2007-03-19 04:40:44 · 6 answers · asked by Bugg 2 in Science & Mathematics ➔ Chemistry
pH = log (1/[H+]) pOH = log (1/[OH-]
at 25°c [H+] [OH-] = 10^-14 else 10^14 = 1/[H+]*1/ [OH-]
and due to log formula pOH+pH =14
since log 10^14 =14 formula of pH and pOH given above
so if pH =2.5 pOH = 14-2.5 =11.5
by definition
pH =log 1/[H+] = 2.5 so pH = -log[H+] ; log[H+] =-2.5
[H+] = 10^-2.5
same for OH pOH = 10^-11.5
2007-03-19 05:03:20 · answer #1 · answered by maussy 7 · 1⤊ 0⤋
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2016-12-19 08:52:27 · answer #2 · answered by Anonymous · 0⤊ 0⤋
in fact these calculations are valid at 25 degree celcius that pH+pOH=14 , the other temperature the sumation has another value
2007-03-19 05:45:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋
pH = 2.5
pH + pOH ==> pOH = 14 - 2.5 = 11.5
pH = -log(H+) = 2.5
[H+] = 10-2.5 = 3.16*10-3
[OH-] = 10-11.5 = 3.16*10-12
2007-03-19 04:56:09 · answer #4 · answered by Dr Dave P 7 · 0⤊ 0⤋
pH = 2.5
pH+pOH=14
pOH=14-2.5=11.5
pH = -log(H+)
(H+) = 10^-2.5=0.00316M
(H+)(OH-) = 1.00 10^-14
(OH-) = 1.00 10^-14 / 0.00316 =3.16 10^-12
2007-03-19 04:52:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋
pH =-log [H+]
pH + pOH = 14
pOH = -log [OH]
2007-03-19 05:19:51 · answer #6 · answered by shiara_blade 6 · 0⤊ 0⤋