School Chemistry — Grades 1 to 12 · Grades 1–12

44Acids and Bases: Ka and pKa

Many ants defend themselves with methanoic acid, and on the skin it burns. Methanoic acid is a small carboxylic acid. Hydrochloric acid at the same concentration would burn far more. Both are acids; both give a solution of pH below 7; yet they are not acidic in the same way. To describe an acid fully, two numbers are needed: how much of it there is, its concentration, and how readily it gives away its hydrogen ion, its acidity constant.

You already know

Acidic, basic and neutral solutions; the pH scale, from 0 to 14, lower when there are more hydrogen ions (Chapter 18). The reaction quotient $Q$ and the equilibrium constant $K$ (Chapter 43). A curly arrow shows a pair of electrons moving (Chapter 40).

Wood ants: like many ants, they defend themselves with methanoic acid.

44.1 Brønsted acids and bases

Definition 44.1 (Brønsted acid and base, acid–base couple)

A Brønsted acid is a species able to give a hydrogen ion $\ce{H+}$ ; a Brønsted base is a species able to take one. An acid HA and the base $\ce{A-}$ it becomes by losing $\ce{H+}$ form an acid–base couple, written $\ce{HA}$ / $\ce{A-}$ : $\ce{HA} \rightleftharpoons \ce{A-} + \ce{H+}$ .

Definition 44.2 (Oxonium ion, ampholyte)

In water a hydrogen ion never stays alone: it is taken by a water molecule, giving the oxonium ion $\ce{H3O+}$ . A species that is the acid of one couple and the base of another is an ampholyte; water is one, acid of the couple $\ce{H2O}$ / $\ce{OH-}$ and base of the couple $\ce{H3O+}$ / $\ce{H2O}$ .

Example 44.3 (An acid in water)

When ethanoic acid dissolves, a lone pair of a water molecule takes the hydrogen of its $\ce{O-H}$ , and the $\ce{O-H}$ pair stays on the acid’s oxygen:

\ce{CH3COOH + H2O <=> CH3COO- + H3O+} .

Two couples exchange a hydrogen ion: $\ce{CH3COOH}$ / $\ce{CH3COO-}$ gives it, $\ce{H3O+}$ / $\ce{H2O}$ takes it. From now on, the hydrogen ions of earlier chapters are written $\ce{H3O+}$ whenever water is the solvent.

A hydrogen ion passes from the acid to water: a lone pair of water forms the new O-H bond (orange), and the old O-H pair stays on the oxygen of the ethanoate ion (blue). — A hydrogen ion passes from the acid to water: a lone pair of water forms the new $\ce{O-H}$ bond (orange), and the old $\ce{O-H}$ pair stays on the oxygen of the ethanoate ion (blue).

History — From Arrhenius to Brønsted, 1887–1923

In 1887 Svante Arrhenius proposed that acids give hydrogen ions in water and bases hydroxide ions. In 1923 Johannes Brønsted, and independently Thomas Lowry, broadened the idea: an acid is any species that gives a hydrogen ion, a base any species that takes one, in water or not. The ammonia molecule, which contains no hydroxide, became a base like any other.

44.2 The pH, quantitatively

Definition 44.4 (The pH, precisely)

In a dilute aqueous solution, the pH is defined by

\mathrm{pH} = -\log\frac{[\ce{H3O+}]}{c^\circ}, \qquad\text{that is}\qquad [\ce{H3O+}] = c^\circ \times 10^{-\mathrm{pH}} ,

where $\log$ is the decimal logarithm.

Proposition 44.5 (Tenfold dilution, one unit)

If $[\ce{H3O+}]$ is divided by 10, the pH increases by 1; if it is multiplied by 10, the pH decreases by 1.

Proof. $-\log\big([\ce{H3O+}]/10c^\circ\big) = -\log\big([\ce{H3O+}]/c^\circ\big) + \log 10 = \mathrm{pH} + 1$ . ∎

Definition 44.6 (Strong and weak acids and bases)

A strong acid reacts totally with water: none of the acid remains as such in the solution. A weak acid reacts partially: its reaction with water reaches an equilibrium. Likewise a strong base reacts totally with water, a weak base partially.

Proposition 44.7 (pH of a strong acid)

A solution of a strong acid of concentration $c$ (not too dilute) has $[\ce{H3O+}] = c$ and $\mathrm{pH} = -\log(c/c^\circ)$ .

