pH — the hydrogen ion concentration scale — Chemistry

Before this

This leads to

Buffer solutions
Indicators and titration curves
Ka, Kb and pH calculations · coming soon

Does boiling make water acidic?

Every school book says that pure water has a pH of 7, and that a pH below 7 means "acidic". Now measure the pH of pure water at 100 °C. It is about 6.14. (Source: Chemguide, "The ionic product for water, Kw".)

So has boiling turned pure water into an acid? Nothing has been added to it. No acid has dissolved in it. Yet its pH is below 7.

The answer is no, and the reason is the real meaning of "neutral". To see it, you need to know what pH actually measures, and where the number 7 comes from. Both come from one tiny reaction that the last lecture met at the end of its Try it yourself: water reacting with itself.

Water reacts with itself, a little

The last lecture showed that water is amphiprotic: it can give a proton and it can take one. So one water molecule can give a proton to another:

2 H_{2} O(l) ⇌ H_{3} O^{+} (aq) + OH^{-} (aq)

This is an equilibrium, and it lies very far to the left. In pure water at 25 °C, the concentration of hydronium ions is only $1.0 \times 1 0^{- 7}$ mol/dm³. Water itself is about 55.5 mol/dm³. So only about one water molecule in every 550 million has given away a proton at any moment. Most chemists shorten H₃O⁺(aq) to H⁺(aq), and this lecture does too.

Like any equilibrium, this one has a constant. The equilibrium constant lecture said that water is left out of the expression when it is the solvent, because its concentration hardly changes. What is left is the product of the two ion concentrations. It is called the ionic product of water, $K_{w}$ .

The ionic product of water

$K_{w} = [H^{+}] [OH^{-}]$

At 25 °C, $K_{w} = 1.0 \times 1 0^{- 14}$ mol²/dm⁶. This is true in pure water and in every dilute solution in water at that temperature.

In pure water, every H⁺ ion came with one OH⁻ ion. So the two concentrations are equal. Call each one $x$ . Then $x^{2} = 1.0 \times 1 0^{- 14}$ , so $x = 1.0 \times 1 0^{- 7}$ mol/dm³.

The second sentence in the box is the powerful one. Add an acid to water, and [H⁺] rises. Because the product must stay at $1.0 \times 1 0^{- 14}$ , [OH⁻] must fall. The two concentrations sit on a seesaw. If one is 1000 times bigger than before, the other is 1000 times smaller.

Numbers too small to say

Hydrogen ion concentrations run from about 1 mol/dm³ in a strong acid down to about 0.000 000 000 000 01 mol/dm³ in a strong alkali. Writing these numbers is awkward, and comparing them is worse. Even in powers of ten, a student has to compare $1 0^{- 3}$ with $1 0^{- 11}$ and remember which is bigger.

In 1909, the Danish chemist Søren Sørensen, working at the Carlsberg Laboratory in Copenhagen, suggested a shortcut. Write the hydrogen ion concentration as a power of ten, and use only the number in the power, without its minus sign. A concentration of $1 0^{- 3}$ mol/dm³ becomes simply "3". He wrote this number as pH. Historians still disagree about what the "p" stood for. It may have meant "power" (Potenz in German, puissance in French), or it may have been a label from his equations. (Sources: W. B. Jensen, Journal of Chemical Education, 2004; J. Nørby, Trends in Biochemical Sciences, 2000.)

"The number in the power" is exactly what a logarithm gives. The Mathematics lecture on logarithms defined $lo g_{10} b$ as the power of 10 that gives $b$ . So:

The definition of pH

$pH = - lo g_{10} [H^{+}]$ , where [H⁺] is in mol/dm³.

Read backwards: $[H^{+}] = 1 0^{- pH}$ mol/dm³.

