Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.
-
Upload
tatum-melling -
Category
Documents
-
view
262 -
download
1
Transcript of Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.
![Page 1: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/1.jpg)
Unit: Acids, Bases, and SolutionsCalculations with Acids and Bases
Day 2 - Notes
![Page 2: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/2.jpg)
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-]
![Page 3: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/3.jpg)
pH ScalepH scale: the measure of acidity of a solution
pH=-log[H+]
acidic
neutral
0 7 14
basic
[H+] = concentration in Molarity
![Page 4: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/4.jpg)
Before we try an example,
you will need to locate the
“log” button on your calculator.
![Page 5: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/5.jpg)
Example:What is the pH of a solution that has an [H+]=1.5x10-4M?
pH=-log[1.5x10-4]pH=3.8
![Page 6: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/6.jpg)
Example:What is the [H+] in a solution with pH=9.42?
9.42=-log[H+]-9.42=log[H+]10-9.42=[H+][H+]=3.80x10-10M
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.
![Page 7: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/7.jpg)
pH ScalepOH scale: the measure of alkalinity (basic-ness) of a solution
pOH=-log[OH-]
acidic
neutral
0 7 14
basic
![Page 8: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/8.jpg)
Example:What is the pOH of a solution that has an [OH-]=3.27x10-
9M?
pOH=-log[3.27x10-9]pOH=8.49
![Page 9: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/9.jpg)
Since the pH and pOH scales are opposite each other:
Example: What is the pH in a solution with a pOH=8.6?
pH + 8.6 = 14pH=5.4
pH + pOH = 14
![Page 10: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/10.jpg)
Summary…
pH=-log[H+]pOH=-
log[OH-]pH + pOH = 14
![Page 11: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/11.jpg)
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
![Page 12: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/12.jpg)
Example:What is the [H+] in a solution with [OH-] = 6.73x10-5M?
[H+][6.73x10-5]=1.0x10-14
[H+] = 1.49x10-10M
![Page 13: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/13.jpg)
Sometimes multiple formulas must be used to carry out these
calculations:
![Page 14: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/14.jpg)
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
![Page 15: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/15.jpg)
Example:What is the [OH-] in a solution with pH=8.1?
(Note: there are multiple ways to do this problem!)
pH + pOH = 148.1 + pOH = 14pOH = 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
![Page 16: Unit: Acids, Bases, and Solutions Calculations with Acids and Bases Day 2 - Notes.](https://reader030.fdocuments.us/reader030/viewer/2022012818/56649c7b5503460f9492f90a/html5/thumbnails/16.jpg)
Questions?Begin WS4