A solution has [H-]=4.2 x 10^-3 M. The OH in this solution is?

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A solution has [H-]=4.2 x 10^-3 M. The OH in this solution is?

how do i find the OH?

2007-10-19 07:50:51 · 4 answers · asked by Dirty S 1 in Science & Mathematics ➔ Chemistry

4 answers

[H+][OH-] = 10^-14
So [OH-] = 10^-14/[H+] = 10^-14/4.2*10^-3 = 2.4*10^-12

2007-10-19 07:57:17 · answer #1 · answered by Alexey V 5 · 1⤊ 0⤋

[OH-] = 10-14/4.2 x 10-3

2007-10-19 14:56:35 · answer #2 · answered by Gervald F 7 · 0⤊ 3⤋

[H+] [OH-] = 1.00 x 10^-14
[OH-] = 1.00 x 10^-14 / 4.2 x 10^-3 = 2.38 x 10^-12

2007-10-19 14:57:20 · answer #3 · answered by Dr.A 7 · 0⤊ 2⤋

In water, the product of the concentration of the hydrogen ions and hydroxide ions is always 1E-14. Therefore, if the hydrogen ion concentration is 4.2E-3 and X is the concentration of hydroxide ions:

(X)(4.2E-3) = 1E-14
X = 1E-14/4.2E-3
X = 2.38E-12

2007-10-19 14:57:57 · answer #4 · answered by Roger S 7 · 0⤊ 2⤋