I tried finding pOH as -log(2x10^-8)That was about 7.7
then I tried finding pH=14-7.7.The answer was 6.3...but I don't think this is right as KOH is a base..how can I get the correct answer?
2007-02-12 16:47:49 · 3 answers · asked by mammaluv 2 in Science & Mathematics ➔ Chemistry
The concentration is so low that you have to take into account the self-dissociation of water.
The amount of OH- coming from the strong base is 2*10^-8 M
Let's see how the self-dissociation of water is affected
.. .. .. .. .. .. H2O <=> H+ + OH-
Initial .. .. .. . . . . .. . . .. .. .. .. 2*10^-8
dissociate.. x
produce .. ... .. .. .. .. . x .. .. .. x
At equil. .. .. .. .. .. .. .. x .. .. x+2*10^-8
Kw=[H+][OH-] = x(x+2*10^-8) =10^-14 =>
x^2+2*10^-8-10^-14 = 0
The only acceptable solution is
x=9.05*10^-8
pH=-log[H+] =-logx =-log(9.05*10^-8) =7.04
2007-02-13 10:01:40 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋
Your formula works in general for strong bases, but the problem is that 10^-8 is an extremely low concentration and that water normally has about 10^-7 M H+ as well as 10^-7 M OH- due to this reaction in equilibrium H+ + OH- <-> H20. Anyways for this problem you need to know that the concentration of H+ and OH- always multiply to 10^-14. Well, since KOH ionizes completely, then OH- will have a concentration of 2*10^-8 greater than the concentration of H+. That means there will be about 1.1 * 10^-7 M OH- and 9* 10^-8 M H+ so you get a pH of 7.05
2007-02-13 01:01:45 · answer #2 · answered by xit_vono 2 · 0⤊ 0⤋
For very dilute solutions, the autoionization of water has to be taken into consideration.
2007-02-13 01:05:03 · answer #3 · answered by Anonymous · 0⤊ 0⤋