8.4 - Acid-Base Reactions and Buffers

Overview

When you mix an acid with a base, they react first, and only then does an equilibrium decide the pH. This topic is about the four kinds of mixtures (strong or weak acid with strong or weak base), which species are left after the reaction, and how those species decide which pH method to use. It's also where buffers come from: they form when a weak acid is only partly neutralized.

The game plan: react first, then find the pH

Any reaction involving a strong acid or a strong base goes essentially to completion. Chemists say it happens quantitatively. You can see why from the equilibrium constants:

Reaction	$K$
$\mathrm{H_3O^{+}(aq) + OH^{-}(aq) \rightarrow 2H_2O(l)}$	$1/K_w = 1.0 \times 10^{14}$
$\mathrm{CH_3COOH(aq) + OH^{-}(aq) \rightarrow CH_3COO^{-}(aq) + H_2O(l)}$	$K_a/K_w = 1.8 \times 10^{9}$
$\mathrm{NH_3(aq) + H_3O^{+}(aq) \rightarrow NH_4^{+}(aq) + H_2O(l)}$	$1/K_a(\mathrm{NH_4^+}) = 1.8 \times 10^{9}$

With constants that large, whichever reactant runs out first is used up completely. So every mixing problem follows the same steps:

Find the moles of each reactant ( $n = MV$ ). Millimoles are fine: mL times mol/L gives mmol.

Write the net ionic equation and do the stoichiometry in a before-change-after table, in moles. Don't use equilibrium here; the reaction goes to completion.

List what's left (the major species), then pick the method: excess strong acid or base, a buffer, or a weak acid or base alone.

If you need concentrations, divide by the total volume of the mixture.

(If you mix two acids instead, nothing gets neutralized. A strong acid sets the pH, and its

\mathrm{H_3O^{+}}

pushes a weak acid's ionization back even further.)

Strong acid with strong base

A strong acid and a strong base react quantitatively:

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

(You'll also see it written

\mathrm{H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)}

, which means the same thing.) The cation of the base and the anion of the acid, like

\mathrm{Na^{+}}

and

\mathrm{Cl^{-}}

, are spectators that don't change the pH. So the pH depends only on which reactant is in excess: divide the leftover moles of

\mathrm{H_3O^{+}}

\mathrm{OH^{-}}

by the total volume. If the moles are exactly equal, only water and spectator ions remain, and the pH is 7.00 at 25 °C.

Example 1: Mixing HCl with barium hydroxide

A student mixes 30.0 mL of 0.150 M HCl with 20.0 mL of 0.100 M

\mathrm{Ba(OH)_2}

. What is the pH of the mixture at 25 °C?

Step 1: Find millimoles. Remember that

\mathrm{Ba(OH)_2}

gives two

\mathrm{OH^{-}}

(Topic 8.2).

n_{\mathrm{H_3O^+}} = (30.0\ \mathrm{mL})(0.150\ \mathrm{M}) = 4.50\ \mathrm{mmol} \qquad n_{\mathrm{OH^-}} = 2(20.0\ \mathrm{mL})(0.100\ \mathrm{M}) = 4.00\ \mathrm{mmol}

Step 2: React them.

(mmol)	$\mathrm{H_3O^{+}}$	$\mathrm{OH^{-}}$
Before	4.50	4.00
Change	$-4.00$	$-4.00$
After	0.50	0

Step 3: The leftover

\mathrm{H_3O^{+}}

is in 30.0 + 20.0 = 50.0 mL of solution.

[\mathrm{H_3O^+}] = \frac{0.50\ \mathrm{mmol}}{50.0\ \mathrm{mL}} = 0.010\ \mathrm{M} \qquad \mathrm{pH} = -\log(0.010) = 2.00

Answer: pH = 2.00. The subtraction leaves only two significant figures (0.50 mmol), so the pH has two decimal places.

Weak acid with strong base

A strong base takes the proton from a weak acid, and this reaction also goes to completion:

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

What's left depends on which reactant was in excess, and each case uses a different method:

After the reaction	Major species	How to find the pH
Weak acid in excess	HA and $\mathrm{A^{-}}$ : a buffer	Henderson-Hasselbalch: pH = p $K_a$ + log([ $\mathrm{A^{-}}$ ]/[HA])
Equal moles	only $\mathrm{A^{-}}$ (a weak base)	$K_b = K_w/K_a$ , then solve like Topic 8.3 (slightly basic)
Strong base in excess	$\mathrm{OH^{-}}$ and $\mathrm{A^{-}}$	moles of excess $\mathrm{OH^{-}}$ over total volume

In the equal-moles case, the conjugate base reacts with water and makes the solution slightly basic:

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

When strong base is in excess, the solution contains

\mathrm{A^{-}}

too, but the

\mathrm{OH^{-}}

it adds is tiny next to the excess strong base, so you can ignore it.

