calculate the [OH1-] in mol/L, the pH, and the pOH

Question

If the [H1+] of a solution at 25oC is 1.7 x 10-4 mol/L, calculate the [OH1-] in mol/L, the 

pH, and the pOH.

Answer 1

There are a few ways to to do this problem. If answering in the order asked:

1. [OH^-] = K_w / [H⁺] = 1.0 x 10^-14/ 1.7 x 10^-4 = 5.9 x10^-11 mol/L
2. pH = -log₁₀ [H⁺] = - log₁₀ [1.7 x 10^-4] = 3.8 (rounded from 3.77, see note below)
3. pOH = 14.0 - pH = 14.0 - 3.8 = 10.2 (this requires you to use the answer from question 2)

OR

pOH = -log₁₀ [OH^-] = - log₁₀ [5.9 x10^-11] = 10.2 (this requires you to use the answer from question 1)

But if you answer the questions in a different order than asked:

Do 2 first as shown above. Do 3 next as shown above with the first method illustrated.

Do part 1 as follows:

pOH = -log₁₀ [OH^-] so rearranging, [OH^-] = 10^-pOH = 10^(-10.2) = 6.3 x1 0^-11 mol/L

Note: The last answer of 6.3 x 10^-11 mol/L differs by the answer 5.9 x 10^-11 mol/L shown above because of rounding. If the pH was used as 3.77 instead of 3.8 and the pOH found by 14 - 3.77 = 10.23 instead of 10.2, then:

[OH^-] = 10^-pOH = 10^(-10.23) = 5.9 x 10^-11 mol/L

I show this last solution for finding [OH^-] as it illustrates how you can answer all three of the questions just making use of the three pH and pOH equations:

pH = -log10 [H+] and pOH = -log10 [OH-] and 14 = pH + pOH

and NOT making any use of:

K_w = 1.0 x 10^-14 = [H+] {OH-]

Answer 2

Hello Mariah,

I posted an answer to your question in the form of a video for two reasons:

1) I wanted to make use of an amazing graphic organizer that is very helpful for these calculations

2) This is the format that online sessions would be in, so I wanted you to see what the Wyzant online platform looks like.

Thanks,

Meagan

Answer 3

Equations to know: Try using these to answer the problem

[H+][OH-] = 1.0x10-14

-log[H+] = pH

-log[OH-] = pOH

pH + pOH = 14.00

Answer attached here: https://drive.google.com/file/d/1fwjPytvdCdcQiyegxibjD7nZerLxJ4jy/view?usp=sharing