Chem 2 - Acid-Base Equilibria III: pH, pOH, and pKw

1. Acid-Base Equilibria (Pt. 3)
pH, pOH, and pKw
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Recall: Autoionization of Water
Water reacts with itself…
𝐇 𝟐 𝐎 𝐥 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)
base 1 conjugate
base 2
conj acid 1acid 2
𝐊 𝐖 = 𝐎𝐇−
[𝐇 𝟑 𝐎+
]

3. Autoionization of Water and Kw
In pure water at 25C, Kw = 1.0  1014
𝟐 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)
𝐊 𝐖 = 𝐎𝐇−
[𝐇 𝟑 𝐎+
]

4. Autoionization of Water and Kw
At 25C, Kw = 1.0  1014
𝐊 𝐖 = 𝐎𝐇−
[𝐇 𝟑 𝐎+
]
𝟏. 𝟎 × 𝟏𝟎−𝟏𝟒
= 𝐎𝐇−
[𝐇 𝟑 𝐎+
]
Therefore, [OH] = [H3O+] = 1.0  107

5. pH, pOH, and pKw
Recall: pH is related to the
concentration of hydronium H3O+.
pH =  log [H3O+]
The pH of pure water is
𝐩𝐇 = −𝐥𝐨𝐠 𝟏 × 𝟏𝟎−𝟕
= 𝟕

6. pH, pOH, and pKw
pOH is related to the concentration
of hydroxide OH
.
pOH =  log [OH
]
The pOH of pure water is also
𝐩𝐎𝐇 = −𝐥𝐨𝐠 𝟏 × 𝟏𝟎−𝟕
= 𝟕

7. pH, pOH, and pKw
Given pH and pOH, how do you think pKw
is calculated?
pKw =  log Kw
The pKw of pure water at 25C is
𝐩𝐊 𝐖 = −𝐥𝐨𝐠 𝟏 × 𝟏𝟎−𝟏𝟒
= 𝟏𝟒

8. A Few More Relationships Between
pH, pOH, and pKw
pH + pOH = pKw
pKw =  log [Kw] = 14 (at 25C)
pH + pOH = 14 (at 25C)

9. Inverse logs for pH, pOH, and pKw
pH =  log [H3O+] 10pH = [H3O+]
pOH =  log [OH
] 10pOH = [OH
]
pKw =  log Kw = 14 10pKw = Kw
Recall: Kw (and pKw) will have different
values at temperatures other than 25C

10. What you Should Be Able to Do
Find pOH, given pH and pKw.
Find pH, given pOH and pKw
Use these relationships to find
[H3O+] and [OH
] for various
solutions.

11. What you Should Be Able to Do
Take the inverse log of the pH to
calculate the hydronium ion
concentration (H3O+).
Use inverse logs to find hydroxide
ion concentration (OH) or the value
for Kw at other temperatures.

12. Example calculations will be posted
separately.
Next up,
Calculating the pH of Strong Acids
versus Weak Acid Equilibria
(Pt 4)