The minus sign is there because the logarithms of these small numbers are negative. $lo g_{10} (1 0^{- 3}) = - 3$ , and the minus sign turns it into a positive 3. In pure water at 25 °C, $pH = - lo g_{10} (1 0^{- 7}) = 7$ . That is where the famous 7 comes from. It is not a choice. It is the result of $K_{w} = 1.0 \times 1 0^{- 14}$ at 25 °C.

The scale below shows both concentrations at 25 °C. As one goes up the scale, the other goes down.

Each step of 1 on the pH scale is a factor of 10 in [H⁺]. At 25 °C, [H⁺] × [OH⁻] = 1.0 × 10⁻¹⁴ at every point, so as [H⁺] falls, [OH⁻] rises by the same factor.

One unit means ten times

Because pH is a power of ten, each step of 1 on the scale is a factor of 10 in [H⁺]. A solution of pH 3 has ten times the H⁺ concentration of a solution of pH 4. A solution of pH 2 has a hundred times as much as one of pH 4. This is the biggest difference between pH and an ordinary scale such as temperature, and it is the most common source of mistakes. Commit to an answer below before you read on.

Predict — diluting an acid ten times

Hydrochloric acid of concentration 0.10 mol/dm³ has a pH of 1.0. You dilute it with water until it is ten times more dilute, 0.010 mol/dm³. What is the new pH? Choose one, then read the explanation.

This box shows what happens to the pH of 0.10 mol/dm³ hydrochloric acid when it is diluted ten times.

The new [H⁺] is 0.010 mol/dm³, which is $1 0^{- 2}$ mol/dm³, so the pH is 2.0. A tenfold dilution raises the pH by exactly 1. It does not multiply the pH by 10, and it does not make the solution neutral.

So pH and [H⁺] move in opposite directions, and on very different scales. A small-looking change in pH is a large change in concentration.

What "neutral" really means

Now return to the boiling water. The reaction that forms H⁺ and OH⁻ from water absorbs energy. So, by Le Chatelier's principle, a higher temperature pushes the equilibrium to the right. More ions form, and $K_{w}$ grows.

Temperature	$K_{w}$ (mol²/dm⁶)	pH of pure water
0 °C	$0.114 \times 1 0^{- 14}$	7.47
25 °C	$1.008 \times 1 0^{- 14}$	7.00
50 °C	$5.476 \times 1 0^{- 14}$	6.63
100 °C	$51.3 \times 1 0^{- 14}$	6.14

(Source: Chemguide, "The ionic product for water, Kw".) At 100 °C there are more H⁺ ions than at 25 °C. But there are exactly as many OH⁻ ions as H⁺ ions, because each H⁺ came with one OH⁻. Boiled water is not acidic. It is neutral, with a lower pH.

Acidic, neutral and alkaline

A solution is neutral when [H⁺] = [OH⁻]. It is acidic when [H⁺] is greater than [OH⁻], and alkaline when [OH⁻] is greater than [H⁺].

Only at 25 °C does neutral mean pH 7.

That is it. That is the whole idea. pH measures [H⁺]. Neutral means [H⁺] equals [OH⁻]. The number 7 joins the two only at 25 °C. Explore this below.

Explore — the H⁺ and OH⁻ seesaw

Move the pH. Watch [H⁺] and [OH⁻] change together. Then change the temperature, and see where "neutral" moves.

Temperature:

pH: 7.0

This table shows [H⁺] and [OH⁻] at several pH values, at 25 °C ( $K_{w} = 1.0 \times 1 0^{- 14}$ ) and at 100 °C ( $K_{w} = 5.13 \times 1 0^{- 13}$ ), with each solution's type.

pH	[H⁺]	[OH⁻] at 25 °C	At 25 °C	At 100 °C
1	10⁻¹	10⁻¹³	acidic	acidic
6.1	7.9 × 10⁻⁷	1.3 × 10⁻⁸	acidic	very slightly acidic
7	10⁻⁷	10⁻⁷	neutral	alkaline
13	10⁻¹³	10⁻¹	alkaline	alkaline

At 25 °C, neutral is pH 7.00. At 100 °C, neutral is pH 6.14, so pH 7 is alkaline there.

