What is the pH of a solution of NaOH 10^-12 M?
2 answers
how did you find OH? we aren't allowed to use a calculator and the answer book is saying the pH is 7, not 12
I don't want that post to stick around on the web so I'm deleting it. But this response will stay so as to answer your question. I don't know why I answered that way because as I was posting I kept thinking things just weren't right. This ranks up there among the top goofs a chemist can make. The excuse, and excuses don't count, is that I used OH^- = 10^-12 and confused that with K. First question first.
How did I do the pOH?
If NaOH is 10^-12, then
NaOH ==> Na^+ + OH^-
10^-12..10^-12..10^-12
So OH^- from NaOH is 10^-12. All of that is good information. Here is where you go from there.
The water equilibrium is
.......HOH ==> H^+ + OH^-
I....liquid....0.....10^-12
C....liquid....x......x
E....liquid....x.....x+10^-12
(H^+)(OH^-) = Kw = 1E-14
(x)(x+10^-12) = 1E-14
To avoid a quadratic we assume x + 10^-12 = x and the equation reduces to
(x)(x) = 1E-14
x = (H^+) = (OH^-) = sqrt(1E-14) = 1E-7
and pH = -log(H^+) = -log(1E-7) = 7
I don't know how my twisted brain got 0.01 out of 10^-12 but the way I arrived at pOH = 2 was -log(0.01) = 2 and you don't need a calculator for that.
We should always check to make sure our assumption is correct; that is, 10^-12 really is negligible compared with x = (H^+) but 1E-7 + 1E-12 really is, for all practical purposes, 1E-7
How did I do the pOH?
If NaOH is 10^-12, then
NaOH ==> Na^+ + OH^-
10^-12..10^-12..10^-12
So OH^- from NaOH is 10^-12. All of that is good information. Here is where you go from there.
The water equilibrium is
.......HOH ==> H^+ + OH^-
I....liquid....0.....10^-12
C....liquid....x......x
E....liquid....x.....x+10^-12
(H^+)(OH^-) = Kw = 1E-14
(x)(x+10^-12) = 1E-14
To avoid a quadratic we assume x + 10^-12 = x and the equation reduces to
(x)(x) = 1E-14
x = (H^+) = (OH^-) = sqrt(1E-14) = 1E-7
and pH = -log(H^+) = -log(1E-7) = 7
I don't know how my twisted brain got 0.01 out of 10^-12 but the way I arrived at pOH = 2 was -log(0.01) = 2 and you don't need a calculator for that.
We should always check to make sure our assumption is correct; that is, 10^-12 really is negligible compared with x = (H^+) but 1E-7 + 1E-12 really is, for all practical purposes, 1E-7