Calculations with Acids and Bases

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Presentation transcript:

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

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-]

[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

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: 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+] 10-9.42=[H+] [H+]=3.80x10-10M

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

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: pH + pOH = 14 Example: What is the pH in a solution with a pOH=8.6? pH + 8.6 = 14 pH=5.4

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:

(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

(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

Questions? Begin WS4