 2.34*10^-7)
pH = 6.63 and likewise the pOH = 6.63 at 50°C
Because the [H+] = [OH-] the water remains neutral , but the neutreal pH of water at 50°C is 6.63. At 50°C. and acidic solution will have pH < 6.63 and a basic solution will have pH > 6.63.
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