### Fifteen conversions between pH, pOH, [H3O+], and [OH¯]

There are six inter-connected equations you should know. Here they are:

 pH = −log [H3O+] pOH = −log [OH¯] [H3O+] = 10¯pH [OH¯] = 10¯pOH [H3O+] [OH¯] = Kw pH + pOH = pKw

Please be aware that many teachers and texts use H+ rather than H3O+. Often, you find them freely mixed within the written materials, sometimes within the same sentence or problem. This is because H+ and H3O+ are taken to mean the exact same thing.

With the exercises below, you are often asked if a given solution is acid, basic, or neutral. Remember this:

pH < 7 ---> acidic
pH > 7 ---> basic
pH = 7 ---> neutral

Another way to look at this is:

[H3O+] > [OH¯] ---> acidic
[H3O+] < [OH¯] ---> basic
[H3O+] = [OH¯] ---> neutral

And, another way:

[OH¯] < [H3O+] ---> acidic
[OH¯] > [H3O+] ---> basic
[OH¯] = [H3O+] ---> neutral

With these types of problems, there are often two paths to the answer. Sometimes, I will mention both and, sometimes I'll mention only one.

The Internet has many images that show conversions between the pH, the pOH, etc. Here is a search that shows some.

Oh yes, remember that Kw equals 1.00 x 10¯14. Sometimes, you will see a unit of M2 (molarity squared) on the value. This is incorrect (for reasons beyond the scope of this website) but, as always, do what you teacher requires.

One weird thing below: the unit of molarity seems to just appear and disappear. When I state a concentration (either in the problem statement or in the answer), the M will appear. When the solution setup is shown, no unit of M will appear.

Lastly, you might not get exactly 1.0 x 10¯14 if you multiply a [H3O+] and an [OH¯], say from the answers in Exercise #3 or #4. This is because of rounding off.

Example #1: Calculate [OH¯] in:

(a) [H3O+] = 7.50 x 10¯5 M
(b) [H3O+] = 1.50 x 10¯9 M
(c) [H3O+] = 1.00 x 10¯7 M

Also, identify each solution as being acidic, basic, or neutral.

Solution:

(a)

The relationship between [H3O+] and [OH¯] is given by:

[H3O+] [OH¯] = Kw

(7.50 x 10¯5) ([OH¯]) = 1.00 x 10¯14

[OH¯] = 1.33 x 10¯10 M

Because [H3O+] > [OH¯], this solution is acidic.

(b)

(1.50 x 10¯9) ([OH¯]) = 1.00 x 10¯14

OH¯ = 1.00 x 10¯14 / 1.50 x 10¯9

OH¯ = 6.67 x 10¯6 M

Because [H3O+] < [OH¯], this solution is basic.

(c)

(1.00 x 10¯5) ([OH¯]) = 1.00 x 10¯14

[OH¯] = 1.00 x 10¯7 M

[H3O+] = [OH¯]

Neutral

Comment: when one concentration is 1.00 x 10¯7 M, you should be able to see via a mental calculation that the other value is also 1.00 x 10¯7 M.

Example #2: Calculate [H3O+] in:

(a) [OH¯] = 3.50 x 10¯4 M
(b) [OH¯] = 8.48 x 10¯10 M
(c) [OH¯] = 1.00 x 10¯7 M

Also, identify each solution as being acidic, basic, or neutral.

Solution:

(a)

([H3O+]) (3.50 x 10¯4) = 1.00 x 10¯14

[H3O+] = 1.00 x 10¯14 / 3.50 x 10¯4

[H3O+] = 2.86 x 10¯11 M

[OH¯] > [H3O+]

Basic.

(b)

([H3O+]) (8.48 x 10¯10) = 1.00 x 10¯14

[H3O+] = 1.18 x 10¯5 M

[OH¯] < [H3O+]

Acidic.

