Applications of Common Logarithms
Examples Involving pH
In chemistry, the pH of a solution is defined as pH=−log[H3O+],where
[H3O+]is the hydronuim ion concentration in moles per liter. The pH value is a measure of the acidity or alkalinity of a solution. Pure water has pH 7.0, substances with pH values greater than 7.0 are alkaline, and substances with pH values less than 7.0 are acidic. It is customary to round pH values to the nearest tenth.
- Find the pH of a solution with
[H3O+]=2.5×10−4.
- Find the hydronium ion concentration of a solution with pH = 7.1.
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Wetlands are classified as bogs, fens, marshes, and swamps based on pH values. A pH value between 6.0 and 7.5 indicates that the wetland is a “rich fen.” When the pH is between 3.0 and 6.0, it is a “poor fen,” and if the pH falls to 3.0 or less, the wetland is a “bog.”
Suppose that the hydronium ion concentration of a sample of water from a wetland is
6.3×10−5. How would this wetland be classified?
Solutions
- Find the pH of a solution with
[H3O+]=2.5×10−4.
pH=−log[H30+]=−log(2.5×10−4)=−(log2.5+log10−4)
=−(0.3979−4)=−0.3979+4≈3.6
Note: We used the equality symbol, =, rather than the approximate equality symbol, ≈, when replacing log 2.5 with 0.3979. This is often done for convenience, despite the fact that most logarithms used in applications are indeed approximations. - Find the hydronium ion concentration of a solution with pH = 7.1.
pH=−log[H30+]
7.1=−log[H30+]
−7.1=log[H30+]
[H30+]=10−7.1
[H30+]≈7.9×10−8
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Suppose that the hydronium ion concentration of a sample of water from a wetland is
6.3×10−5. How would this wetland be classified?
pH=−log[H3O+]=−log(6.3×10−5)
=−(log6.3+log10−5)=−log6.3−(−5)=−log6.3+5≈4.2
Example Involving the Loudness of Sound
The loudness of sounds is measured in decibels. We first assign an intensity of I0 to a very faint threshold sound. If a particular sound has intensity
I, then the decibel rating
d of this louder sound is given by the following formula.
d=10logII0.
Find the rating in decibels of a sound with intensity 10,000I0.
Solution
d=10log10,000I0I0=10log10,000=10(4)=40
The sound has a decibel rating of 40.