Buffers and the Henderson-Hasselbalch equation

A buffer is a solution with large amounts of both members of a conjugate acid-base pair, like

\mathrm{CH_3COOH}

and

\mathrm{CH_3COO^{-}}

. Buffers resist changes in pH, because the weak acid can neutralize added base and the conjugate base can neutralize added acid (Topics 8.8 to 8.10 go deeper). You can make one two ways: dissolve the weak acid and a salt of its conjugate base together, or partly neutralize a weak acid with a strong base, which is what happens in this case.

The pH of a buffer comes from the Henderson-Hasselbalch equation (Topic 8.9):

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

Both members of the pair are in the same volume, so you can use their mole ratio in place of the concentration ratio. You don't even need the total volume.

Example 2: Three mixtures of acetic acid and NaOH (AP-style)

A student adds 0.100 M NaOH to 50.0 mL of 0.100 M

\mathrm{CH_3COOH}

(

K_a = 1.8 \times 10^{-5}

) at 25 °C. Identify the major species and calculate the pH after adding (a) 30.0 mL, (b) 50.0 mL, and (c) 60.0 mL of NaOH.

The acid starts with (50.0 mL)(0.100 M) = 5.00 mmol. The NaOH adds 3.00, 5.00, and 6.00 mmol of

\mathrm{OH^{-}}

(mmol)	$\mathrm{CH_3COOH}$	$\mathrm{OH^{-}}$	$\mathrm{CH_3COO^{-}}$
(a) 30.0 mL: before	5.00	3.00	0
(a) after	2.00	0	3.00
(b) 50.0 mL: after	0	0	5.00
(c) 60.0 mL: after	0	1.00	5.00

(a) Acetic acid is in excess, so the mixture is a buffer of

\mathrm{CH_3COOH}

and

\mathrm{CH_3COO^{-}}

\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74 \qquad \mathrm{pH} = 4.74 + \log\frac{3.00}{2.00} = 4.74 + 0.18 = 4.92

There's more

\mathrm{CH_3COO^{-}}

than

\mathrm{CH_3COOH}

, so the pH is a little above the p

K_a

(b) Equal moles: only acetate is left, 5.00 mmol in 50.0 + 50.0 = 100.0 mL, so

[\mathrm{CH_3COO^-}] = 0.0500

M. Acetate is a weak base, with

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

[\mathrm{OH^-}] = \sqrt{(5.6 \times 10^{-10})(0.0500)} = 5.3 \times 10^{-6}\ \mathrm{M} \qquad \mathrm{pOH} = 5.28 \qquad \mathrm{pH} = 14.00 - 5.28 = 8.72

The solution is basic even though the acid and base were mixed in exactly equal amounts.

(c) 1.00 mmol of

\mathrm{OH^{-}}

is left over in 50.0 + 60.0 = 110.0 mL.

[\mathrm{OH^-}] = \frac{1.00\ \mathrm{mmol}}{110.0\ \mathrm{mL}} = 9.09 \times 10^{-3}\ \mathrm{M} \qquad \mathrm{pOH} = 2.041 \qquad \mathrm{pH} = 14.00 - 2.041 = 11.96

Answer: (a) pH 4.92, a buffer of

\mathrm{CH_3COOH}

and

\mathrm{CH_3COO^{-}}

. (b) pH 8.72, mostly

\mathrm{CH_3COO^{-}}

. (c) pH 11.96, excess

\mathrm{OH^{-}}

plus

\mathrm{CH_3COO^{-}}

. Plotting the pH as more base is added gives the curve below. Each part of this example sits in a different region of it (titration curves are Topic 8.5).

Weak base with strong acid

This is the mirror image. A strong acid gives its proton to a weak base, quantitatively:

\mathrm{B(aq) + H_3O^{+}(aq) \rightarrow HB^{+}(aq) + H_2O(l)}

After the reaction	Major species	How to find the pH
Weak base in excess	B and $\mathrm{HB^{+}}$ : a buffer	Henderson-Hasselbalch with p $K_a$ of $\mathrm{HB^{+}}$ : pH = p $K_a$ + log([B]/[ $\mathrm{HB^{+}}$ ])
Equal moles	only $\mathrm{HB^{+}}$ (a weak acid)	$K_a = K_w/K_b$ , then solve like Topic 8.3 (slightly acidic)
Strong acid in excess	$\mathrm{H_3O^{+}}$ and $\mathrm{HB^{+}}$	moles of excess $\mathrm{H_3O^{+}}$ over total volume

In the equal-moles case, the conjugate acid gives a proton to water and makes the solution slightly acidic:

\mathrm{HB^{+}(aq) + H_2O(l) \rightleftharpoons B(aq) + H_3O^{+}(aq)}

Example 3: Ammonia with hydrochloric acid

A student starts with 40.0 mL of 0.200 M

\mathrm{NH_3}

(

K_b = 1.8 \times 10^{-5}

). Find the pH at 25 °C after adding (a) 15.0 mL and (b) 40.0 mL of 0.200 M HCl.

