1. pOH and [OH-] Calculations

2. Determining the pOH of Strong Bases
Determining the pH is a useful calculation, but is limited to acids because acids give of H+ ions.
If we want to consider bases, we must do something different. When bases dissolve, they give off [OH-] ions instead of [H+] ions.
To solve for the concentration of [OH-] ions we use:
pOH = - log [OH-]
It’s the same exact thing as pH, except this time it’s the hydroxide ion concentration!

3. pOH Scale
For any solution, we can examine the concentration of hydroxide ions and put that on a logarithmic scale.
It is the exact inverse of the pH scale where bases are low numbers and acids are high numbers.

4. pH + pOH = ???
There is a relationship between the pH and pOH that is important to know, and water can help us. What do we know about the pH of water?
Water is right in the middle of the pH scale because it has the exact same [H+] and [OH-] concentrations. They are…
[H+] = 1.0 x 10-7
[OH-] = 1.0 x 10-7

5. Convert both of these to their respective pH and pOH by taking their negative log:
- pH = 7 & pOH = 7
It makes sense that water is neutral on both the pH and pOH scale because it has equal concentrations of H+ and OH- ions.
The sum of pH and pOH will always represent the total H+ and OH- ions in solution. So what is the sum of pH and pOH going to be?
14!!! – This will always be true of ANY substance and can be helpful in calculations.

6. pH Scale and Concentrations

7. Problem #1
Determine the pOH of a 3.45 x 10-2 M sodium hydroxide solution? How would you describe the strength of this basic solution?
pOH = -log[OH-]
pOH = -log[3.45 x 10-2]
pOH = – Strong BASE on the pOH scale.

8. Using pOH to get [H+]
What is the hydrogen ion concentration of a M NaOH solution?
*Note we cannot simply take the –log because it’s not an acid. We must go through the pOH because NaOH gives off hydroxide ions [OH-].
pOH = -log[OH-]  pOH = -log[0.001] = 3.0
But we want the [H+] so pH + pOH = ????

9. Using pOH to get [H+] pH + pOH = 14.0
pH = 14.0 – pOH  pH = 14.0 – 3.0 = 11.0
Note: This makes sense because 11 is a base on the pH scale and NaOH is a base b/c of the hydroxide ion.
Almost there: pH = -log[H+]
11.0 = -log[H+]  = [H+]
[H+] = 1 X 10-11

10. Problem #2
The pH of a sample of human blood was measured to be 7.41 at 25 °C. Calculate pOH, [H+], and [OH-] for the sample.
pH = 7.41
7.41 = -log[H+]  = [H+]  3.9 M x 10-8 [H+]
pOH = 14.0 – 7.41 = 6.59
6.59 = -log[OH-]  = [H+]  2.6 x 10-7 M [H+]

11. Problem #3
A chemist dilutes concentrated hydrochloric acid to 0.25 M. Calculate the [H3O+], pH, [OH-], and pOH of the solution at 25°C.