AP Chemistry Unit 8 Notes: Acids, Bases, Buffers, pH/pKa, and Titrations

Acid-Base Reactions and Buffers

What counts as an acid-base reaction (and why AP Chem cares)

In AP Chemistry, most acid-base work uses the Brønsted-Lowry definition: an acid is a proton donor (gives H^+) and a **base** is a proton acceptor (takes H^+). This matters because it unifies lots of reactions under one idea: whenever you see H^+ being transferred from one species to another, you’re looking at acid-base chemistry.

A Brønsted-Lowry acid and base always come in conjugate acid-base pairs—two species that differ by exactly one proton. When an acid donates H^+, it becomes its **conjugate base**. When a base accepts H^+, it becomes its conjugate acid. Recognizing these pairs is a core skill because it helps you:

• predict products of acid-base reactions
• decide which direction a reaction favors (reactants vs products)
• connect reaction “strength” to equilibrium constants like K_a and K_b

Example of conjugate pairs:

HF + H_2O \rightleftharpoons F^- + H_3O^+

Here, HF is the acid and F^- is its conjugate base. H_2O is the base and H_3O^+ is its conjugate acid.

A key relationship to remember conceptually: the stronger the acid, the weaker its conjugate base (because strong acids donate protons so completely that their conjugate bases have little tendency to grab them back).

Neutralization: the “reaction engine” behind many problems

A very common acid-base reaction type is neutralization, where an acid reacts with a base to form water and (usually) a salt. In AP Chem, you often write a net ionic equation to focus on what actually changes.

For strong acid–strong base neutralization, the net ionic equation is:

H_3O^+ + OH^- \rightarrow 2H_2O

This is important because it is essentially a “driving force” reaction—it goes to completion. Many titration calculations start by doing mole accounting with this stoichiometry before you ever compute a pH.

Strong vs weak: why equilibrium matters

Not all acids and bases behave the same way in water.

• Strong acids (like HCl, HNO_3) essentially fully ionize in water.
• Weak acids (like HF, CH_3COOH) partially ionize and establish an equilibrium.

For a weak acid HA in water:

HA + H_2O \rightleftharpoons A^- + H_3O^+

The acid dissociation constant is:

K_a = \frac{[H_3O^+][A^-]}{[HA]}

A larger K_a means a stronger weak acid (more products at equilibrium). Similarly, for a weak base B:

B + H_2O \rightleftharpoons BH^+ + OH^-

K_b = \frac{[BH^+][OH^-]}{[B]}

These equilibria are why weak-acid and weak-base problems often involve ICE tables (or approximations) rather than simple stoichiometry alone.

Predicting the direction of an acid-base reaction using K

A powerful AP-style idea: acid-base reactions tend to favor formation of the weaker acid/base pair.

If you have a reaction of the form:

HA + B^- \rightleftharpoons A^- + HB

You can relate its equilibrium constant to K_a values:

K = \frac{K_a(HA)}{K_a(HB)}

Interpretation: if HA is the stronger acid (larger K_a) than HB, then K is greater than 1 and products are favored.

This shows up when you’re asked whether a weak acid can “protonate” the conjugate base of another weak acid, or whether a salt will react with water.

Buffers: what they are and how they resist pH change

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers matter because many real systems (blood, lakes, cells, industrial processes) only function in a narrow pH range—and because AP Chem titration curves and equilibrium problems frequently include buffer regions.

A buffer requires a conjugate acid-base pair present in comparable amounts:

• a weak acid HA and its conjugate base A^-
• or a weak base B and its conjugate acid BH^+

The key mechanism is “consumption” of added strong acid or base:

• If you add H_3O^+ (strong acid), the base component A^- consumes it:

A^- + H_3O^+ \rightarrow HA + H_2O

• If you add OH^- (strong base), the acid component HA consumes it:

HA + OH^- \rightarrow A^- + H_2O

The buffer does not magically prevent pH change; it reduces it by converting strong acid/base into weak acid/base.

Henderson-Hasselbalch equation (what it means, not just how to use it)

For an HA/A^- buffer, the buffer equilibrium is:

K_a = \frac{[H_3O^+][A^-]}{[HA]}

Rearranging and taking negative logs gives the Henderson-Hasselbalch equation:

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

This equation matters because it links pH directly to the ratio of conjugate base to weak acid. Two big conceptual takeaways:

1. When [A^-] = [HA], the log term is 0, so pH = pK_a. This “equal concentrations” point is extremely important in titrations (it happens at the half-equivalence point for a weak acid titrated with strong base).
2. pH changes slowly when the ratio [A^-]/[HA] changes slowly—which is why buffers are most effective when you have significant amounts of both components.

Important caution: in many buffer calculations (especially after adding strong acid/base), you should use moles in the ratio rather than concentrations, because dilution often cancels if both are in the same total volume. The ratio \frac{[A^-]}{[HA]} equals \frac{n(A^-)}{n(HA)} as long as both are in the same solution volume.

Buffer capacity and when buffers “fail”

Buffer capacity is how much acid or base a buffer can absorb before its pH changes significantly. Capacity increases when you have larger total amounts of HA and A^- (higher concentrations).

