# Calculate the [OH–] if the hydrogen ion concentration is 4.12 x 10–9 M.?

[H+][OH-] = 1.00 x 10^-14
[OH-] = 1.00 x 10^-14 / [H+] = 2.43 x 10^-6 M

There is another way :
pH + pOH = 14
pH = - log 4.12 x10-9 = 8.38
pOH = 14 - pH = 5.61
[OH-] = 10^-5.61 = 2.43 x 10^-6 M
[OH-]=kw/[H+]

[OH]=1e-14/4.12e-9

[OH-]= 2.43e-6 M
Multiplying the [H+] concentration and the [OH-] concentration always yields 1.0X10^-14.
So...
[H+] [OH-] = 1.0x10^14
[4.12x10^-9] [OH-] = 1.0x10^-14
[OH-] = 2.43x10^-6
The previous answers are true..only if the solution is pure H20.

The answers post by the user, for information only, FunQA.com does not guarantee the right.

More Questions and Answers:
• Cu + hcl ->?
• Chem qustion dealing with pH?
• Which of the following is not an example of an intermolecular attraction?
• The density of zinc is 455 lbs/cu.ft. Find the mass of 9 cu. cm. of zinc in grams.?
• How many molecules are in a mole of ammonia NH3?
• Which is the most rearest elemental metal on earth?
• What 1/2 reaction would be reduced by Cr(NO3)3 being reduced by 3 electrons?
• Do spontaneous processes have K<1?
• Using tabulated values of S°, calculate the standard entropy per mole of atoms for Kr.?