Proof. The reaction $\ce{HA + H2O -> A- + H3O+}$ is total: each molecule of acid gives one oxonium ion. The ions coming from water itself are negligible as long as $c$ is much larger than $10^{-7}\,\mathrm{mol}/\mathrm{L}$ . ∎

Example 44.8 (Hydrochloric acid)

Hydrochloric acid at $0.010\,\mathrm{mol}/\mathrm{L}$ : $[\ce{H3O+}] = 0.010\,\mathrm{mol}/\mathrm{L}$ and $\mathrm{pH} = -\log 0.010 = 2.0$ . Diluted ten times, pH 3.0.

44.3 Water and the ionic product

Definition 44.9 (Ionic product of water)

Water reacts very slightly with itself, $\ce{2H2O <=> H3O+ + OH-}$ . The equilibrium constant of this reaction is the ionic product of water:

K_e = \frac{[\ce{H3O+}]\,[\ce{OH-}]}{(c^\circ)^2}, \qquad pK_e = -\log K_e .

Proposition 44.10 (Neutral water and strong bases)

At $25\,{}^{\circ}\mathrm{C}$ , $K_e = 1.0 \times 10^{-14}$ and $pK_e = 14.0$ . In pure water $[\ce{H3O+}] = [\ce{OH-}] = 1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ : pH 7.0. A strong base of concentration $c$ gives $[\ce{OH-}] = c$ , so $[\ce{H3O+}] = K_e (c^\circ)^2 / c$ and $\mathrm{pH} = 14.0 + \log(c/c^\circ)$ .

Proof. In pure water the two ions come in pairs, so they are equal, and their product is $1.0 \times 10^{-14}$ : each is $1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ . For the base, $K_e$ still holds in the solution: $-\log([\ce{H3O+}]/c^\circ) = -\log K_e + \log(c/c^\circ)$ . ∎

Example 44.11 (Sodium hydroxide)

Sodium hydroxide at $0.010\,\mathrm{mol}/\mathrm{L}$ : $[\ce{OH-}] = 0.010$ , so $[\ce{H3O+}] = 1.0 \times 10^{-14}/0.010 = 1.0 \times 10^{-12}\,\mathrm{mol}/\mathrm{L}$ and pH 12.0.

44.4 The acidity constant

Definition 44.12 (Acidity constant, $pK_a$ )

The acidity constant $K_a$ of a couple $\ce{HA}$ / $\ce{A-}$ is the equilibrium constant of the reaction of the acid with water, $\ce{HA + H2O <=> A- + H3O+}$ :

K_a = \frac{[\ce{A-}]\,[\ce{H3O+}]}{[\ce{HA}]\,c^\circ} .

The pKa of the couple is $pK_a = -\log K_a$ .

Proposition 44.13 (The smaller the $pK_a$ , the stronger the acid)

Of two weak acids at the same concentration, the one with the larger $K_a$ , that is the smaller $pK_a$ , reacts more with water and gives the lower pH; its conjugate base is the weaker.

Proof. A larger $K_a$ means more $\ce{A-}$ and $\ce{H3O+}$ at equilibrium for the same $\ce{HA}$ . The base $\ce{A-}$ of a couple that gives up its hydrogen ion easily takes it back reluctantly. ∎

Some couples placed by their pK_a at 25\, C, acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base. — Some couples placed by their $pK_a$ at $25\,{}^{\circ}\mathrm{C}$ , acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base.

Method 44.14 (pH of a weak acid)

For a weak acid of concentration $c$ and constant $K_a$ :

Write the progress table of $\ce{HA + H2O <=> A- + H3O+}$ per litre, with $x = [\ce{H3O+}]$ : $[\ce{HA}] = c - x$ , $[\ce{A-}] = x$ (water’s own ions neglected).
Solve $\dfrac{x^2}{(c - x)\,c^\circ} = K_a$ , the quadratic $x^2 + K_a c^\circ x - K_a c^\circ c = 0$ , keeping the positive root.
$\mathrm{pH} = -\log(x/c^\circ)$ ; check that $x$ is much larger than $10^{-7}\,\mathrm{mol}/\mathrm{L}$ .

Example 44.15 (Ethanoic acid at $0.010\,\mathrm{mol}/\mathrm{L}$ )

$pK_a = 4.76$ , so $K_a = 10^{-4.76} = 1.74 \times 10^{-5}$ . Then $x^2 + 1.74 \times 10^{-5}\,x - 1.74 \times 10^{-7} = 0$ gives $x = 4.1 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ and pH 3.39. Only $4.1 \times 10^{-4}/0.010 = 4.1\,\%$ of the acid has reacted with water; hydrochloric acid at the same concentration, pH 2.0, has reacted entirely.

