Working with
+
–
[H3O ], [OH ], pH,
and pOH
Introduction
Here, we’ll introduce some useful relationships that exist among hydronium
concentration, hydroxide concentration, pH, and pOH
From Math:
Here’s something that we should be aware of from Math, concerning logs
From Math:
If:
If A TIMES B is equal to c
a
×
b
= c
From Math:
If:
a × b
= c
Then: log(a) + log(b) = log(c)
Then the log of (a) PLUS the log of (b) is equal to the log of (c).
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
We already know that in any aqueous solution the concentration of hydronium
times the concentration of hydroxide is equal to Kw.
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
If we take the log of everything, it follows that the log of H3O+ concentration plus
the log of OH minus concentration is equal to the log of Kw.
 H 3O  



 OH  







–1 log  H 3O   log  OH 

 


Kw

log K w





 log  H 3O    log  OH     log K w 
pH

pOH

pK w
Now, we’ll multiply everything by negative 1
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
And we get that the negative log of hydronium ion concentration plus the negative
log of hydroxide ion concentration is equal to the negative log of Kw.
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
The negative log of the hydronium ion concentration is the pH,
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
the negative log of the hydroxide ion concentration is pOH,
 H 3O  




 OH  







log  H 3O   log  OH 
 

Kw

log K w






 log  H 3O    log  OH     log K w 
pH

pOH

pK w
And the negative log of Kw is equal to something called pKw.
 H 3O  



 OH  



Kw
pH

pOH

pK w
So we have two important equations: the concentration of hydronium times the
concentration of hydroxide is equal to Kw.
 H 3O  



 OH  



Kw
pH

pOH

pK w
And the pH PLUS the pOH is equal to pKw.
 H 3O  



 OH  



Kw
pH

pOH

pK w
Where
pKw = –logKw
Where the pKw is defined as the –logKw
 H 3O    OH   K w

 



