The student possibly has not heard of the term logarithms or know what they are. The simple answer is they are exponents of base numbers. In this discussion, the base number is 10. What is the value of the number 100 ? It is 1. What is the value of the number 101? It is 10. So if 100 is 1 and 101 is 10, What would be the exponents for numbers 2 to 9?. The following table gives these values.
|Number||Exponent||Number & exponent||Log of number|
|1||0||1 = 100||log 1 = 0|
|2||0.3010||2 = 100.3010||log 2 = 0.3010|
|3||0.4771||3 = 100.4771||log 3 = 0.4771|
|4||0.6020||4 = 100.6020||log 4 = 0.6020|
|5||0.6990||5 = 100.6990||log 5 = 0.6990|
|6||0.7782||6 = 100.7782||log 6 = 0.7782|
|7||0.8451||7 = 100.8451||log 7= 0.8451|
|8||0.9031||8 = 1009031||log 8 = 0.9031|
|9||0.9542||9 = 100.9542||log 9 = 0.9542|
|10||1.0||10 =101||log 10 = 1|
The exponents of 10 that equal a particular number are called the logarithms (usually shortened to log) of that number. So one can say the log of 2 = 0.3010 and the log of 8 = 0.9031
In the section titled Acids-Bases Intro, pH was defined as the negative log of the hydrogen concentration. Hopefully, this discussion will show the student how to manipulate logarithms and be able to convert molar concentration of H+ ions and OH– to pH and pOH.
First example: [H+] = 1 x 10-4
The definition is: pH = -log [H+]. In this example the [H+] equals 1 x 10 -4. Substitute this concentration into the pH equation
pH = – log [H+] = -log [1 x 10-4]
The 10-4 is expressed as an exponent, but what about the number 1. From the discussion in the introductory paragraph, the number 1 can also be expressed as 100 . So we can write the pH equation as:
pH = – log[100 x 10-4]
When the base is the same for the two exponents, the exponents can be added to simplify the expression.
pH = – log [10-4]
The log of 10-4 is -4. The pH can be expressed as pH = -[-4] = 4
Therefore the pH of this solution 4 which is an acidic solution.
What is the pOH of this solution? Recall from the Acid-Base intro section
14 = pH + pOH . The equation for this solution becomes: 14 = 4 + pOH Subtracting 4 from both sides gives pOH = 10.
Example 2. What is the pH of a solution whose [H+] = 3.5 x 10-3 ?
Substitute the [H+], 3.5 x 10-3 into the pH equation to get:
pH = -log [3.5 x 10-3]
How does one get the logarithm of the number 3.5? One could use a hand held calculator that has a log function. The log value could also be obtained by using a slide rule . The easiest way is to find an online calculator such as one found by conducting a Google search. Open a new browser window. Type in the address bar: What is the log of 3.5 (or any number) In our example the log of 3.5 is 0.54.
Next place this in the pH equation above.
pH = -log[ 100.54 x 10-3] = -log[10-2.46]
pH = 2.46
Example 3. What is the pH of a solution whose [OH–] =7.5 x 10-4
In this example, the pOH is first determined and then the pH. The pOH equation is:
pOH = -log[OH–]
Substitute 7.5 x 10 -4 into the equation to get:
pOH = -log[7.5 x 10-4]
The first step is to get the log 7.5. Open a browser window and enter log 7.5 in the address bar. The value is 0.88 (rounded)
pOH = -log[100.88 x 10-4 = -log[10-3.12 ] = -[ -3.12] = 3.12
The water equilibrium constant in pValues is; 14 = pH + pOH
So entering the value 3.12 in for the pOH, the value of pH is:
pH = 14 – 3.12 = 10.88
These calculations seem daunting at first, but with practice they become easier. The student will probably not encounter these types of pH calculations until the second semester of a chemistry class.