(c)

When [OH¯] = 1.00 x 10¯7 M, you should be able to mentally calculate that the [H3O+] is also 1.00 x 10¯7 M. The two values being equal means:

Neutral.

That being said, be prepared for a teacher to insist on you writing down the proper calculation, which is:

([H3O+]) (1.00 x 10¯7) = 1.00 x 10¯14

Example #3: pH = 5.75. Calculate:

(a) pOH
(b) [H3O+]
(c) [OH¯]
(d) acid, base, neutral?

Solution:

(a)

pH + pOH = 14.00

pOH = 14.00 − 5.75 = 8.25

(b)

[H3O+] = 10¯pH

[H3O+] = 10¯5.75

[H3O+] = 1.78 x 10¯6 M

(c)

[OH¯] = 10¯pOH

[OH¯] = 10¯8.25

[OH¯] = 5.62 x 10¯9 M

You could also have done this:

(1.78 x 10¯6) ([OH¯]) = 1.00 x 10¯14

(d)

pH < 7, therefore acidic.

Example #4: pOH = 3.20. Calculate:

(a) pH
(b) [H3O+]
(c) [OH¯]
(d) acid, base, neutral?

Solution:

(a)

pH + pOH = 14.00

pH = 14.00 − 3.20 = 10.80

(b)

[H3O+] = 10¯10.80 = 1.58 x 10¯11 M

(c)

[OH¯] = 1.00 x 10¯14 / 1.58 x 10¯11 = 6.33 x 10¯4 M

(d)

Basic. Because [OH¯] > [H3O+]. Also could say because pH > pOH.

Example #5: [H3O+] = 5.65 x 10¯8 M. Calculate:

(a) pH
(b) pOH
(c) [OH¯]
(d) acid, base, neutral?

Solution:

(a)

pH = −log [H3O+]

pH = − log 5.65 x 10¯8

pH = 7.248 (and yes, that is three sig figs)

(b)

pH + pOH = pKw

7.248 + pOH = 14.000

pOH = 6.752

Example #6: [OH¯] = 2.84 x 10¯6 M. Calculate:

(a) pH
(b) pOH
(c) [H3O+]
(d) acid, base, neutral?

Solution:

(b)

pOH = −log [OH¯]

pOH = −log 2.84 x 10¯6

pOH = 5.547

(a)

pH + pOH = pKw

pH = 14.000 − 5.547 = 8.453

Note how I did (a) and (b) out of order.

(c)

[H3O+] [OH¯] = Kw

([H3O+]) (2.84 x 10¯6) = 1.00 x 10¯14

[H3O+] = 3.52 x 10¯9 M

Note: you could have also used 10¯8.453 to get the [H3O+]

(d)

Basic. One way (of several) to tell this is that the pH is greater than 7.

Example #7: Calculate the pH of the following:

(a) [H+] = 1 x 10¯3 M
(b) [OH¯] = 1 x 10¯5 M
(c) [H+] = 1 x 10¯7 M
(d) [H+] = 1 x 10¯13 M
(e) [OH¯] = 1 x 10¯0 M

Also, identify the neutral pH in the above list.

Solution:

1) The solution process is the same for (a), (c), and (d). Here is (d):

pH = −log [H+]

pH = −log 1 x 10¯13

pH = 13.0

2) When the hydroxide concentration is given, an additional step is required to get the pH. Here is (b):

pOH = −log [OH¯]

pOH = −log 1 x 10¯5 = 5.0

pH + pOH = pKw

pH = 14.0 − 5.0 = 9.0

3) (e) is an interesting one due to the exponent of zero:

pOH = −log [OH¯]

pOH = −log 1 x 10¯0

Remember, anything raised to the zero power equals 1, so 1 x 10¯0 becomes 1 x 1 which equals 1.

pOH = −log 1 = 0

pH = 14.0 (with one sig fig, even though I wrote 0 just above. Think of it as 0.0)

4) The neutral pH is (c). When [H+] equals 1 x 10¯7 M, the pH equals 7.0. That, by definition, is neutral.