pH  pOH  pK w
True for any temperature at which water is a liquid
Both of these equations are true for ANY temperature at which water is a liquid.
25°C
Now, we’ll zoom into a temperature of 25°C.
At 25°C
Kw = 1.0 × 10-14
pKw = –logKw = –log(1.0 × 10–14) = 14.00
At 25°C, Kw = 1.0 × 10-14
At 25°C
Kw = 1.0 × 10-14
pKw = –logKw = –log(1.0 × 10–14) = 14.00
So at 25°C, the pKw…
At 25°C
Kw = 1.0 × 10-14
pKw = –logKw = –log(1.0 × 10–14) = 14.00
Which is the negative log of Kw
At 25°C
Kw = 1.0 × 10-14
pKw = –logKw = –log(1.0 × 10–14) = 14.00
Is the –log of 1.0 × 10–14 …
At 25°C
Kw = 1.0 × 10-14
pKw = –logKw = –log(1.0 × 10–14) = 14.00
Which is equal to 14.00
At 25°C
pKw = 14.00
So we can say that specifically at 25°C
At 25°C
pKw = 14.00
The pKw is equal to 14.00
At 25°C
pKw = 14.00
pKw = pH + pOH = 14.00
Remember, we had recently determined that pKw is equal to pH + pOH
At 25°C
pKw = 14.00
pKw = pH + pOH = 14.00
And at 25° pkw = 14.00
At 25°C
pH + pOH = 14.00
Therefore we can say that at 25°C, pH + pOH = 14.00
At 25°C
This is
true ONLY
at 25°C!
pH + pOH = 14.00
You’ll be using this equation a lot. Just make sure you use caution. Remember, this
is true ONLY at 25°C.
At 25°C
If temperature is not
mentioned, assume that
it is 25°C
Remember that if temperature is not mentioned in a problem, we can assume that
it is 25°C
At 25°C
If temperature is not
mentioned, assume that
pH + pOH = 14.00
And we can assume that pH + pOH is equal to 14.
The pOH of a solution is 3.49.
What is the pH?
pH + pOH = 14.00
Here’s an example. We’re told that the pOH of a solution is 3.49 and we’re asked
what the pH is?
The pOH of a solution is 3.49.
What is the pH?
pH  pOH  14.00
pH  14.00  pOH
 14.00  3.49
pH  10.51
pH + pOH = 14.00
We are not given the temperature, so we can assume its 25°C and that pH + pOH is
equal to 14
The pOH of a solution is 3.49.
What is the pH?
pH  pOH  14.00
pH  14.00  pOH
 14.00  3.49
pH  10.51
pH + pOH = 14.00
We want to find the pH, so we rearrange the blue equation to solve for pH, and we
get the yellow equation: pH = 14 minus pOH.
The pOH of a solution is 3.49.
What is the pH?
pH  pOH  14.00
pH  14.00  pOH
 14.00  3.49
pH  10.51
pH + pOH = 14.00
Which is 14 minus 3.49
The pOH of a solution is 3.49.
What is the pH?
pH  pOH  14.00
pH  14.00  pOH
 14.00  3.49
pH  10.51
pH + pOH = 14.00
And that equals 10.51. So the pH is 10.51.
At ANY Temperature
ONLY at 25°C
Now we’ll review the things we know are true at any temperature and things we
know are true ONLY at 25°C.
At ANY Temperature
ONLY at 25°C
We’ll start with equations that are true at ANY temperature
At ANY Temperature
[H3O+][OH–] = Kw
[H+][OH–] = Kw
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH + pOH = pKw
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
pH = –log[H3O+]
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
[H3O+] = 10–pH
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
pOH = –log[OH–]
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
[OH–] = 10–pOH
ONLY at 25°C
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
pKw = –log(Kw)
ONLY at 25°C
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
We can solve the previous equation for Kw, we get Kw = 10–pKw
At ANY Temperature
[H3O+][OH–] = Kw
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
Now we’ll review what is true ONLY at 25°C
ONLY at 25°C
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
[H+][OH–] = 1.0 × 10–14
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH + pOH = 14.00
pH = –log[H3O+]
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
pH + pOH = 14.00
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH + pOH = 14.00
pH = –log[H3O+]
Kw = 1.0 × 10–14
[H3O+] = 10–pH
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
Kw = 1.0 × 10–14
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH + pOH = 14.00
pH = –log[H3O+]
Kw = 1.0 × 10–14
[H3O+] = 10–pH
pKw = 14.00
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
pKw = 14.00.
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH + pOH = 14.00
pH = –log[H3O+]
Kw = 1.0 × 10–14
[H3O+] = 10–pH
pKw = 14.00
pOH = –log[OH–]
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
So we see that any equations that contain the number 14, are ONLY true at 25°C.
At ANY Temperature
ONLY at 25°C
[H3O+][OH–] = Kw
[H3O+][OH–] =1.0 × 10–14
pH + pOH = pKw
pH + pOH = 14.00
pH = –log[H3O+]
Kw = 1.0 × 10–14
[H3O+] = 10–pH
pKw = 14.00
pOH = –log[OH–]
You REALLY
need to KNOW
all these
equations!
[OH–] = 10–pOH
pKw = –log(Kw)
Kw = 10–pKw
In order to succeed in the rest of this unit You REALLY need to KNOW all these equations!
Pause and make a screen capture of this, save it, and go over it periodically.
[H3
O +]
[H3O+][OH–] = 1.00 × 10–14
pH =
–log[H3O+]
pOH =
–log[OH–]
[H3O+] =
10–pH
pH
[OH–]
[OH–] =
10–pOH
pH + pOH = 14.00
pOH
Here’s the “square” at 25°C. It shows all the formulas you can use to make one step or
two step conversions among [H3O+], [OH-], pH, and pOH.
[H3
O +]
[H3O+][OH–] = 1.00 × 10–14
pH =
–log[H3O+]
pOH =
–log[OH–]
[H3O+] =
10–pH
pH
[OH–]
[OH–] =
10–pOH
pH + pOH = 14.00
pOH
It would be good if you could draw something similar to this from memory. It will help
you with the calculations you’ll be required to do.
[H3
O +]
[H3O+][OH–] = 1.00 × 10–14
pH =
–log[H3O+]
pOH =
–log[OH–]
[H3O+] =
10–pH
pH
[OH–]
[OH–] =
10–pOH
pH + pOH = 14.00
pOH
For now, you may want to pause the video, take a screen shot and print yourself a copy
of this to work with.
[H3O+]
[OH–]
pH
pOH
Here’s a simpler version we can use to help us come up with plans for calculations
[H3O+]
[OH–]
Given
?
pH
pOH
For example, Let’s say we’re given the hydronium ion concentration and we want to find
the pOH
[H3O+]
[OH–]
Given
?
pH
pOH
We can do this in two steps. We could start (click) by converting hydronium ion
concentration of pH…
[H3O+]
[OH–]
Given
?
pH
And in the second step (click), we’ll convert pH to pOH.
pOH
[H3O+]
[OH–]
Given
?
pH
pOH
Alternately, we could have started by converting (click) hydronium concentration to
hydroxide concentration
[H3O+]
[OH–]
Given
?
pH
pOH
And then (click) hydroxide ion concentration to pOH. This would give us the same
answer as the other method.
[H3O+]
[OH–]
?
Given
pH
pOH
In another example, let’s say we’re given the pH and we want to find hydroxide ion
concentation.
[H3O+]
[OH–]
Given
pH
We could start (click) by converting pH to pOH
pOH
?
[H3O+]
[OH–]
Given
pH
And then (click) pOH to hydroxide concentration
pOH
?
[H3O+]
[OH–]
?
Given
pH
pOH
Or alternately, we could have started with pH (click) and converted to hydronium
concentration
[H3O+]
[OH–]
Given
pH
pOH
And then from (click) hydronium concentration to hydroxide concentration.
?