This post answers two questions
- How to obtain the concentration knowing the pH?
- How to obtain the pH knowing the concentration
If you are not interested in the theory, jump to the examples at the end of the post.
As you well know the pH is the negative of the log in
base 10 of the H+/H3O+ ion concentration. If we use [H+] to represent
that concentration, then pH=-log[H+]
To obtain the [H+] you need to calculate the antilog. You write the definition in the form log[H+] =-pH
and then calculate 10 to the power of each member. The equality remains valid as both members are treated similarly. Thus
10^( log[H+] ) = 10^(-pH)
raising 10 to a power is the inverse function of taking the log in base
10, 10^(log(x))=log(10^(x)) = x (they are inverse of one another), you
are left with[H+]=10^(-pH)
Your calculator has a function [10 to x] accessed by pressing the [2nd] function key. To use it you must enter the negative value of the pH, press the
[2nd] function key then the [10 to x], then the = key to get the result (concentration)Example
1. Let the pH=5.5, what is the H+ concentration?
With [(-)] being the change sign key, then
[H+]:[ (-) ] 5.5 [2nd][10 to x] [=]
The result is 0.000003162 or 3.16 x 10^(-6)
Calculating the pH
For all H+/H3O+ concentrations of the form 1.*10^(a)
where a is an integer number between 0 and -14
, the pH is the negative value of the exponent.
Concentration =10^(-3), pH=3
For other concentrations such as 3.567*10^(-8), one cannot use the shortcut above, but have to calculate the log of the concentration
[H+/H3O+] = 3.567*10^(-8)
pH= - log(3.567*10^(-8))
This is keyed in as follows (to minimize the number of parentheses)8 (-) [2nd][10 to x] [*] 3.567 [LOG] [=] (-)
Here you have two (-) change sign, the first is entered after the exponent of 10, the other at the end of the calculation to take the negative of the displayed result. You may notice that it is entered in the reverse order of the defining relation
To verify your calculation, the result is 7.447696891 or just 7.45
If you have a problem with the first (-) try entering it before you type in 8.
Hope it helps
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