ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: dekoyl on February 22, 2009, 05:13:47 pm
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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!
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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...
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^No, it did make sense =] Thanks RandmAzn