You should have seen the two bars move in opposite directions by exactly the same amount, at either temperature. You should also have seen pH 7 change from neutral to alkaline when the water was hot. Nothing about the solution changed. Only the meaning of "neutral" moved.

Stop and think

At body temperature, 37 °C, $K_{w} = 2.4 \times 1 0^{- 14}$ mol²/dm⁶. A solution at 37 °C has a pH of exactly 7.00. Is it acidic, neutral or alkaline?

Show the answer

Slightly alkaline. At pH 7.00, $[H^{+}] = 1.0 \times 1 0^{- 7}$ mol/dm³. Then $[OH^{-}] = 2.4 \times 1 0^{- 14} \div 1.0 \times 1 0^{- 7} = 2.4 \times 1 0^{- 7}$ mol/dm³, more than [H⁺]. Neutral water at 37 °C has $[H^{+}] = 2.4 \times 1 0^{- 14} \approx 1.55 \times 1 0^{- 7}$ mol/dm³, which is pH 6.81. (The value of $K_{w}$ is from the University of Waterloo, Chem 13 News, 2015.) In a 2014 exam that asked this kind of question, only about one student in three found the neutral pH correctly.

pOH, and why pH + pOH = 14

The same shortcut works for hydroxide ions: $pOH = - lo g_{10} [OH^{-}]$ . Take $- lo g_{10}$ of both sides of $K_{w} = [H^{+}] [OH^{-}]$ . The logarithm of a product is the sum of the logarithms, so:

- lo g_{10} K_{w} = pH + pOH

At 25 °C, $- lo g_{10} (1.0 \times 1 0^{- 14}) = 14$ , so $pH + pOH = 14$ . This is not a separate fact to remember. It is $K_{w}$ , written in the language of logarithms. At other temperatures, the 14 changes, as $K_{w}$ does.

Strong acids and strong bases

The last lecture described hydrochloric acid as a strong acid: in water, essentially every HCl molecule gives its proton to water. So for a strong acid such as HCl that gives one H⁺ per molecule, [H⁺] equals the concentration of the acid. In the same way, a strong base such as NaOH separates completely into its ions, so [OH⁻] equals the concentration of the base. The examples below are all at 25 °C.

Worked Example 1 — A strong acid

Find the pH of 0.050 mol/dm³ hydrochloric acid.

Think first. HCl is strong, so $[H^{+}] = 0.050$ mol/dm³.

$pH = - lo g_{10} (0.050) \approx 1.30$ (2 decimal places).

Check. 0.050 lies between 0.10 (pH 1) and 0.010 (pH 2), so the pH must lie between 1 and 2. ✓ It is closer to 1 because 0.050 is closer to 0.10 on a ten-times scale: it is half of 0.10, but five times 0.010. ✓

Worked Example 2 — A strong base, two ways

Find the pH of 0.010 mol/dm³ sodium hydroxide.

Think first. NaOH gives OH⁻, not H⁺. So find [H⁺] from $K_{w}$ .

Way 1. $[OH^{-}] = 0.010$ mol/dm³, so $[H^{+}] = \frac{1.0 \times 1 0 ^{- 14}}{0.010} = 1.0 \times 1 0^{- 12}$ mol/dm³, and $pH = 12$ .

Way 2. $pOH = - lo g_{10} (0.010) = 2$ , so $pH = 14 - 2 = 12$ .

Check. Both ways agree. ✓ An alkali must have a pH above 7 at 25 °C, and 12 is. ✓ The answer 2 is the pOH, not the pH. That is Trap 3.

Worked Example 3 — Diluting an acid

25.0 cm³ of 0.10 mol/dm³ hydrochloric acid is diluted with water to a total volume of 250 cm³. Find the pH before and after.

Think first. The amount of HCl stays the same. Only the volume changes, ten times.

