Working with [H3O+], [OH–], pH, and pOHIntroduction

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: a × b = c

If A TIMES B is equal to 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).

We already know that in any aqueous solution the concentration of hydronium times the concentration of hydroxide is equal to Kw.

3

3

w

3 w

w

w

log H O log OH log

H O O

log H O log OH log

pH pO p

H

H

K

K

K

K

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.

3

3

3 w

w

w

w

log H O log OH log

H O OH

log H O log OH log

pH pOH p

K

K

K

K

Now, we’ll multiply everything by negative 1

3

3

3 w

w

w

w

log H O log OH log

H O OH

log H O log OH log

pH pOH p

K

K

K

K

–1

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.

3

3 w

3

w

w

w

H O OH

log H O log OH log

log H O log OH l

p p

o

H p H

g

O

K

K

K

K

The negative log of the hydronium ion concentration is the pH,

3

3 w

3 w

w

w

H O OH

log H O log OH log

log OH llog H og

pOH pH

O

p

K

K

K

K

the negative log of the hydroxide ion concentration is pOH,

3 w

3 w

3 w

w

H O OH

log H O log OH log

log H O log OH lo

p

g

HpH pO

K

K

K

K

And the negative log of Kw is equal to something called pKw.

3 w

3 w

3 w

w

H O OH

log H O log OH log

log H O log OH

pH pOH p

log

K

K

K

K

So we have two important equations: the concentration of hydronium times the concentration of hydroxide is equal to Kw.

3 w

w

H O OH

pH pOH p

K

K

And the pH PLUS the pOH is equal to pKw.

w

3 wH O OH

pH pOH pK

K

Where the pKw is defined as the –logKw

w

3 wH O OH

pH pOH pK

K

Where pKw = –logKw

Both of these equations are true for ANY temperature at which water is a liquid.

3 w

w

H O OH

pH pOH p

K

K

True for any temperature at which water is a

liquid

Now, we’ll zoom into a temperature of 25°C.

25°C

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 Kw = 1.0 × 10-

14

pKw = –logKw = –log(1.0 × 10–14) = 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

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

pKw = 14.00

pKw = pH + pOH = 14.00

Therefore we can say that at 25°C, pH + pOH = 14.00

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

pH + pOH = 14.00

This is true ONLY at 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 it is 25°C

And we can assume that pH + pOH is equal to 14.

At 25°C If temperature is not mentioned, assume

that

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

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

14.0

pH pOH 14

0 3.49

pH 10.51

.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

14.0

pH pOH 14

0 3.49

pH 10.51

.00

Which is 14 minus 3.49

The pOH of a solution is 3.49. What is the pH?

pH + pOH = 14.00

pH pOH 14.00

pH 14.00 p

14.00

pH 10

3.

.

OH

1

49

5

And that equals 10.51. So the pH is 10.51.

The pOH of a solution is 3.49. What is the pH?

pH + pOH = 14.00

pH pOH 14.00

pH 14.00 pOH

14.00 3.49

pH 10.51

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 ONLY at 25°C

[H+][OH–] = Kw

At ANY Temperature ONLY at 25°C

[H3O+][OH–] = Kw

pH + pOH = pKw

At ANY Temperature ONLY at 25°C

[H3O+][OH–] = Kw

pH + pOH = pKw

pH = –log[H3O+]

At ANY Temperature ONLY at 25°C

[H3O+][OH–] = Kw

pH + pOH = pKw

pH = –log[H3O+]

[H3O+] = 10–pH

At ANY Temperature ONLY at 25°C

[H3O+][OH–] = Kw

pH + pOH = pKw

pH = –log[H3O+]

[H3O+] = 10–pH

pOH = –log[OH–]

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

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)

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)

We can solve the previous equation for Kw, we get Kw = 10–pKw

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

Now we’ll review what is true 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

[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 = –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+]

[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

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–]

[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.

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

You REALLY need to

KNOW all these

equations!

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.

[H3O+

][OH–

]

pH pOH

pH =–log[H3O+]

[OH–] =10–pOH

pOH =–

log[OH–][H3O+]

=10–pH

[H3O+][OH–] = 1.00 × 10–14

pH + pOH = 14.00

The Squareat 25°C

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.

[H3O+

][OH–

]

pH pOH

pH =–log[H3O+]

[OH–] =10–pOH

pOH =–

log[OH–][H3O+]

=10–pH

[H3O+][OH–] = 1.00 × 10–14

pH + pOH = 14.00

The Squareat 25°C

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

pH =–log[H3O+]

[OH–] =10–pOH

pOH =–

log[OH–][H3O+]

=10–pH

[H3O+][OH–] = 1.00 × 10–14

pH + pOH = 14.00

The Squareat 25°C

Here’s a simpler version we can use to help us come up with plans for calculations

[H3O+

][OH–

]

pH pOH

For example, Let’s say we’re given the hydronium ion concentration and we want to find the pOH

[H3O+

][OH–

]

pH pOH

?

Given

We can do this in two steps. We could start (click) by converting hydronium ion concentration of pH…

[H3O+

][OH–

]

pH pOH

Given

?

And in the second step (click), we’ll convert pH to pOH.

[H3O+

][OH–

]

pH pOH

Given

?

Alternately, we could have started by converting (click) hydronium concentration to hydroxide concentration

[H3O+

][OH–

]

pH pOH

Given

?

And then (click) hydroxide ion concentration to pOH. This would give us the same answer as the other method.

[H3O+

][OH–

]

pH pOH

Given

?

In another example, let’s say we’re given the pH and we want to find hydroxide ion concentation.

[H3O+

][OH–

]

pH pOH

?

Given

We could start (click) by converting pH to pOH

[H3O+

][OH–

]

pH pOH

?

Given

And then (click) pOH to hydroxide concentration

[H3O+

][OH–

]

pH pOH

?

Given

Or alternately, we could have started with pH (click) and converted to hydronium concentration

[H3O+

][OH–

]

pH pOH

?

Given

And then from (click) hydronium concentration to hydroxide concentration.

[H3O+

][OH–

]

pH pOH

?

Given