8.3 - Weak Acid and Base Equilibria

Overview

Most acids and bases are weak: only a small fraction of their molecules react with water. That makes their solutions equilibrium systems, so Unit 7's tools (equilibrium expressions, ICE tables, and the small-

x

approximation) are exactly what you need here. This topic covers

K_a

and

K_b

, how to get the pH of a weak acid or base, percent ionization, and how an acid's strength connects to its conjugate base's.

Weak acids ionize only a little

A weak acid reacts with water to make hydronium ions, but only a small percentage of its molecules do it at any moment, so the reaction gets an equilibrium arrow:

\mathrm{HA(aq) + H_2O(l) \rightleftharpoons H_3O^{+}(aq) + A^{-}(aq)}

Because so few molecules ionize,

[\mathrm{H_3O^+}]

is much less than the starting concentration of the acid, and the vast majority of the acid stays as intact HA molecules. In a weak acid solution, the major species are HA and

\mathrm{H_2O}

\mathrm{H_3O^{+}}

and

\mathrm{A^{-}}

are minor species, present in equal, small amounts.

Each box started with 5 solute particles (water isn't shown). In the strong acid, every HBr has ionized. In the weak acid, only 1 of 5 HA molecules has given up its proton. In the weak base, only 1 of 5 B molecules has taken a proton from water, making one

\mathrm{HB^{+}}

and one

\mathrm{OH^{-}}

. Real weak acids and bases are usually even less ionized than this (often 1% or less), but then the ions would be too rare to draw.

The acid ionization constant, $K_a$

The equilibrium constant for a weak acid ionizing is called

K_a

, the acid ionization constant. Water is the solvent, so it's left out:

K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \qquad \mathrm{p}K_a = -\log K_a

A smaller

K_a

means a weaker acid: less ionization, fewer ions at equilibrium. Because of the minus sign, a weaker acid has a larger p

K_a

. Values at 25 °C for some common weak acids:

Acid	Formula	$K_a$	p $K_a$
hydrofluoric acid	HF	$6.3 \times 10^{-4}$	3.20
nitrous acid	$\mathrm{HNO_2}$	$4.5 \times 10^{-4}$	3.35
acetic acid	$\mathrm{CH_3COOH}$	$1.8 \times 10^{-5}$	4.74
hypochlorous acid	HOCl	$3.0 \times 10^{-8}$	7.52
hydrocyanic acid	HCN	$4.9 \times 10^{-10}$	9.31

To find the pH of a weak acid from its starting concentration and

K_a

(or p

K_a

; use

K_a = 10^{-\mathrm{p}K_a}

), set up an ICE table, write the

K_a

expression, and solve for

x

[\mathrm{H_3O^+}]

The small- $x$ approximation and the 5% check

Because a weak acid barely ionizes,

x

is usually tiny compared with the starting concentration

C_0

, so

C_0 - x \approx C_0

. That turns the equation into a simple square root:

K_a \approx \frac{x^2}{C_0} \quad \Rightarrow \quad x = [\mathrm{H_3O^+}] \approx \sqrt{K_a C_0}

Always check the approximation afterward: if

x

is less than 5% of

C_0

, it's fine. If not, solve the full quadratic,

x^2 + K_a x - K_a C_0 = 0

. AP questions are usually built so the approximation works, but you're expected to know when it does. A quick test before you start: the approximation works whenever

C_0/K_a

is greater than 400.

Example 1: The pH of vinegar

Vinegar is about 0.83 M acetic acid,

\mathrm{CH_3COOH}

(

K_a = 1.8 \times 10^{-5}

). Find the pH of vinegar and the concentrations of all the species in it at 25 °C.

Step 1: Write the reaction and an ICE table. The

\mathrm{H_3O^{+}}

from water (

10^{-7}

M) is negligible, so start it at 0.

\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons H_3O^{+}(aq) + CH_3COO^{-}(aq)}

	$\mathrm{CH_3COOH}$ (M)	$\mathrm{H_3O^{+}}$ (M)	$\mathrm{CH_3COO^{-}}$ (M)
Initial	0.83	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.83 - x$	$x$	$x$

Step 2: Write

K_a

and use the small-

x

approximation.