Before: $pH = - lo g_{10} (0.10) = 1.0$ . After: the concentration is $0.10 \times \frac{25.0}{250} = 0.010$ mol/dm³, so $pH = 2.0$ .

Check. A tenfold dilution raised the pH by exactly 1, as the predict question showed. ✓

Worked Example 4 — From pH back to concentration

A solution has a pH of 3.70 at 25 °C. Find [H⁺] and [OH⁻].

Think first. Run the definition backwards: $[H^{+}] = 1 0^{- pH}$ .

$[H^{+}] = 1 0^{- 3.70} \approx 2.0 \times 1 0^{- 4}$ mol/dm³. $[OH^{-}] = \frac{1.0 \times 1 0 ^{- 14}}{2.0 \times 1 0 ^{- 4}} \approx 5.0 \times 1 0^{- 11}$ mol/dm³.

Check. pH 3.70 lies between 3 and 4, so [H⁺] must lie between $1 0^{- 4}$ and $1 0^{- 3}$ . ✓ The pOH is $14 - 3.70 = 10.30$ , and $1 0^{- 10.30} \approx 5.0 \times 1 0^{- 11}$ . ✓

Two more facts complete the picture. First, the scale does not stop at 0 and 14. By the simple definition, 2.0 mol/dm³ HCl has a pH of about −0.30, and 2.0 mol/dm³ NaOH a pH of about 14.30. Real solutions this concentrated behave less simply, but pH below 0 and above 14 are both measured. The range 0 to 14 covers most solutions you will meet. It is not a wall.

Second, not every acid is strong. A weak acid, such as ethanoic acid (the acid in vinegar), gives its proton to only a small fraction of its molecules. So [H⁺] is much smaller than the acid's concentration. For example, 0.100 mol/dm³ ethanoic acid has a pH of about 2.88 at 25 °C, while 0.100 mol/dm³ HCl has a pH of 1.00. Only about 1.3% of the ethanoic acid molecules have given away a proton. (This uses ethanoic acid's acid dissociation constant, $1.74 \times 1 0^{- 5}$ mol/dm³, from Chemguide. The Pre-University lecture on Ka shows how to calculate it.) So the pH of an acid depends on its strength as well as its concentration.

Three traps

Trap 1 — "pH 7 always means neutral"

Neutral means [H⁺] = [OH⁻]. That is pH 7 only at 25 °C. Boiled pure water has a pH of about 6.14 and is still neutral. Pure water at body temperature has a pH of about 6.81. Always say the temperature when you call pH 7 neutral.

Trap 2 — "pH 2 is twice as acidic as pH 4"

A difference of 2 pH units is a factor of $1 0^{2} = 100$ in [H⁺]. A solution of pH 2 has a hundred times the H⁺ concentration of one of pH 4. Turn a pH difference into a power of ten before you compare.

Trap 3 — Giving the pOH as the pH of an alkali

For 0.010 mol/dm³ NaOH, $- lo g_{10} (0.010) = 2$ . That is the pOH. An answer of pH 2 for an alkali is the warning sign: an alkali cannot have a pH below 7 at 25 °C. Use $K_{w}$ , or $pH = 14 - pOH$ , to finish.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Assume 25 °C unless a question says otherwise.

Find the pH of 0.0025 mol/dm³ nitric acid, HNO₃, a strong acid that gives one H⁺ per molecule.

Hint 1
For a strong acid of this kind, what is [H⁺]?

Hint 2
$[H^{+}] = 0.0025$ mol/dm³. Now use $pH = - lo g_{10} [H^{+}]$ .

Full answer
$pH = - lo g_{10} (0.0025) \approx 2.60$ . Check: 0.0025 lies between 0.01 (pH 2) and 0.001 (pH 3), so the pH lies between 2 and 3. ✓
Find the pH of 0.20 mol/dm³ potassium hydroxide, KOH, a strong base.

Hint 1
KOH gives OH⁻. Which concentration do you know?

