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8.1 - Introduction to Acids and Bases

Overview

Acid-base chemistry comes down to two ions: hydronium, H3O+\mathrm{H_3O^{+}}, and hydroxide, OH−\mathrm{OH^{-}}. This topic gives you the tools to measure them: the pH and pOH scales, and the equilibrium constant for water itself, KwK_w. Every calculation in Unit 8 starts with these conversions, so practice them until they're automatic.

pH and pOH

In water solutions, [H3O+][\mathrm{H_3O^+}] can be anywhere from about 1 M down to 10−1410^{-14} M. Numbers that span 14 powers of ten are awkward to compare, so chemists use a logarithmic scale:
pH=−log⁡[H3O+]pOH=−log⁡[OH−]\mathrm{pH} = -\log[\mathrm{H_3O^+}] \qquad \mathrm{pOH} = -\log[\mathrm{OH^-}]
The "p" means "take the negative log of." Because of the minus sign, more H3O+\mathrm{H_3O^{+}} means a lower pH. Because it's a log, each change of 1 pH unit is a tenfold change in [H3O+][\mathrm{H_3O^+}]. A solution at pH 3 has 10 times the hydronium concentration of one at pH 4, and 100 times that of one at pH 5.
To go backward, undo the log:
[H3O+]=10−pH[OH−]=10−pOH[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}} \qquad [\mathrm{OH^-}] = 10^{-\mathrm{pOH}}
Hydrogen ion or hydronium ion? The terms "hydrogen ion" and "hydronium ion", and the symbols H+(aq)\mathrm{H^{+}(aq)} and H3O+(aq)\mathrm{H_3O^{+}(aq)}, all mean the same aqueous ion here. A bare proton never floats free in water (Topic 4.8), so H3O+(aq)\mathrm{H_3O^{+}(aq)} is preferred, but the AP exam accepts H+(aq)\mathrm{H^{+}(aq)} too.

Significant figures with logs

When you take a log, the digits after the decimal point in pH (or pOH) match the significant figures in the concentration. For example, [H3O+]=4.7×10−3[\mathrm{H_3O^+}] = 4.7 \times 10^{-3} M has two significant figures, so its pH is 2.33, with two decimal places. The 2 in front of the decimal only tells you the power of ten. Going backward works the same way: a pH with two decimal places gives a concentration with two significant figures.

Example 1: Converting between concentration and pH

(a) A solution has [H3O+]=4.7×10−3[\mathrm{H_3O^+}] = 4.7 \times 10^{-3} M. Find its pH.
pH=−log⁡(4.7×10−3)=2.33\mathrm{pH} = -\log(4.7 \times 10^{-3}) = 2.33
(b) A solution has pOH = 4.15. Find [OH−][\mathrm{OH^-}].
[OH−]=10−4.15=7.1×10−5 M[\mathrm{OH^-}] = 10^{-4.15} = 7.1 \times 10^{-5}\ \mathrm{M}
(c) How many times greater is [H3O+][\mathrm{H_3O^+}] in a solution with pH 3.0 than in one with pH 5.0?
10−3.010−5.0=102.0=100\frac{10^{-3.0}}{10^{-5.0}} = 10^{2.0} = 100
Answer: (a) pH = 2.33. (b) [OH−]=7.1×10−5[\mathrm{OH^-}] = 7.1 \times 10^{-5} M. (c) The pH 3.0 solution has 100 times more H3O+\mathrm{H_3O^{+}}. A difference of 2 pH units is a factor of 10210^2, not a factor of 2.
On a calculator, put the whole concentration in parentheses, like -log(4.7E-3), and use 10^(-4.15) to go backward.