Buffers fail in two common ways:

• You add enough strong acid to consume nearly all of A^-, leaving mostly HA (the buffer can no longer neutralize added acid).
• You add enough strong base to consume nearly all of HA, leaving mostly A^- (the buffer can no longer neutralize added base).

A practical rule: buffers work best when [A^-] and [HA] are within about a factor of 10, because then the log term stays between -1 and +1.

Worked example: buffer pH and response to added acid

You prepare a buffer with 0.200\text{ mol} CH_3COOH (acetic acid, HA) and 0.150\text{ mol} CH_3COO^- (acetate, A^-) in 1.00\text{ L}. Given K_a = 1.8\times 10^{-5}, find the pH. Then you add 0.0100\text{ mol} of HCl. Find the new pH.

First compute pK_a:

pK_a = -\log(K_a)

pK_a = -\log(1.8\times 10^{-5})

pK_a \approx 4.74

Initial pH using Henderson-Hasselbalch:

pH = 4.74 + \log\left(\frac{0.150}{0.200}\right)

pH = 4.74 + \log(0.750)

pH \approx 4.74 - 0.125 = 4.62

Now add strong acid: H_3O^+ reacts with A^- to form HA. Do stoichiometry in moles:

• n(A^-) decreases by 0.0100
• n(HA) increases by 0.0100

New moles:

• n(A^-) = 0.150 - 0.0100 = 0.140
• n(HA) = 0.200 + 0.0100 = 0.210

New pH:

pH = 4.74 + \log\left(\frac{0.140}{0.210}\right)

pH = 4.74 + \log(0.667)

pH \approx 4.74 - 0.176 = 4.56

Notice the point: adding 0.0100\text{ mol} of strong acid to a 1 L solution would drastically lower pH if unbuffered, but here pH changed only about 0.06.

Common misconceptions to avoid (built into buffer thinking)

• “A buffer has a pH of 7.” False—buffers can be acidic or basic. Buffer means resists change, not “neutral.”
• “Adding water changes buffer pH a lot.” Dilution decreases buffer capacity, but the ratio [A^-]/[HA] can stay the same, so pH may change only slightly.
• “Henderson-Hasselbalch always works.” It works best when you truly have a buffer (both components present in significant amount) and when activities can be approximated by concentrations (typical AP setting).

Exam Focus

• Typical question patterns:
  • Given a conjugate pair and K_a or pK_a, calculate buffer pH before and after adding a small amount of strong acid/base.
  • Identify which mixtures form a buffer (weak acid plus its salt, weak base plus its salt) and explain why.
  • Compare buffer capacity or effectiveness for different concentrations/ratios.
• Common mistakes:
  • Forgetting to do the reaction stoichiometry first when strong acid/base is added (you must update moles of HA and A^- before using Henderson-Hasselbalch).
  • Using initial concentrations instead of moles after mixing solutions of different volumes.
  • Treating a solution with only a weak acid (or only its conjugate base) as a buffer.

pH and pKa

pH: what it measures and why it’s logarithmic

pH is a measure of the hydronium ion concentration in solution. It’s defined as:

pH = -\log([H_3O^+])

The log scale matters because [H_3O^+] can vary over many powers of ten. A change of 1 pH unit corresponds to a factor of 10 change in [H_3O^+].

Because pH is logarithmic, “small” pH differences can reflect big chemical differences. For example, pH 3 has ten times more H_3O^+ than pH 4.

Similarly:

pOH = -\log([OH^-])

At 25^\circ\text{C}, water’s ion-product constant is:

K_w = [H_3O^+][OH^-] = 1.0\times 10^{-14}

Taking negative logs gives:

pK_w = 14.00

and therefore:

pH + pOH = 14.00

This relationship is used constantly to move between acidity (hydronium) and basicity (hydroxide).

pKa: translating acid strength into a more usable number

For a weak acid, K_a can be a very small number, which is hard to compare quickly. So we define:

pK_a = -\log(K_a)

Interpretation:

• Smaller pK_a means larger K_a, which means a stronger acid.
• Larger pK_a means a weaker acid.

You use pK_a constantly in buffers and titrations because it connects directly to pH via Henderson-Hasselbalch and because the half-equivalence point condition pH = pK_a is a major shortcut.

Relationship between conjugates: connecting K_a and K_b

If HA is an acid and A^- is its conjugate base, then the base reaction is:

A^- + H_2O \rightleftharpoons HA + OH^-

The equilibrium constant for that base reaction is K_b. The key relationship (at a given temperature, typically 25^\circ\text{C}) is:

K_a K_b = K_w

This matters because it lets you compute basicity from acidity (and vice versa). In pKa terms, taking negative logs yields:

pK_a + pK_b = pK_w

At 25^\circ\text{C}, this becomes:

pK_a + pK_b = 14.00

How to choose an approach for pH problems

AP Chemistry pH problems typically fall into a few “structures,” and you choose the math based on what controls [H_3O^+].

