pH and how to calculate it | pOH and how to calculate it | Return to the Acid Base menu |

Acid base calculations will involve one or more of these __four__ variables and __two__ constants.

[H _{3}O^{+}]pH [OH¯] pOH K _{w}pK _{w}<--- some discussion on the numerical value of K_{w}and pK_{w}will be found below.

There are __six__ inter-connected equations you should memorize.

pH = −log [H _{3}O^{+}]pOH = −log [OH¯] [H _{3}O^{+}] = 10¯^{pH}[OH¯] = 10¯ ^{pOH}[H _{3}O^{+}] [OH¯] = K_{w}pH + pOH = pK _{w}

Please be aware that many teachers and texts use H^{+} rather than H_{3}O^{+}. Often, you find them freely mixed within the written materials, sometimes within the same sentence or problem. This is because H^{+} and H_{3}O^{+} are taken to mean the exact same thing.

In the future (where we will spend the rest of our lives), there will be four more constants introduced:

K _{a}pK _{a}K _{b}pK _{b}

With the exercises below, you are often asked if a given solution is acid, basic, or neutral. Here is some guidance on how to determine if a solution is acidic, basic, or neutral:

**I. Acidic:** If we find that any __one__ of these six conditions is true, then the solution __must__ be acidic.

pH < 7 pOH > 7 [H _{3}O^{+}] > [OH¯][OH¯] < [H _{3}O^{+}][H _{3}O^{+}] > 10¯^{7}M[OH¯] < 10¯ ^{7}M

**II. Basic:** If we find that any __one__ of these six conditions is true, then the solution __must__ be basic.

pH > 7 pOH < 7 [H _{3}O^{+}] < [OH¯][OH¯] > [H _{3}O^{+}][H _{3}O^{+}] < 10¯^{7}M[OH¯] > 10¯ ^{7}M

**III. Neutral:** If we find that any __one__ of these six conditions is true, then the solution __must__ be neutral.

pH = 7 pOH = 7 [H _{3}O^{+}] = [OH¯][OH¯] = [H _{3}O^{+}][H _{3}O^{+}] = 10¯^{7}M[OH¯] = 10¯ ^{7}M

**Several Comments**

(a) With the problems below, there are often two paths to the answer. Sometimes, I will mention both and, sometimes I'll mention only one.(b) The Internet has many images that show conversions between the pH, the pOH, etc. Here is a search that shows some.

(c) K

_{w}equals 1 x 10¯^{14}. Sometimes, you will see a unit of M^{2}(molarity squared) on the value. This is incorrect (for reasons beyond the scope of this website) but, as always, do what you teacher requires.(d) The value used for K

_{w}might be presented with a differing number of decimal places: 1, 1.0, 1.00, 1.000 being the most common. Do not ever let the value of the K_{w}determine the number of significant figures. Use the information given about the solution determine sig figs.(e) As you look at the examples, 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 a pH, a pOH, or K

_{w}is shown, no unit of M will appear.(f) You might not get exactly 1.0 x 10¯

^{14}if you multiply a [H_{3}O^{+}] and an [OH¯] from the same problem, say from the answers in Exercise #3 or #4. This is because of rounding off.(g) This equation:

[H_{3}O^{+}] [OH¯] = K_{w}= 1 x 10¯^{14}is sometimes called the governing equation of acid base behavior. Notice the "see-saw" nature of the equation. When one variable increases in value, the other value MUST decrease. This is because the product of the two variables MUST always equal 1 x 10¯

^{14}. Indeed, we can never have a situation where both variables are greater than 10¯^{7}M or where both variables are less than 10¯^{7}M.

**Example #1:** Calculate [OH¯] in:

(a) [H_{3}O^{+}] = 7.50 x 10¯^{5}M

(b) [H_{3}O^{+}] = 1.50 x 10¯^{9}M

(c) [H_{3}O^{+}] = 1.00 x 10¯^{7}M

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

**Solution:**

(a)

The relationship between [H_{3}O^{+}] and [OH¯] is given by:[H

_{3}O^{+}] [OH¯] = K_{w}(7.50 x 10¯

^{5}) ([OH¯]) = 1.00 x 10¯^{14}[OH¯] = 1.33 x 10¯

^{10}MBecause [H

_{3}O^{+}] > [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}MBecause [H

_{3}O^{+}] < [OH¯], this solution is basic.

(c)

(1.00 x 10¯

^{7}) ([OH¯]) = 1.00 x 10¯^{14}[OH¯] = 1.00 x 10¯

^{7}M[H

_{3}O^{+}] = [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 [H_{3}O^{+}] 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)

([H_{3}O^{+}]) (3.50 x 10¯^{4}) = 1.00 x 10¯^{14}[H

_{3}O^{+}] = 1.00 x 10¯^{14}/ 3.50 x 10¯^{4}[H

_{3}O^{+}] = 2.86 x 10¯^{11}M[OH¯] > [H

_{3}O^{+}]Basic.

(b)

([H_{3}O^{+}]) (8.48 x 10¯^{10}) = 1.00 x 10¯^{14}[H

_{3}O^{+}] = 1.18 x 10¯^{5}M[OH¯] < [H

_{3}O^{+}]Acidic.

