They don't...
If you are seeing this experimentally, then you are doing something wrong (e.g. with the calibration of the pH-meter) or simply the carbon dioxide dissolved in your solution from the air is much more than the base you have.
If you are calculating the pH of a strong base, e.g. NaOH, then when it is concentrated enough you don't consider the self-dissocaiton of water. However for very dilute solutions (<10^-6M) you cannot overlook the self-dissociation of water and you have to solve the problem considering both the dilute base and water.
E.g. for 10^-8 M NaOH you can solve it as an equilibrium that is perturbed:
The initial equilibrium is that of pure water where [OH-]= [H+]=10^-7
When you add NaOH you will have x mole/lt OH- reacting with H+ to give water (Le Chatelier's principle forces the H2O <=> H+ + OH- to the left)
so you have Kw = [H+][OH-]= (10^-7-x) *(10^-7 +10^-8 -x)=10^-14
You solve for x and then pH=-log(10^-7-x)
You can also solve it if you say that in the presence of NaOH, less water will dissociate. So if x mole/lt dissociate you will have
Kw = [H+][OH-]= x *(10^-8+x)=10^-14
Solve and then pH=-logx
By the way, all of these are for room temperature. At higher temperatures the pH scale changes. E.g. at 60 C, I think, you have neutral solutions at pH=6.5 so then pH=6.7 would be alkaline...
2006-10-20 23:42:10
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answer #1
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answered by bellerophon 6
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If a solution has a pH less than 7 then, by definition, is not basic.
If you know that a solution is basic, but you are measuring a pH less than 7 then you may be seeing an expression of an experimental or calibration error.
2006-10-19 20:08:13
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answer #2
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answered by Dr. J. 6
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they dont! if a solution has pH below 7 then by definition its not basic anymore!
2006-10-19 19:48:56
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answer #3
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answered by impeachrob 3
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because there is an increase in concentration of H+ ions. pH measures the concentration of pH by the formula pH = -log*(concentration of H+).
2006-10-19 19:39:24
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answer #4
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answered by Anonymous
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