1.8 \times 10^{-5} = \frac{x^2}{0.83 - x} \approx \frac{x^2}{0.83} \quad \Rightarrow \quad x = \sqrt{(1.8 \times 10^{-5})(0.83)} = 3.9 \times 10^{-3}\ \mathrm{M}

Step 3: Check the approximation.

\frac{3.9 \times 10^{-3}}{0.83} \times 100\% = 0.47\% \quad (\text{less than } 5\%, \text{ so it's fine})

Step 4: Find the pH and the remaining species.

\mathrm{pH} = -\log(3.9 \times 10^{-3}) = 2.41 \qquad [\mathrm{OH^-}] = \frac{1.0 \times 10^{-14}}{3.9 \times 10^{-3}} = 2.6 \times 10^{-12}\ \mathrm{M}

Answer: pH = 2.41.

[\mathrm{CH_3COOH}] = 0.83

M (essentially unchanged),

[\mathrm{H_3O^+}] = [\mathrm{CH_3COO^-}] = 3.9 \times 10^{-3}

M, and

[\mathrm{OH^-}] = 2.6 \times 10^{-12}

M. Fewer than 1 in 200 acetic acid molecules are ionized. That's what "weak" means.

Example 2: Finding $K_a$ from a measured pH

A student uses a calibrated pH meter to measure a 0.100 M solution of an unknown monoprotic acid at 25 °C and reads pH = 2.19. Calculate

K_a

and identify the acid from the table above.

Step 1: Convert pH to

[\mathrm{H_3O^+}]

. Keep an extra digit until the end.

[\mathrm{H_3O^+}] = 10^{-2.19} = 6.46 \times 10^{-3}\ \mathrm{M}

Step 2: Each

\mathrm{H_3O^{+}}

came with one

\mathrm{A^{-}}

from one HA, so

[\mathrm{A^-}] = 6.46 \times 10^{-3}

M and

[\mathrm{HA}] = 0.100 - 0.00646 = 0.0935

Step 3: Substitute into

K_a

. You already know

x

, so there's no reason to approximate.

K_a = \frac{(6.46 \times 10^{-3})^2}{0.0935} = \frac{4.17 \times 10^{-5}}{0.0935} = 4.5 \times 10^{-4}

Answer:

K_a = 4.5 \times 10^{-4}

, which matches nitrous acid,

\mathrm{HNO_2}

. Here the acid is 6.5% ionized, so dropping the subtraction (dividing by 0.100) would have given

4.2 \times 10^{-4}

, a noticeably different value.

Weak bases and $K_b$

A weak base reacts with water by taking a proton from it, which makes hydroxide ions. Again, only a small percentage of the base molecules react at any moment:

\mathrm{B(aq) + H_2O(l) \rightleftharpoons HB^{+}(aq) + OH^{-}(aq)}

[\mathrm{OH^-}]

is much less than the starting concentration of the base, and most of the base stays as B. The equilibrium constant is

K_b

, the base ionization constant:

K_b = \frac{[\mathrm{OH^-}][\mathrm{HB^+}]}{[\mathrm{B}]} \qquad \mathrm{p}K_b = -\log K_b

The most common weak bases are nitrogen compounds like ammonia (

K_b = 1.8 \times 10^{-5}

) and the amines, plus the conjugate bases of weak acids (like acetate,

\mathrm{CH_3COO^{-}}

). The method is the same as for a weak acid, except that

x

[\mathrm{OH^-}]

, so you find pOH first and then pH = 14.00

-

pOH.

Example 3: The pH of an ammonia solution

Find the pH of 0.20 M

\mathrm{NH_3}

at 25 °C. For ammonia, p

K_b

= 4.74.