What becomes of a strong and a weak acid at the same concentration: the strong acid is entirely turned into ions, the weak acid hardly at all. Hence the pH difference, 2.0 against 3.4.

Safety

Concentrated methanoic acid is flammable, causes severe burns of the skin and the eyes, and its vapour is toxic. It is handled only by the teacher, under a fume hood; the dilute solutions of the exercises still need goggles.

44.5 Exercises

Exercise 44.1 ★

Give the conjugate base of: $\ce{HCOOH}$ , $\ce{NH4+}$ , $\ce{H2O}$ , $\ce{HCO3-}$ ; and the conjugate acid of: $\ce{NH3}$ , $\ce{OH-}$ , $\ce{H2O}$ , $\ce{HCO3-}$ . Which species are ampholytes?

Solution

Solution of Exercise 44.1.

Conjugate bases: $\ce{HCOO-}$ , $\ce{NH3}$ , $\ce{OH-}$ , $\ce{CO3^{2-}}$ . Conjugate acids: $\ce{NH4+}$ , $\ce{H2O}$ , $\ce{H3O+}$ , $\ce{CO2,H2O}$ . Ampholytes: $\ce{H2O}$ and $\ce{HCO3-}$ .

Exercise 44.2 ★

Compute the pH of hydrochloric acid at $0.10\,\mathrm{mol}/\mathrm{L}$ , $0.0010\,\mathrm{mol}/\mathrm{L}$ and $5.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ .

Solution

Solution of Exercise 44.2.

pH 1.0; 3.0; $-\log(5.0 \times 10^{-3}) = 2.30$ .

Exercise 44.3 ★

A solution has pH 4.5. Compute $[\ce{H3O+}]$ . A solution has pH 11.2: compute $[\ce{H3O+}]$ and $[\ce{OH-}]$ at $25\,{}^{\circ}\mathrm{C}$ .

Solution

Solution of Exercise 44.3.

$10^{-4.5} = 3.2 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$ . At pH 11.2: $[\ce{H3O+}] = 6.3 \times 10^{-12}\,\mathrm{mol}/\mathrm{L}$ and $[\ce{OH-}] = 1.0 \times 10^{-14}/6.3 \times 10^{-12} = 1.6 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ .

Exercise 44.4 ★

Write the reaction with water and the expression of $K_a$ for methanoic acid and for the ammonium ion.

Solution

Solution of Exercise 44.4.

$\ce{HCOOH + H2O <=> HCOO- + H3O+}$ , $K_a = \dfrac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]\,c^\circ}$ ; $\ce{NH4+ + H2O <=> NH3 + H3O+}$ , $K_a = \dfrac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]\,c^\circ}$ .

Exercise 44.5 ★

Which is the stronger acid: methanoic acid ( $pK_a = 3.74$ ) or ethanoic acid ( $pK_a = 4.76$ )? Which is the stronger base: $\ce{HCOO-}$ or $\ce{CH3COO-}$ ?

Solution

Solution of Exercise 44.5.

Methanoic acid (smaller $pK_a$ ). The stronger base is $\ce{CH3COO-}$ , the base of the weaker acid.

Exercise 44.6 ★★

A solution of a weak acid HA at $0.050\,\mathrm{mol}/\mathrm{L}$ has pH 3.0. Compute $[\ce{A-}]$ , $[\ce{HA}]$ , then $K_a$ and $pK_a$ .

Solution

Solution of Exercise 44.6.

$[\ce{A-}] = [\ce{H3O+}] = 1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ ; $[\ce{HA}] = 0.050 - 0.001 = 0.049\,\mathrm{mol}/\mathrm{L}$ ; $K_a = 1.0 \times 10^{-6}/0.049 = 2.0 \times 10^{-5}$ ; $pK_a = 4.69$ .

Exercise 44.7 ★★

Compute the pH of benzoic acid at $0.020\,\mathrm{mol}/\mathrm{L}$ ( $pK_a = 4.20$ ).

Solution

Solution of Exercise 44.7.

$K_a = 6.31 \times 10^{-5}$ ; $x^2 + 6.31 \times 10^{-5}\,x - 1.26 \times 10^{-6} = 0$ gives $x = 1.09 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ : pH 2.96.

Exercise 44.8 ★★

Compute the pH of a sodium hydroxide solution at $2.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ .

Solution

Solution of Exercise 44.8.

