2007-10-17 16:51:48 · 4 answers · asked by Lostwithchem 1 in Science & Mathematics ➔ Chemistry
[OH-] [H+] = Kw =1.00 x 10^-14
[H+] = 1.00 x 10^-14 / [OH-] = 1.00 x 10^-14 / 1.10^-5 =
= 1.0 x10^-9
pH = - log [H+] = - log 1.0 x 10^-9 = 9.0
There is another way :
pOH = - log [OH-] = - log 1.0 x 10^-5 = 5.0
pH + pOH = 14
pH = 14 - pH = 14 - 5.0 = 9.0
2007-10-17 16:58:33 · answer #1 · answered by Dr.A 7 · 0⤊ 1⤋
I don't have a calculator.
pOH= log [OH-]
log [1.0Ã10-5 M] = pOH
pH20 = pOH +pH
pH = pH2O - pOH
pH = 14 - 5
9.... That is assuming my mental math is with log 1 x 10-5 =5.
just remember ph and pOH add to 14.
2007-10-17 23:58:55 · answer #2 · answered by ucenigma 3 · 0⤊ 0⤋
pOH then = 5.0
pH = 14-5=9.0
2007-10-17 23:59:01 · answer #3 · answered by vv 6 · 0⤊ 0⤋
9
[OH-]+[H+]=14
your [OH-] is 5.
2007-10-17 23:59:40 · answer #4 · answered by christigmc 5 · 0⤊ 0⤋