Step 1: Convert p

K_b

K_b

K_b = 10^{-4.74} = 1.8 \times 10^{-5}

\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^{+}(aq) + OH^{-}(aq)}

Step 2: With

[\mathrm{NH_4^+}] = [\mathrm{OH^-}] = x

and

[\mathrm{NH_3}] \approx 0.20

1.8 \times 10^{-5} \approx \frac{x^2}{0.20} \quad \Rightarrow \quad x = [\mathrm{OH^-}] = \sqrt{(1.8 \times 10^{-5})(0.20)} = 1.9 \times 10^{-3}\ \mathrm{M}

Step 3: Check:

1.9 \times 10^{-3}/0.20 = 0.95\%

, under 5%. Then convert.

\mathrm{pOH} = -\log(1.9 \times 10^{-3}) = 2.72 \qquad \mathrm{pH} = 14.00 - 2.72 = 11.28

Answer: pH = 11.28. The major species are

\mathrm{NH_3}

and

\mathrm{H_2O}

. The minor species are

\mathrm{NH_4^{+}}

and

\mathrm{OH^{-}}

(

1.9 \times 10^{-3}

M each) and

\mathrm{H_3O^{+}}

(

5.3 \times 10^{-12}

M).

Percent ionization

Percent ionization tells you what fraction of a weak acid (or base) has reacted with water:

\%\ \text{ionization} = \frac{[\mathrm{A^-}]_{eq}}{[\mathrm{HA}]_0} \times 100\% = \frac{[\mathrm{H_3O^+}]_{eq}}{[\mathrm{HA}]_0} \times 100\%

You can get it two ways. From p

K_a

and the initial concentration, solve for

x

as in Example 1. Or, from the initial concentration and the measured equilibrium concentration of any species, as in Example 2:

[\mathrm{H_3O^+}]

[\mathrm{A^-}]

gives

x

directly, and the leftover HA gives

x = [\mathrm{HA}]_0 - [\mathrm{HA}]_{eq}

. For a weak base, use

[\mathrm{OH^-}]

[\mathrm{HB^+}]

over the starting concentration of B. The 5% check from earlier is the same thing as asking whether the percent ionization is under 5%.

Percent ionization depends on concentration: diluting a weak acid increases its percent ionization. Adding water lowers every concentration by the same factor. The top of the

K_a

expression has two concentrations multiplied together, and the bottom has one, so diluting tenfold makes

Q

ten times smaller than

K_a

. To get back to equilibrium, more HA ionizes. (Le Châtelier's principle says the same thing: the side with more dissolved particles is favored on dilution.)

For acetic acid, a 0.10 M solution is 1.3% ionized, but a 0.0010 M solution is 12.5% ionized. Below about 0.0072 M, the small-

x

shortcut gives more than 5%, and the more dilute the solution, the further the red dashed curve drifts above the exact one. Even though a larger fraction ionizes, the diluted solution still has a lower

[\mathrm{H_3O^+}]

, so its pH is higher.

How $K_a$ and $K_b$ are related

Every weak acid HA has a conjugate base

\mathrm{A^{-}}

, and that base reacts with water too. Add the two reactions and you get the autoionization of water:

\mathrm{HA(aq) + H_2O(l) \rightleftharpoons H_3O^{+}(aq) + A^{-}(aq)}

\mathrm{A^{-}(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^{-}(aq)}

\mathrm{2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)}

When you add reactions, you multiply their equilibrium constants (Unit 7). So for any conjugate acid-base pair:

K_w = K_a \times K_b \qquad \mathrm{p}K_w = \mathrm{p}K_a + \mathrm{p}K_b = 14.00 \text{ at } 25\ ^\circ\mathrm{C}

This puts a number on a rule from Topic 4.8: the weaker the acid, the stronger its conjugate base. Since

K_a \times K_b

is fixed, a small

K_a

forces a large

K_b

Acid	$K_a$	Conjugate base	$K_b = K_w/K_a$
HF	$6.3 \times 10^{-4}$	$\mathrm{F^{-}}$	$1.6 \times 10^{-11}$
$\mathrm{HNO_2}$	$4.5 \times 10^{-4}$	$\mathrm{NO_2^{-}}$	$2.2 \times 10^{-11}$
$\mathrm{CH_3COOH}$	$1.8 \times 10^{-5}$	$\mathrm{CH_3COO^{-}}$	$5.6 \times 10^{-10}$
HOCl	$3.0 \times 10^{-8}$	$\mathrm{OCl^{-}}$	$3.3 \times 10^{-7}$
HCN	$4.9 \times 10^{-10}$	$\mathrm{CN^{-}}$	$2.0 \times 10^{-5}$