Hint 2
$[OH^{-}] = 0.20$ mol/dm³. Find the pOH first, or find [H⁺] from $K_{w}$ .

Full answer
$pOH = - lo g_{10} (0.20) \approx 0.70$ , so $pH = 14 - 0.70 = 13.30$ . Or: $[H^{+}] = 1.0 \times 1 0^{- 14} \div 0.20 = 5.0 \times 1 0^{- 14}$ , and $- lo g_{10} (5.0 \times 1 0^{- 14}) \approx 13.30$ . ✓ Check: a concentrated alkali, so the pH should be well above 7. ✓
The water in a lake has a pH of 6.5. After years of acid rain, it has a pH of 5.0. By what factor has [H⁺] changed?

Hint 1
Each unit of pH is a factor of 10. How many units has the pH fallen?

Hint 2
The pH fell by 1.5 units, so the factor is $1 0^{1.5}$ .

Full answer
[H⁺] has risen by a factor of $1 0^{6.5 - 5.0} = 1 0^{1.5} \approx 31.6$ , so about 32 times. Check: $1 0^{- 5.0} \div 1 0^{- 6.5} = 1 0^{1.5}$ . ✓ A fall of "only 1.5" in pH is more than thirty times as much H⁺. This kind of change can harm the fish and plants in the lake.
A stretch. 50.0 cm³ of 0.10 mol/dm³ HCl is mixed with 50.0 cm³ of 0.080 mol/dm³ NaOH. Find the pH of the mixture.

Hint 1
H⁺ and OH⁻ react 1 : 1 to make water. Which one is in excess, and by how many moles?

Hint 2
Moles of H⁺ = 0.0050 and moles of OH⁻ = 0.0040. The excess is now in a total volume of 100 cm³.

Full answer
Moles of H⁺: $0.10 \times \frac{50.0}{1000} = 0.0050$ mol. Moles of OH⁻: $0.080 \times \frac{50.0}{1000} = 0.0040$ mol. They react 1 : 1, leaving 0.0010 mol of H⁺ in excess. The total volume is 100 cm³ = 0.100 dm³, so $[H^{+}] = 0.0010 \div 0.100 = 0.010$ mol/dm³, and $pH = 2.0$ . Check: the acid was in excess, so the pH must be below 7. ✓ A common mistake is to forget the doubled volume, which gives 0.020 mol/dm³ and pH 1.70.

The bigger idea

A logarithmic scale lets one short number line cover a range of a hundred million million. The price is that small differences in pH hide large differences in concentration. Your blood shows why this matters. Its normal pH is kept between 7.35 and 7.45. (Source: StatPearls, "Physiology, Acid Base Balance".) That range looks tiny, but it is [H⁺] between about 45 and 35 nanomoles per dm³, a factor of about 1.26. A blood pH of 7.0 would mean about 2.5 times the normal H⁺ concentration of pH 7.4.

Your body holds [H⁺] steady even though cells keep producing acid, mostly as carbon dioxide. It does this with mixtures that resist changes in pH. The next lecture explains how such a mixture, a buffer, works.

Key concepts in this lesson

Water reacts with itself a little, and $K_{w} = [H^{+}] [OH^{-}] = 1.0 \times 1 0^{- 14}$ mol²/dm⁶ at 25 °C, in pure water and in every dilute solution in water.
$pH = - lo g_{10} [H^{+}]$ , so each step of 1 in pH is a factor of 10 in [H⁺], and a higher pH means fewer H⁺ ions.
A solution is neutral when [H⁺] = [OH⁻], which is pH 7 only at 25 °C, because $K_{w}$ grows with temperature.
For a strong acid that gives one H⁺ per molecule, [H⁺] equals its concentration, and for a strong base, [OH⁻] equals its concentration, with [H⁺] found from $K_{w}$ .
$pH + pOH = 14$ at 25 °C is $K_{w}$ written with logarithms, and a weak acid has a higher pH than a strong acid of the same concentration.