What is [ OH - ] for a solution where pH = 6.80?

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2007-03-10 15:04:31 · 5 answers · asked by Mononkeneo 1 in Science & Mathematics ➔ Chemistry

pOH=14-6.8=7.2
-log[OH-]=7.2
10^-7.2=6.3095e-8

2007-03-10 15:13:36 · answer #1 · answered by trish 1 · 0⤊ 0⤋

The pOH would be 7.20, and since pOH is the negative logarithm of the [OH-] the [OH-] would be 6.3 x 10-8 M

2007-03-10 23:18:18 · answer #2 · answered by chemcook 4 · 0⤊ 0⤋

pOH = 14 - 6.8 = 7.2
[OH -] = 10^(-7.2) M
= 6.31 x 10^(-8) M

2007-03-10 23:17:22 · answer #3 · answered by Newbody 4 · 0⤊ 0⤋

pH = -log [H+]

pOH = -log [OH-]

pH + pOH = 14

pOH = 7.20

[OH-] = 6.3 e -8

2007-03-10 23:13:38 · answer #4 · answered by Hk 4 · 0⤊ 0⤋

That would mean it was an acid.

2007-03-10 23:08:04 · answer #5 · answered by ACTiNGisLiFE 3 · 0⤊ 0⤋