Unit: Acids, Bases, and SolutionsCalculations with Acids and Bases

Day 2 - Notes

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-]

pH ScalepH scale: the measure of acidity of a solution

pH=-log[H+]

acidic

neutral

0 7 14

basic

[H+] = concentration in Molarity

Before we try an example,

you will need to locate the

“log” button on your calculator.

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

pH=-log[1.5x10-4]pH=3.8

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.

pH ScalepOH scale: the measure of alkalinity (basic-ness) of a solution

pOH=-log[OH-]

acidic

neutral

0 7 14

basic

Example:What is the pOH of a solution that has an [OH-]=3.27x10-

9M?

pOH=-log[3.27x10-9]pOH=8.49

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

Summary…

pH=-log[H+]pOH=-

log[OH-]pH + pOH = 14

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

Example:What is the [H+] in a solution with [OH-] = 6.73x10-5M?

[H+][6.73x10-5]=1.0x10-14

[H+] = 1.49x10-10M

Sometimes multiple formulas must be used to carry out these

calculations:

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

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

Questions?Begin WS4