Example #8: Calculate the pOH for the following:

(a) [H+] = 1 x 10¯8 M
(b) [OH¯] = 1 x 10¯11 M
(c) [H+] = 1 x 10¯2 M
(d) [H+] = 1 x 10¯0 M
(e) [OH¯] = 1 x 10¯7 M

Also, identify the neutral pOH in the above list.

Solution:

1) The solution process is the same for (b) and (e). Here's (b):

pOH = −log [OH¯]

pOH = −log 1 x 10¯11

pOH = 11.0

2) When the hydrogen ion concentration is given, an additional step is required to get the pOH. Here is (a):

pH = −log [H+]

pH = −log 1 x 10¯8 = 8.0

pH + pOH = pKw

pOH = 14.0 − 8.0 = 6.0

3) (d) is an interesting one due to the exponent of zero:

pH = −log [H+]

pH = −log 1 x 10¯0

Remember, anything raised to the zero power equals 1, so 1 x 10¯0 becomes 1 x 1 which equals 1.

pH = −log 1 = 0

pOH = 14.0 (with one sig fig, even though I wrote 0 just above. Think of it as 0.0)

4) The neutral pOH is (e).

Example #9: What would the hydrogen ion concentration be if the pOH was 4.92?

Solution:

pH + pOH = pKw

pH = 14.00 − 4.92 = 9.08

[H+] = 10¯pH

[H+] = 10¯9.08 = 8.3 x 10¯10 M

Example #10: What would the hydroxide ion concentration be if the hydrogen ion concentration was 4.7 x 10¯5 M?

Solution:

[H3O+] [OH¯] = Kw

(4.7 x 10¯5) ([OH¯]) = 1.0 x 10¯14

[OH¯] = 2.1 x 10¯10 M

Example #11: What would the pH of a solution that had a hydroxide ion concentration of 8.2 x 10¯9 M be?

Solution #1:

pOH = −log [OH¯]

pOH = −log 8.2 x 10¯9 = 8.09

pH + pOH = pKw

pH = 14.00 − 8.1 = 5.91

Solution #2:

[H3O+] [OH¯] = Kw

([H3O+]) (8.2 x 10¯9) = 1.0 x 10¯14

[H3O+] = 1.2195 x 10¯6 M (kept some guard digits)

pH = −log 1.2195 x 10¯6 = 5.91

Example #12: What would the pH of a solution that had a hydrogen ion concentration of 2.3 x 10¯7 M be?

Solution:

pH = −log 2.3 x 10¯7 = 6.64

Example #13: What would the hydroxide ion concentration be if the hydrogen ion concentration was 6.8 x 10¯5 M?

Solution:

[H3O+] [OH¯] = Kw

(6.8 x 10¯5) ([OH¯]) = 1.0 x 10¯14

[OH¯] = 1.5 x 10¯10 M

Example #14: Given [H+] = 1.50 M, calculate the pH and the pOH.

Solution:

pH = −log 1.50 = −0.176

Yes, negative pH values are possible. They just don't show up very often.

pOH = pKw − pH = 14.176

Yes, pOH values above 14 are possible, just not very common.

Example #15: Given [OH¯] = 1.50 M, calculate the pH and the pOH.

Solution:

pOH = −log 1.50 = −0.176

Yes, negative pOH values are possible. They just don't show up very often.

pH = pKw − pOH = 14.176

Yes, pH values above 14 are possible, just not very common.

Bonus Example: Fill in all the blanks.

 [H+] [OH¯] pH pOH acidicbasic A 0.00038 M B 6.5 x 10¯6 M C 9.28 D 12.29 E neutral

Solution:

1) Remember those six inter-connected equations? Here they are again:

 pH = −log [H3O+] pOH = −log [OH¯] [H3O+] = 10¯pH [OH¯] = 10¯pOH [H3O+] [OH¯] = Kw pH + pOH = pKw