1. Calculations with Acids and Bases
Day 2 - Notes
Unit: Acids and Bases
Calculations with Acids and Bases

2. After today you will be able to…
Explain the correlation to strength of acids and bases to pH and pOH scale
Calculate pH, pOH, [H+], and [OH-]

3. [H+] = concentration in Molarity
pH Scale
pH scale: the measure of acidity of a solution
neutral
acidic
basic
Strong acid 7
Weak Acid/base 14
Strong base
pH=-log[H+]
[H+] = concentration in Molarity

4. Before we try an example, you will need to locate the “log” button on your calculator.

5. Example:
What is the pH of a solution that has an [H+]=1.5x10-4M? pH=-log[1.5x10-4] pH=3.8

6. Example:
To do this calculation you will need to use the inverse log. Locate the “10x” button. Usually it is the second function of the log button.
What is the [H+] in a solution with pH=9.42? 9.42=-log[H+] -9.42=log[H+] =[H+] [H+]=3.80x10-10M

7. pOH=-log[OH-] pH Scale
pOH scale: the measure of alkalinity (basic-ness) of a solution
neutral
basic
acidic 7 14
pOH=-log[OH-]

8. Example:
What is the pOH of a solution that has an [OH-]=3.27x10-9M? pOH=-log[3.27x10-9] pOH=8.49

9. Since the pH and pOH scales are opposite each other:
pH + pOH = 14
Example: What is the pH in a solution with a pOH=8.6? pH = 14 pH=5.4

10. Summary…
pH=-log[H+]
pOH=-log[OH-]
pH + pOH = 14

11. Ion-Product Constant for Water
Water will self-ionize to a certain extent into its individual ions. Because of this, the following relationship can be used:
[H+][OH-]=1.0x10-14M Kw

12. Example:
What is the [H+] in a solution with [OH-] = 6.73x10-5M? [H+][6.73x10-5]=1.0x10-14 [H+] = 1.49x10-10M

13. Sometimes multiple formulas must be used to carry out these calculations:

14. (Note: there are multiple ways to do this problem!)
Example:
What is the pOH in a solution with an [H+] = 2.17x10-5M?
(Note: there are multiple ways to do this problem!)
[H+][OH-]=1.0x10-14
[2.17x10-5][OH-]=1.0x10-14
[OH-]=4.61x10-10M
pOH=-log[OH-]
pOH=-log[4.61x10-10]
pOH=9.34
pH=-log[H+]
pOH=-log[OH-]
pH + pOH = 14
[H+][OH-]=1.0x10-14
pH=-log[H+]
pOH=-log[OH-]
pH + pOH = 14
[H+][OH-]=1.0x10-14

15. (Note: there are multiple ways to do this problem!)
Example:
What is the [OH-] in a solution with pH=8.1?
(Note: there are multiple ways to do this problem!)
pH + pOH = 14
8.1 + pOH = 14
pOH = 5.9
pOH=-log[OH-]
5.9=-log[OH-]
[OH-]=1.3x10-6M
pH=-log[H+]
pOH=-log[OH-]
pH + pOH = 14
[H+][OH-]=1.0x10-14
pH=-log[H+]
pOH=-log[OH-]
pH + pOH = 14
[H+][OH-]=1.0x10-14

16. Questions? Begin WS4