fredag 21 oktober 2011

The autoprotolysis of water explained.

Calculate the molarity of OH- in solutions by the following H3O+ concentrations!

a. 0,02 mol/L H3O+ . → 1 · 10-14 / 0,02 = 5 · 10-13 M/L OH-

b. 3,1 mM/L H3O+ → 1 · 10-14 / 0,0031M = 3,22 · 10-12 M/L OH-


The calculations above lean on the knowledge of the autoprotolysis of water.


Describe the autoprotolysis of water!

The autoprotolysis of water referes to the fact that the H2O molecule is amphoprotic – it can operate both as a Broenstedt base and as an acid. Proton transfer occurs in pure water as well as in other aqueous solutions. This means that there will always be an exchange of protons between water molecules with H3O+ and OH- molecules as the result. This reaction (just as any chemical reaction) leads to an equilibrium. The equilibrium constant (KW) is wiritten as follows :

H2O H3O+ +OH- KW = [H3O+]·[OH- ]/[H2O]

The activity of H2O is close to 1 and hence excluded from the equation. KW is thus written as :

KW = [H3O+]·[OH- ] and always equals : 1 · 10-14 M at 25 º C!


In pure water [H3O+] [OH- ] respectively is...?

1 · 10-7 , both of them.


Does KW increase or decrease with temperature?

It increases.


The most important thing to understand about KW is that it, just as all equilibrium constants, always remains the same! That is, if we increase [H3O+] , [OH- ] will decrease enough to preserve

KW = 1 · 10-14 M!!!


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