1. Unit 5: Acid-Base Calculations Lesson 3: pH and pOH

2. Logarithms “What is the logarithm of 1000?”
This is the same as asking “Ten to the power of what is 1000?”
Hence, log(1000) = 3
Try this: What is log(0.01)?
Practice: Do #47 on pg. 134.
10 min

3. Antilogs An antilog is the opposite of a log, so...
“What is the antilog of 4?” is the same as asking “What is ten to power of 4?”
Hence, antilog(4) =
Try this: What is antilog(-4)?
Practice: Do #48 on pg. 135.
10 min

4. Key Points on Logs
Since log and antilog are opposites, they cancel each other out, in the same way that divide and multiply do.
E.g. log(antilog(4)) = 4
log(A x B) = log(A) + log(B)
When the log of a number changes by 1, the number itself changes by a factor of 10. (Hmm, what value in chemistry does this...?)
5 min

5. Brain Break!
Fun Fact: Many types of flower petals contain indicator dyes that change colour depending on the pH. One gardener in Vancouver with a unique sense of humour planted a row of hydrangea bushes and made the soil acidic on the left set of bushes by adding aluminum sulphate. He made the soil on the right set of bushes basic by adding powdered limestone (calcium carbonate). When the bushes flowered, the plants in acidic soil had pink flowers and the plants in the basic soil had blue flowers. The next year, he reversed the acidity on the left and right sides, thereby creating different coloured flowers from those of the previous year. Puzzled neighbours thought he had dug up the bushes and rearranged them.

6. Calculating pH and pOH pH = -log[H3O+] pOH = -log[OH-] Try this:
If [H3O+] = 3.94x10-4 M, what is the pH?
If [OH-] = 9.51x10-12 M, what is the pOH?
5 min

7. Calculating [H3O+] and [OH-]
Example: If pH = 1.59, what is [H3O+]?
Your turn: If pOH = 11.68, what is [OH-]?
5 min

8. Don’t Forget Sig Figs!
In a pH or pOH, only the digits after the decimal place are significant!
E.g. In a solution with [H3O+] = 5.28x10-5 M, the pH is 4.277

9. pKw
At 25°C, pKw = -log(Kw) = -log(1.00x10-14) = pKw = -log(Kw) = -log([H3O+][OH-]) = -log[H3O+] + -log[OH-] = pH + pOH Hence: pH + pOH = 14

10. Putting It All Together
Kw = [H3O+][OH-]
[H3O+]
[OH-]
pOH = -log[OH-]
pH = -log[H3O+] pH
pOH
pH + pOH = 14

11. Homework:
Pg. 139 #49-50 (a, c, e, etc.) and #51