$\mathrm{pH} = 14.0 + \log(2.0 \times 10^{-3}) = 14.0 - 2.70 = 11.3$ .

Exercise 44.9 ★★

Using the $pK_a$ scale, list the acids of the couples shown from the strongest to the weakest. Where would a strong acid such as $\ce{HCl}$ lie?

Solution

Solution of Exercise 44.9.

$\ce{HF}$ , $\ce{HCOOH}$ , ascorbic acid, $\ce{C6H5COOH}$ , $\ce{CH3COOH}$ , $\ce{CO2,H2O}$ , $\ce{NH4+}$ , $\ce{HCO3-}$ . A strong acid lies far below, beyond the end of the scale: it reacts totally with water.

Exercise 44.10 ★★

Carbon dioxide dissolved in water makes it slightly acidic (couple $\ce{CO2,H2O}$ / $\ce{HCO3-}$ , $pK_a = 6.37$ ). A sample of rain water in contact with the air is found to have pH 5.6. Compute $[\ce{H3O+}]$ , and compare with pure water.

Solution

Solution of Exercise 44.10.

$[\ce{H3O+}] = 10^{-5.6} = 2.5 \times 10^{-6}\,\mathrm{mol}/\mathrm{L}$ , 25 times more than in pure water ( $1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ ).

Exercise 44.11 ★★

Show that for the couple $\ce{NH4+}$ / $\ce{NH3}$ , $pK_a = pK_e - pK_b$ , where $K_b = 1.8 \times 10^{-5}$ is the constant of $\ce{NH3 + H2O <=> NH4+ + OH-}$ . Compute the $pK_a$ .

Solution

Solution of Exercise 44.11.

$K_a K_b = \dfrac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]c^\circ} \times \dfrac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]c^\circ} = K_e$ , so $pK_a = pK_e - pK_b = 14.0 - 4.74 = 9.26$ .

Exercise 44.12 ★★★

Ethanoic acid at $0.010\,\mathrm{mol}/\mathrm{L}$ has pH 3.39. Compute the pH after a tenfold dilution. Does it rise by one unit, as for a strong acid? Explain.

Solution

Solution of Exercise 44.12.

At $0.0010\,\mathrm{mol}/\mathrm{L}$ : $x^2 + 1.74 \times 10^{-5}\,x - 1.74 \times 10^{-8} = 0$ , $x = 1.24 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ , pH 3.91: a rise of 0.52, not 1. Diluted, a weak acid reacts in a larger proportion with water (here $12\,\%$ against $4\,\%$ ), which partly makes up for the dilution.

Exercise 44.13 ★★★

The hydrogencarbonate ion is an ampholyte. Write its two reactions with water and their constants, using the $pK_a$ scale. Which is larger? What does that suggest for the pH of a sodium hydrogencarbonate solution?

Solution

Solution of Exercise 44.13.

As an acid: $\ce{HCO3- + H2O <=> CO3^{2-} + H3O+}$ , $K = 10^{-10.33}$ . As a base: $\ce{HCO3- + H2O <=> H2CO3 + OH-}$ ( $\ce{H2CO3}$ standing for $\ce{CO2,H2O}$ ), $K = K_e/K_a = 10^{-14.0+6.37} = 10^{-7.63}$ . The base reaction has the larger constant: the solution is slightly basic.

Exercise 44.14 ★★★

Hydrochloric acid at $1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ is diluted a thousand times, then a thousand times again. A careless rule gives pH 6, then pH 9. Why is pH 9 impossible for a diluted acid? What pH does the very dilute solution approach?

Solution

Solution of Exercise 44.14.

An acid, however dilute, cannot make a solution basic. When the acid’s own oxonium ions become fewer than the $1 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ that water provides, the water’s ions dominate: the pH approaches 7 from below and never passes it.

Exercise 44.15 ★★★

Ascorbic acid (vitamin C, $pK_a = 4.04$ , $M = 176.0\,\mathrm{g}/\mathrm{mol}$ ): a tablet of $500\,\mathrm{mg}$ is dissolved in $200\,\mathrm{mL}$ of water. Compute the pH (consider only the first acidity).

Solution

Solution of Exercise 44.15.

$n = 0.500 / 176.0 = 2.84 \times 10^{-3}\,\mathrm{mol}$ , $c = 0.0142\,\mathrm{mol}/\mathrm{L}$ ; $K_a = 9.12 \times 10^{-5}$ ; $x^2 + 9.12 \times 10^{-5}\,x - 1.30 \times 10^{-6} = 0$ gives $x = 1.09 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ : pH 2.96.

