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**Unit: Acids, Bases, and Solutions**

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

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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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**[H+] = concentration in Molarity**

pH Scale pH scale: the measure of acidity of a solution neutral acidic basic 7 14 pH=-log[H+] [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+]=1.5x10-4M?**

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

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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

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**pOH=-log[OH-] pH Scale**

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

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

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

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**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

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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-]=1.0x10-14M Kw

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

[H+][6.73x10-5]=1.0x10-14 [H+] = 1.49x10-10M

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

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**(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

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**(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

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Questions? Begin WS4

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Calculating pH and pOH. pH pH = - log [H + ] [H + ] = the hydrogen ion concentration pH: “potential of hydrogen” - A way of expressing the hydrogen ion.

Calculating pH and pOH. pH pH = - log [H + ] [H + ] = the hydrogen ion concentration pH: “potential of hydrogen” - A way of expressing the hydrogen ion.

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