The ammonia is (40.0 mL)(0.200 M) = 8.00 mmol. For the conjugate acid,

K_a(\mathrm{NH_4^+}) = (1.0 \times 10^{-14})/(1.8 \times 10^{-5}) = 5.6 \times 10^{-10}

, and p

K_a

= 14.00

-

4.74 = 9.26.

(a) The HCl adds 3.00 mmol of

\mathrm{H_3O^{+}}

, which turns 3.00 mmol of

\mathrm{NH_3}

into

\mathrm{NH_4^{+}}

. That leaves 5.00 mmol

\mathrm{NH_3}

and 3.00 mmol

\mathrm{NH_4^{+}}

: ammonia is in excess, so it's a buffer.

\mathrm{pH} = 9.26 + \log\frac{5.00}{3.00} = 9.26 + 0.22 = 9.48

(b) The HCl adds 8.00 mmol, exactly enough to convert all the ammonia. Only

\mathrm{NH_4^{+}}

is left: 8.00 mmol in 80.0 mL, or 0.100 M. It's a weak acid.

[\mathrm{H_3O^+}] = \sqrt{(5.6 \times 10^{-10})(0.100)} = 7.48 \times 10^{-6}\ \mathrm{M} \qquad \mathrm{pH} = -\log(7.48 \times 10^{-6}) = 5.13

Answer: (a) pH 9.48 (a buffer of

\mathrm{NH_3}

and

\mathrm{NH_4^{+}}

). (b) pH 5.13. Mixing equal moles of a weak base and a strong acid gives a slightly acidic solution.

Weak acid with weak base

When both partners are weak, the reaction doesn't necessarily go to completion. It reaches an equilibrium:

\mathrm{HA(aq) + B(aq) \rightleftharpoons A^{-}(aq) + HB^{+}(aq)}

Its equilibrium constant compares the two acids on opposite sides, HA and

\mathrm{HB^{+}}

K = \frac{K_a(\mathrm{HA})}{K_a(\mathrm{HB^+})} = \frac{K_a(\mathrm{HA}) \times K_b(\mathrm{B})}{K_w}

If HA is the stronger acid,

K > 1

and products are favored. This is the Topic 4.8 rule in numbers: an acid-base equilibrium favors the side with the weaker acid and weaker base.

Example 4: Which side is favored?

Calculate

K

and decide which side is favored. (a) Acetic acid with ammonia. (b) Acetic acid with fluoride ion.

(a)

\mathrm{CH_3COOH(aq) + NH_3(aq) \rightleftharpoons CH_3COO^{-}(aq) + NH_4^{+}(aq)}

K = \frac{1.8 \times 10^{-5}}{5.6 \times 10^{-10}} = 3.2 \times 10^{4}

K \gg 1

, so products are favored. Acetic acid is a much stronger acid than

\mathrm{NH_4^{+}}

(b)

\mathrm{CH_3COOH(aq) + F^{-}(aq) \rightleftharpoons CH_3COO^{-}(aq) + HF(aq)}

K = \frac{K_a(\mathrm{CH_3COOH})}{K_a(\mathrm{HF})} = \frac{1.8 \times 10^{-5}}{6.3 \times 10^{-4}} = 2.9 \times 10^{-2}

K < 1

, so reactants are favored. The left side has the weaker acid (acetic acid) and the weaker base (

\mathrm{F^{-}}

K_b = 1.6 \times 10^{-11}

), so most of the

\mathrm{F^{-}}

stays unreacted.

Common mistakes

Doing equilibrium before stoichiometry. React the strong acid or base completely first, in moles. Only then think about equilibrium.

Forgetting the total volume. After mixing, divide by the combined volume, not the volume of one solution.

Assuming equal moles means pH 7. That's true only for strong acid with strong base. Weak acid with strong base ends basic; weak base with strong acid ends acidic.

Using Henderson-Hasselbalch in the wrong place. It needs both HA and $\mathrm{A^{-}}$ present in large amounts. Don't use it at the equal-moles point or when strong base is in excess.

Treating weak acid plus strong base as an equilibrium. It goes to completion ( $K \approx 10^9$ ).

Forgetting the 2 in group 2 hydroxides. 1 mol of $\mathrm{Ba(OH)_2}$ neutralizes 2 mol of $\mathrm{H_3O^{+}}$ .

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