pH and pOH
When working with dilute solutions of strong or weak acids and likewise with
bases, the [H3O+] or [OH-] is very small. With such small quantities, it is difficult
to compare concentrations, hence the pH and pOH scales were developed.
The term “pH” was named by the Danish chemist S.P.L Sorenson: the “p” is an
abbreviation for “potenz”, which means “potency” or “power” in German. Hence,
pH represents “power of 10 of​ ​ [H3O+].
Background information:
What are logs?
Log10(X) is a logarithm to the base of 10 ( it is the power of 10 used to represent a
number).
Example:
What is the logarithm of 1000?
= log (1000)
* 1000= 10​3
therefore, the
= log(1000) = 3
You try!
Calculate the logs of each of the following:
a. 100
b. 10​-7
c. 5.59 x 10​-6
The reverse procedure of taking the log is the antilog
Example:
What is the antilog of 4?
antilog(4)= 10​4​ = 10 000
You try!
Calculate the antilog of each of the following:
a. 2
b. -3
c. -6.155
Now that you know how to use your calculator! Let’s get back to pH and pOH
IV. 8 pH and pOH
The pH of a solution is the negative logarithm of the molar concentration of
hydronium ion and pOH is the negative logarithm of the molar concentration of
hydroxide ion.
pH= -log [H3O​+​]
pOH= - log [OH​-​]
[H3O​+​] = antilog (-pH)
[OH​-​]= antilog (-pOH)
Let’s try!
[ H+] = 5.57 x 10​-6​ M
[OH-] = 6.117 x10​-2​ M
pH = 12.761
pOH= 9.62
Special rule with sig figs!
There is a relationship between pH, pOH and Kw.
K​w​= [H​+​][OH​-​] = 1.0 x 10​-14
If we take the -log of both sides of the equation
-log K​w​ = -log([H​+​][OH​-​])
Therefore, if pH = 6.00 the pOH must be?
●
●
The pH and the pOH scales are logarithmic scales, a difference in one pH or
pOH unit is equivalent to a tenfold (10x) difference in concentration.
The pH and pOH are the negative of the exponent therefore, low pH and pOH
values mean relatively high values of [H+] and [OH-] and vice versa for high
pH and pOH values.
Example:
Using what we know about strong acids and bases:
What is the pH that results when 30.0 mL of 0.340 M HCl is mixed with 20.0 mL of
0.460 M of NaOH?
2. Calculate the pOH resulting from the mixture of 50.0 mL of 0.0185M Sr(OH)2
with 35.0 mL of a solution containing 0.130g of HCl
3. What is the final pH when 25.0 mL of pH 4.745 solution is mixed with 50.0 mL
of pH 9.150 solution?
● Identify which is the acid and which is the base
● Work in concentrations units, not pH
+​
-​
● Work with [H​ ] if the solution is acidic and with [OH​ ] if it is basic.