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

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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 - ]

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pH Scale pH scale: the measure of acidity of a solution pH=-log[H + ] acidic neutral 0714 basic [H+] = concentration in Molarity

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

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

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

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pH Scale pOH scale: the measure of alkalinity (basic-ness) of a solution pOH=-log[OH - ] acidic neutral 0714 basic

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

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

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

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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 - ]=1.0x M KwKw

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

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

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

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

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Questions? Begin WS4

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