Water's autoionization and KwK_w

Even perfectly pure water contains a few ions, because water molecules pass protons to each other (Topic 4.8 introduced this autoionization):
2H2O(l)⇌H3O+(aq)+OH−(aq)\mathrm{2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)}
Like any equilibrium (Unit 7), this one has an equilibrium constant. Liquid water is left out of the expression, so the constant, written KwK_w, is just the product of the two ion concentrations:
Kw=[H3O+][OH−]=1.0×10−14at 25 ∘CK_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} \quad \text{at } 25\ ^\circ\mathrm{C}
KwK_w is tiny, so the reaction barely happens. In pure water at 25 °C, only about 2 of every billion water molecules are ionized at any moment.
KwK_w applies to every aqueous solution, not only pure water. That links the two ions like a seesaw: add acid and [H3O+][\mathrm{H_3O^+}] goes up, so [OH−][\mathrm{OH^-}] must go down to keep the product at 1.0×10−141.0 \times 10^{-14}. An acidic solution still contains some OH−\mathrm{OH^{-}}, and a basic solution still contains some H3O+\mathrm{H_3O^{+}}.
Taking the negative log of both sides turns the product into a sum:
−log⁡Kw=−log⁡[H3O+]+(−log⁡[OH−])⇒pKw=pH+pOH-\log K_w = -\log[\mathrm{H_3O^+}] + \left(-\log[\mathrm{OH^-}]\right) \quad \Rightarrow \quad \mathrm{p}K_w = \mathrm{pH} + \mathrm{pOH}pKw=14.00=pH+pOHat 25 ∘C\mathrm{p}K_w = 14.00 = \mathrm{pH} + \mathrm{pOH} \quad \text{at } 25\ ^\circ\mathrm{C}
So if you know any one of [H3O+][\mathrm{H_3O^+}], [OH−][\mathrm{OH^-}], pH, or pOH, you can find the other three.
8.1 - Introduction to Acids and Bases
Reading the scale: pH and [H3O+][\mathrm{H_3O^+}] move in opposite directions, and [H3O+][\mathrm{H_3O^+}] and [OH−][\mathrm{OH^-}] always multiply to 10−1410^{-14}. The everyday markers are typical values (lemon juice about 2, black coffee about 5, seawater about 8.1, household ammonia about 11.5), and real samples vary.

Example 2: The pH of seawater

Seawater has a pH of 8.10 at 25 °C. Find its pOH, [H3O+][\mathrm{H_3O^+}], and [OH−][\mathrm{OH^-}]. Is seawater acidic or basic?
Step 1: Use pH + pOH = 14.00.
pOH=14.00−8.10=5.90\mathrm{pOH} = 14.00 - 8.10 = 5.90
Step 2: Undo each log.
[H3O+]=10−8.10=7.9×10−9 M[OH−]=10−5.90=1.3×10−6 M[\mathrm{H_3O^+}] = 10^{-8.10} = 7.9 \times 10^{-9}\ \mathrm{M} \qquad [\mathrm{OH^-}] = 10^{-5.90} = 1.3 \times 10^{-6}\ \mathrm{M}
Step 3: Check with KwK_w: (7.9×10−9)(1.3×10−6)=1.0×10−14(7.9 \times 10^{-9})(1.3 \times 10^{-6}) = 1.0 \times 10^{-14}. It works.
Answer: pOH = 5.90, [H3O+]=7.9×10−9[\mathrm{H_3O^+}] = 7.9 \times 10^{-9} M, and [OH−]=1.3×10−6[\mathrm{OH^-}] = 1.3 \times 10^{-6} M. Seawater is slightly basic, because [OH−][\mathrm{OH^-}] is greater than [H3O+][\mathrm{H_3O^+}].

Neutral solutions

A solution is neutral when [H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}]. Pure water is neutral, because autoionization makes one of each ion every time it happens. Setting the two equal at 25 °C:
[H3O+]=[OH−]=1.0×10−14=1.0×10−7 M⇒pH=pOH=7.00[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\ \mathrm{M} \quad \Rightarrow \quad \mathrm{pH} = \mathrm{pOH} = 7.00
At 25 °C
Ion concentrations
pH
pOH
Acidic
[H3O+][\mathrm{H_3O^+}] > [OH−][\mathrm{OH^-}]
below 7.00
above 7.00
Neutral
[H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}]
7.00
7.00
Basic
[OH−][\mathrm{OH^-}] > [H3O+][\mathrm{H_3O^+}]
above 7.00
below 7.00

