What is the OH- concentration of a solution whose pH is 12.40?

Trinity,

from the pH you can determine the H3O^+ concentration by using ph=-log[H3O^+]. Then you can use Kw=KaxKb to solve for OH- concentration

10^-12.40 = 10^log[h3o^+] =
4.0x10^-13 M = [h3o^+]

Now use 4.0x10^-13[OH^-] = 1.00x10^-14 =

[OH^-]= 2.5X10^-2 M

thanks@ Michael

pH + pOH = pKw = 14.

You know pH, solve for pOH.
Then pOH = -log(OH^-)
Substitute and solve for OH. You should get a little less than 2 but that's just an estimate.

Thanks @ DrBob222

I hit the wrong key. The answer should be a little less than 3.

If [H3O+] = 1.7 x 10–3 M, what is the pH of the solution?

To find the OH- concentration of a solution, we need to use the relationship between the pH and pOH. The pOH is the negative logarithm of the OH- concentration.

The formula to convert pH to pOH is:
pOH = 14 - pH

First, we can find the pOH of the solution by using the given pH value of 12.40:
pOH = 14 - 12.40
pOH = 1.60

Now that we have the pOH, we can find the OH- concentration by taking the antilogarithm of the pOH value.

The antilogarithm is the inverse operation of the logarithm, and in this case, we have a base 10 logarithm. To find the antilogarithm, we raise 10 to the power of the pOH:

OH- concentration = 10^(-pOH)
OH- concentration = 10^(-1.60)
OH- concentration = 0.0251 M

Therefore, the OH- concentration of the solution with a pH of 12.40 is approximately 0.0251 M.