It works in the other direction too. For ammonia,

K_b = 1.8 \times 10^{-5}

, so its conjugate acid,

\mathrm{NH_4^{+}}

, has

K_a = (1.0 \times 10^{-14})/(1.8 \times 10^{-5}) = 5.6 \times 10^{-10}

. Constants rounded to two significant figures can make the last digit of a p

K

disagree by 1:

-\log(5.6 \times 10^{-10}) = 9.25

, but 14.00

-

4.74 = 9.26. That's rounding, not an error. Carry one extra digit through the steps and both routes give 9.255. These notes use 9.26.

Example 4: A bleach solution (AP-style)

Sodium hypochlorite, NaOCl, is the active ingredient in household bleach. A student prepares 0.20 M NaOCl at 25 °C. For HOCl,

K_a = 3.0 \times 10^{-8}

. (a) Calculate

K_b

for

\mathrm{OCl^{-}}

. Is

\mathrm{OCl^{-}}

a stronger or weaker base than

\mathrm{F^{-}}

? (b) Calculate the pH of the solution. (c) Calculate the percent of

\mathrm{OCl^{-}}

that reacts with water. (d) The student dilutes the solution tenfold, to 0.020 M. Does the percent that reacts increase, decrease, or stay the same? Justify your answer.

(a) Use

K_a \times K_b = K_w

K_b = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.3 \times 10^{-7}

\mathrm{OCl^{-}}

is the stronger base. Its

K_b

is much larger than that of

\mathrm{F^{-}}

(

1.6 \times 10^{-11}

), because HOCl is a much weaker acid than HF.

(b)

\mathrm{Na^{+}}

is a spectator ion. The hypochlorite ion is a weak base:

\mathrm{OCl^{-}(aq) + H_2O(l) \rightleftharpoons HOCl(aq) + OH^{-}(aq)}

3.3 \times 10^{-7} \approx \frac{x^2}{0.20} \quad \Rightarrow \quad x = [\mathrm{OH^-}] = \sqrt{(3.3 \times 10^{-7})(0.20)} = 2.6 \times 10^{-4}\ \mathrm{M}

\mathrm{pOH} = -\log(2.6 \times 10^{-4}) = 3.59 \qquad \mathrm{pH} = 14.00 - 3.59 = 10.41

(c)

\frac{2.6 \times 10^{-4}}{0.20} \times 100\% = 0.13\%

That's well under 5%, so the approximation in (b) was fine.

(d) It increases. Right after dilution, every concentration is one tenth as large, so

Q = \frac{[\mathrm{HOCl}][\mathrm{OH^-}]}{[\mathrm{OCl^-}]}

drops to one tenth of

K_b

. Since

Q < K_b

, the reaction shifts right and a larger fraction of

\mathrm{OCl^{-}}

reacts (0.41% at 0.020 M). The pH still goes down, to 9.91, because

[\mathrm{OH^-}]

is lower overall.

Common mistakes

Treating a weak acid like a strong one. For a weak acid, $[\mathrm{H_3O^+}]$ is much smaller than the acid's concentration. Setting them equal is the most common error in this unit.

Skipping the 5% check. If $x$ is more than 5% of $C_0$ , the approximation isn't valid.

Forgetting to convert for bases. The square root gives $[\mathrm{OH^-}]$ , so it leads to pOH. Finish with pH = 14.00 $-$ pOH.

Mixing up $K$ and p $K$ . A larger $K_a$ means a stronger acid, but a larger p $K_a$ means a weaker one.

Approximating when you don't need to. If you know $[\mathrm{H_3O^+}]$ from a measured pH, subtract it from $C_0$ exactly.

Saying dilution lowers percent ionization. It's the reverse: dilution raises percent ionization, even though $[\mathrm{H_3O^+}]$ falls.

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