Determine the pH and pOH, and H+ of a solution where the OH- = 2.8 x 10(-4)M.
pH= 3.6x10(-19)??
pOH= 18.4??
Are these the correct answers?
2006-11-24 06:32:44 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Nope, :s sorry
Its the same as before. The K(w) of water is 1.0 *10^-14
K(w) = [H+][OH-]
1.0*10^-14=[H+][2.8*10^-4]
[H+]= 3.57*10^-11
pH=-log[H+]
the pH is 10.4
and pOH =-log[OH-] u have that in the ques
Happy trails
2006-11-24 06:45:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Take the negative logarithm of 2.8 x 10-4, and you should get pOH as 3 point something. Then subtract from 14 to get pH.
2006-11-24 14:37:05 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋
pOH = -log OH concentration = 3.6
pH + pOH = 14 so
pH = 14 - 3.6 = 10.4
The OH(-) x H+ concentrations = 1.00 x 10^(-14) the H+ would be:
1.00 x 10^(-14) divided by 2.8 x 10^(-4) = 3.6 x 10^(-11)
2006-11-24 14:40:31 · answer #3 · answered by The Old Professor 5 · 0⤊ 0⤋
use the pH paper and match it to the color on the box
2006-11-24 15:34:57 · answer #4 · answered by Lauren N 2 · 0⤊ 0⤋