44.6 Problem: The Ant’s Venom

Problem 44.1

Weekend problem — what pH does the methanoic acid of an ant’s venom give at 0.010 mol/L, and how does baking soda soothe its burn?

Methanoic acid, $\ce{HCOOH}$ , is found in the venom of ants. Its $pK_a$ at $25\,{}^{\circ}\mathrm{C}$ is 3.74. We study a solution at $0.010\,\mathrm{mol}/\mathrm{L}$ and compare it with hydrochloric acid.

Part I — The acid.

Draw the Lewis structure of methanoic acid and circle the hydrogen it gives.
Write its couple and its reaction with water.
Write $K_a$ and compute its value.
Place the couple on the $pK_a$ scale: is methanoic acid stronger or weaker than ethanoic acid?
Compute the mass of methanoic acid in $1.0\,\mathrm{L}$ of the solution.

Part II — The pH.

Fill the progress table per litre, with $x = [\ce{H3O+}]$ .
Write the equation $Q_{eq} = K_a$ .
Solve it.
Compute the pH.
What fraction of the acid has reacted with water?
Check that the ions coming from water can be neglected.

Part III — Comparison.

What is the pH of hydrochloric acid at $0.010\,\mathrm{mol}/\mathrm{L}$ ?
How many times more oxonium ions does it contain than the methanoic acid solution?
Explain the difference, using the words strong and weak.

Part IV — Baking soda.

Baking soda is sodium hydrogencarbonate. Write the reaction between methanoic acid and the hydrogencarbonate ion (it gives methanoate and $\ce{CO2,H2O}$ ).
Show that its constant is $K = K_a(\ce{HCOOH}) / K_a(\ce{CO2,H2O})$ , and compute it.
Is the reaction favourable? What gas is seen?
Why is a base much weaker than hydroxide, such as hydrogencarbonate, preferred on the skin?
State the final answer: what is the pH of a $0.010\,\mathrm{mol}/\mathrm{L}$ methanoic acid solution?

Solution

Solution of Problem 44.1.

1. $\ce{H-C(=O)-O-H}$ : the hydrogen given is that of $\ce{O-H}$ .

2. $\ce{HCOOH}$ / $\ce{HCOO-}$ ; $\ce{HCOOH + H2O <=> HCOO- + H3O+}$ .

3. $K_a = \dfrac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]c^\circ} = 10^{-3.74} = 1.82 \times 10^{-4}$ .

4. Below ethanoic acid ( $3.74 < 4.76$ ): stronger.

5. $M = 46.0\,\mathrm{g}/\mathrm{mol}$ : $0.010 \times 46.0 = 0.46\,\mathrm{g}$ .

6. $\ce{HCOOH}$ : $0.010 - x$ ; $\ce{HCOO-}$ : $x$ ; $\ce{H3O+}$ : $x$ .

7. $\dfrac{x^2}{0.010 - x} = 1.82 \times 10^{-4}$ .

8. $x^2 + 1.82 \times 10^{-4}\,x - 1.82 \times 10^{-6} = 0$ : $x = 1.26 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ .

9. $\mathrm{pH} = -\log(1.26 \times 10^{-3}) = 2.90$ .

10. $1.26 \times 10^{-3}/0.010 = 12.6\,\%$ .

11. $1.26 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ is more than ten thousand times $1 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ : water’s own ions are negligible.

12. pH 2.0.

13. $0.010 / 1.26 \times 10^{-3} \approx 8$ times more.

14. Hydrochloric acid is a strong acid, entirely turned into ions; methanoic acid is weak: its reaction with water stops at an equilibrium where most of it remains as $\ce{HCOOH}$ .

15. $\ce{HCOOH + HCO3- -> HCOO- + CO2 + H2O}$ .

16. Multiplying top and bottom of

K = \frac{[\ce{HCOO-}]\,[\ce{CO2}]}{[\ce{HCOOH}]\,[\ce{HCO3-}]}

by $[\ce{H3O+}]$ gives $K = K_a(\ce{HCOOH}) / K_a(\ce{CO2,H2O}) = 1.82 \times 10^{-4}/4.3 \times 10^{-7} \approx 420$ .

17. Yes, $K \gg 1$ ; carbon dioxide bubbles out.

18. Hydroxide is a strong base and would itself attack the skin; hydrogencarbonate is a weak base that neutralises the acid and is mild in any excess.

19. pH 2.9.

Terms defined in this chapter

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