(c)

When [OH¯] = 1.00 x 10¯^{7}M, you should be able to mentally calculate that the [H_{3}O^{+}] 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:

([H

_{3}O^{+}]) (1.00 x 10¯^{7}) = 1.00 x 10¯^{14}

**Example #3:** pH = 5.75. Calculate:

(a) pOH

(b) [H_{3}O^{+}]

(c) [OH¯]

(d) acid, base, neutral?

**Solution:**

(a)

pH + pOH = 14.00pOH = 14.00 − 5.75 = 8.25

(b)

[H_{3}O^{+}] = 10¯^{pH}[H

_{3}O^{+}] = 10¯^{5.75}[H

_{3}O^{+}] = 1.78 x 10¯^{6}M

(c)

[OH¯] = 10¯^{pOH}[OH¯] = 10¯

^{8.25}[OH¯] = 5.62 x 10¯

^{9}MYou 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) [H_{3}O^{+}]

(c) [OH¯]

(d) acid, base, neutral?

**Solution:**

(a)

pH + pOH = 14.00pH = 14.00 − 3.20 = 10.80

(b)

[H_{3}O^{+}] = 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¯] > [H_{3}O^{+}]. You could also say it is basic because pH > pOH.

**Example #5:** [H_{3}O^{+}] = 5.65 x 10¯^{8} M. Calculate:

(a) pH

(b) pOH

(c) [OH¯]

(d) acid, base, neutral?

**Solution:**

(a)

pH = −log [H_{3}O^{+}]pH = − log 5.65 x 10¯

^{8}pH = 7.248 (and yes, that is three sig figs)

(b)

pH + pOH = pK_{w}7.248 + pOH = 14.000

pOH = 6.752

**Example #6:** [OH¯] = 2.84 x 10¯^{6} M. Calculate:

(a) pH

(b) pOH

(c) [H_{3}O^{+}]

(d) acid, base, neutral?

**Solution:**

(b)

pOH = −log [OH¯]pOH = −log 2.84 x 10¯

^{6}pOH = 5.547

(a)

pH + pOH = pK_{w}pH = 14.000 − 5.547 = 8.453

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

(c)

[H_{3}O^{+}] [OH¯] = K_{w}([H

_{3}O^{+}]) (2.84 x 10¯^{6}) = 1.00 x 10¯^{14}[H

_{3}O^{+}] = 3.52 x 10¯^{9}MNote: you could have also used 10¯

^{8.453}to get the [H_{3}O^{+}]

(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 <--- often seen in problems as 1 M (or 1.0 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

Note: the answer for (a) is 3.0 and the answer for (c) is given below.

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}= 5.0pH + pOH = pK

_{w}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 <--- this is where you would start if 1.0 M was the way the concentration was given in the problem

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}MAlso, 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}= 8.0pH + pOH = pK

_{w}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 = pK_{w}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:**

[H_{3}O^{+}] [OH¯] = K_{w}(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}= 8.09pH + pOH = pK

_{w}pH = 14.00 − 8.1 = 5.91

**Solution #2:**

[H_{3}O^{+}] [OH¯] = K_{w}([H

_{3}O^{+}]) (8.2 x 10¯^{9}) = 1.0 x 10¯^{14}[H

_{3}O^{+}] = 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:**

[H_{3}O^{+}] [OH¯] = K_{w}(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 = −0.176Yes, negative pH values are possible. They just don't show up very often.

pOH = pK

_{w}− pH = 14.176Yes, 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 = −0.176Yes, negative pOH values are possible. They just don't show up very often.

pH = pK

_{w}− pOH = 14.176Yes, pH values above 14 are possible, just not very common.

**Bonus Example:** Fill in all the blanks. Try to fill in the last colum "by inspection." Try to fill in all of row E "by inspection." There are a couple more that can be done by inspection (without touching your calculator). Hint: it's where the two values will add up to 14.

[H ^{+}][OH¯] pH pOH acidic

basicA 0.00038 M B 6.5 x 10¯ ^{6}MC 9.28 D 12.29 E neutral

**Solution:**

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

pH = −log [H _{3}O^{+}]pOH = −log [OH¯] [H _{3}O^{+}] = 10¯^{pH}[OH¯] = 10¯ ^{pOH}[H _{3}O^{+}] [OH¯] = K_{w}pH + pOH = pK _{w}

2) Here are the answers:

[H ^{+}][OH¯] pH pOH acidic

basicA 0.00038 M 2.6 x 10¯ ^{11}M3.42 10.58 acidic B 1.5 x 10¯ ^{9}M6.5 x 10¯ ^{6}M8.81 5.19 basic C 5.2 x 10¯ ^{10}M1.9 x 10¯ ^{5}M9.28 4.72 basic D 0.019 M 5.1 x 10¯ ^{13}M1.71 12.29 acidic E 1.0 x 10¯ ^{7}M1.0 x 10¯ ^{7}M7.00 7.00 neutral

pH and how to calculate it | pOH and how to calculate it | Return to the Acid Base menu |