KwK_w changes with temperature

KwK_w is an equilibrium constant, and every equilibrium constant depends on temperature. Autoionization is endothermic (ΔH∘=+55.8 kJ/molrxn\Delta H^\circ = +55.8\ \mathrm{kJ/mol_{rxn}}), so heating water shifts the equilibrium toward the ions, and KwK_w gets larger.
Temperature
KwK_w
pKwK_w
pH of pure water
0 °C
1.13×10−151.13 \times 10^{-15}
14.95
7.47
25 °C
1.0×10−141.0 \times 10^{-14}
14.00
7.00
37 °C (approximate)
2.4×10−142.4 \times 10^{-14}
13.62
6.81
50 °C
5.5×10−145.5 \times 10^{-14}
13.26
6.63
100 °C
5.6×10−135.6 \times 10^{-13}
12.25
6.13
In pure water, [H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}] = Kw\sqrt{K_w}. A larger KwK_w means more of both ions, so the pH of pure water drops as it warms. At 50 °C it's 6.63, and the water is still perfectly neutral, because [OH−][\mathrm{OH^-}] went up by exactly the same amount. That's why the pH of pure, neutral water is 7.00 only at 25 °C.
8.1 - Introduction to Acids and Bases
The data points come from the CRC Handbook of Chemistry and Physics and from Bandura and Lvov (2006). The 37 °C row is read from a smooth curve through those points, so it's approximate. Two takeaways: neutral means [H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}], not "pH 7". And pH + pOH = 14.00 only at 25 °C. At other temperatures, pH + pOH = pKwK_w for that temperature. Unless a problem says otherwise, assume 25 °C.

Example 3: Water at 50 °C (AP-style)

At 50 °C, Kw=5.5×10−14K_w = 5.5 \times 10^{-14}. (a) Calculate [H3O+][\mathrm{H_3O^+}] and the pH of pure water at 50 °C. (b) A student says this water is acidic because its pH is below 7. Do you agree? (c) Is the autoionization of water endothermic or exothermic? Justify your answer. (d) A different solution at 50 °C has pH 7.00. Is it acidic, basic, or neutral?
(a) In pure water the two ions are equal, so Kw=x2K_w = x^2:
[H3O+]=5.5×10−14=2.35×10−7 MpH=−log⁡(2.35×10−7)=6.63[\mathrm{H_3O^+}] = \sqrt{5.5 \times 10^{-14}} = 2.35 \times 10^{-7}\ \mathrm{M} \qquad \mathrm{pH} = -\log(2.35 \times 10^{-7}) = 6.63
(Two significant figures in KwK_w gives 2.3×10−72.3 \times 10^{-7} M; keep the extra digit until you've taken the log.)
(b) No. The water is neutral, because [H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}] = 2.35×10−72.35 \times 10^{-7} M. A pH below 7 means acidic only at 25 °C.
(c) Endothermic. KwK_w is larger at 50 °C (5.5×10−145.5 \times 10^{-14}) than at 25 °C (1.0×10−141.0 \times 10^{-14}). Raising the temperature shifts an endothermic reaction toward products, which increases KK.
(d) At 50 °C, pKw=−log⁡(5.5×10−14)=13.26K_w = -\log(5.5 \times 10^{-14}) = 13.26, so:
pOH=13.26−7.00=6.26\mathrm{pOH} = 13.26 - 7.00 = 6.26
The pOH is lower than the pH, so [OH−][\mathrm{OH^-}] is greater than [H3O+][\mathrm{H_3O^+}]. At 50 °C, a pH of 7.00 is slightly basic.

Common mistakes

  • Saying pH 7 always means neutral. Neutral means [H3O+][\mathrm{H_3O^+}] = [OH−][\mathrm{OH^-}]. That's pH 7.00 only at 25 °C.
  • Using 14.00 at the wrong temperature. pH + pOH = pKwK_w, which equals 14.00 only at 25 °C.
  • Treating pH as linear. pH 2 and pH 4 differ by a factor of 100 in [H3O+][\mathrm{H_3O^+}], not a factor of 2.
  • Thinking an acidic solution has no OH−\mathrm{OH^{-}}. KwK_w means both ions are always present. Acidic just means there's more H3O+\mathrm{H_3O^{+}} than OH−\mathrm{OH^{-}}.
  • Wrong decimal places. The decimals in pH match the significant figures in the concentration.
  • Calculator slips. Forgetting the parentheses in -log(4.7E-3) or the negative sign in 10^(-4.15).

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