if the ph is 2.2 and the poh is 11.8 what is the h+ and oh-
is it 6.3 x 10^-3 and 1.58x 10^-12 could you explain why
2007-05-15 15:58:47 · 4 answers · asked by Adam d 1 in Science & Mathematics ➔ Chemistry
pH is the negative log (base 10) of the hydrogen ion concentration. therefore
-log(6.3x10E-3)= - log.0063= -(-2.20 65) = 2.2
The pH and the pOH when added together always =,s 14 If you know the pH then you know the pOH
2007-05-15 16:11:12 · answer #1 · answered by fred52041 1 · 0⤊ 0⤋
Looks good to me. From the definition of pH,
H+ = 10^-2.2. This is equivalent to (10^-3)(10^0.8), which yields 6.3x10^-3.
As for the rest, from the water ionization constant,
pOH+pH= 14. Also, [H+][OH-]= 10x10^-15. This is often given as 1x10^-14, but as you can see in your calc, the other format is more useful.
2007-05-15 23:05:15 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
To find the H+ hit second, 10x, -2.2
OH is calculated by taking 1.0 x 10^-14 which is the kw for water and divide by H+. hope that helps.
2007-05-15 23:05:10 · answer #3 · answered by Heather I 2 · 0⤊ 0⤋
pH = -log [H+]
pOH = -log [OH-]
pH + pOH = 14
pH = 2.2
2.2 = -log [H+]
[H+] = 10^-2.2
[H+] = 6.3 * 10^-3
pOH = 11.8
11.8 = -log [OH-]
[OH-] = 10^-11.8
[OH-] = 1.58 * 10^-12
2007-05-15 23:04:08 · answer #4 · answered by coconutty beanz xD 4 · 0⤊ 0⤋