Diluting for pH decrease

[dekoyl]

How much water must be added to 1.0L of a solution of a strong base of pH 13.0 in order to decrease the pH to 12.0?

The book gives 9.0L and its reasoning is "to decrease pH by one, a ten fold dilution of the base is needed."

Could someone explain how to answer was obtained and why a ten fold dilution is needed? :S It would greatly help my base knowledge.


Thanks!

[IntoTheNewWorld]

As the concentration of the base is -log base 10 of [H+], reducing the concentration of OH- by 10 fold (through the dilution of the base), which increases the concentration of the H+ (as one goes down, the other goes up as PH is out of 14) by a factor of 10 consequently decreases the pH by 1, as pH is a logarithmic scale.

concentration of H+ ions = 10^-13 = pH of 13       concentration of OH- ions = 10^-1=pOH of 1
                                           10^-12 = pH of 12       concentration of OH- ions = 10^-2=pOH of 2

ten fold decrease in concentration of OH- ions, and ten fold increase in concentration of H+ ions, resulting from the dilution of the base by 10 fold.

I don't think that made sense. Anyway...

[dekoyl]

^No, it did make sense =] Thanks RandmAzn