1. Strong acid/strong base solutions: assume complete dissociation; pH comes from direct concentration (after dilution if mixed).
2. Weak acid/weak base solutions: use equilibrium with K_a or K_b (often ICE tables; sometimes the “x is small” approximation).
3. Salts that hydrolyze: if a salt contains a conjugate base of a weak acid (like F^-) it makes solution basic; if it contains a conjugate acid of a weak base (like NH_4^+) it makes solution acidic.
4. Buffers: use Henderson-Hasselbalch (after stoichiometry if acid/base is added).

Worked example: pH of a weak acid (equilibrium approach)

Find the pH of 0.100\text{ M} HF. Given K_a(HF)=6.8\times 10^{-4}.

Start with the equilibrium:

HF + H_2O \rightleftharpoons F^- + H_3O^+

Set up an ICE-style variable x for how much dissociates:

• Initial: [HF]=0.100, [F^-]=0, [H_3O^+]=0
• Change: -x, +x, +x
• Equilibrium: [HF]=0.100-x, [F^-]=x, [H_3O^+]=x

Plug into K_a:

6.8\times 10^{-4} = \frac{x^2}{0.100-x}

For many AP problems, you test the approximation 0.100-x \approx 0.100 if dissociation is small:

6.8\times 10^{-4} \approx \frac{x^2}{0.100}

x^2 \approx 6.8\times 10^{-5}

x \approx 8.25\times 10^{-3}

So [H_3O^+] \approx 8.25\times 10^{-3} and:

pH = -\log(8.25\times 10^{-3})

pH \approx 2.08

Sanity check: x is about 8% of 0.100 M, which is borderline but often acceptable at AP level; if required, you can solve the quadratic to refine.

Notation reference (common equivalences)

Exam Focus

• Typical question patterns:
  • Compute pH from [H_3O^+] or [OH^-] and convert between pH and pOH using K_w.
  • Given K_a or pK_a, rank acids by strength or predict relative positions on a titration curve (buffer region, half-equivalence).
  • Find pH of weak acids/bases or salts via equilibrium expressions.
• Common mistakes:
  • Taking the log of the wrong quantity (using [H^+] without recognizing AP typically uses [H_3O^+]; numerically same, but conceptually hydronium in water).
  • Mixing up the direction: larger K_a means stronger acid, but larger pK_a means weaker acid.
  • Forgetting temperature context: pH+pOH=14.00 assumes 25^\circ\text{C}.

Acid-Base Titrations

What a titration is actually doing

An acid-base titration is a lab method where you gradually add a solution of known concentration (the titrant) to a solution of unknown concentration (the analyte) while tracking pH. Conceptually, a titration is a controlled neutralization reaction combined with equilibrium chemistry.

The titration curve (pH vs volume added) tells you:

• where neutralization is complete (the equivalence point)
• how the pH changes as the relative amounts of acid and base change
• for weak acids/bases, information related to pK_a or pK_b

A crucial distinction:

• Equivalence point: moles acid and base have reacted in the exact stoichiometric ratio.
• Endpoint: the indicator’s color change (an experimental signal that should be close to equivalence).

The consistent strategy for titration pH calculations

No matter the titration type, the most reliable approach is two-stage thinking:

1. Stoichiometry stage (reaction goes to completion): Use moles and the neutralization reaction to find what remains after the strong reaction finishes.
2. Equilibrium stage (what sets pH now?): Decide whether pH is controlled by leftover strong acid/base, a buffer mixture, or hydrolysis of a conjugate species.

Skipping stage 1 is the most common reason students get titration pH problems wrong.

Strong acid–strong base titration (shape and key features)

In a strong acid–strong base titration (like HCl with NaOH), both dissociate completely.

• Before equivalence: excess strong acid controls pH.
• At equivalence: solution is approximately neutral at 25^\circ\text{C} because the salt (like NaCl) does not hydrolyze.
• After equivalence: excess strong base controls pH.

The curve has a very steep vertical region near equivalence, which is why many indicators work well here.

Weak acid–strong base titration (buffer region and half-equivalence)

This is one of the most tested AP scenarios. Suppose you titrate weak acid HA with strong base OH^-.

Key regions:

1. Initial pH (before any base): determined by weak-acid equilibrium (use K_a).
2. Buffer region (before equivalence): mixture of HA and A^- forms a buffer; use Henderson-Hasselbalch after doing stoichiometry to find moles of HA and A^-.
3. Half-equivalence point: exactly half the initial acid has been converted to A^-, so [A^-]=[HA] and:

pH = pK_a

4. Equivalence point: all HA is converted to A^-. The pH is **basic** because A^- hydrolyzes water to make OH^-.
5. After equivalence: excess strong base dominates pH.

This titration curve has an equivalence point above 7 (at 25^\circ\text{C}) because the conjugate base makes the solution basic.

Choosing an indicator (connecting chemistry to the lab)

Indicators are weak acids/bases whose conjugate forms have different colors. An indicator changes color around its transition range (roughly pK_a \pm 1 for many indicators).

Good indicator choice: pick one whose transition range falls within the steep part of the titration curve near the equivalence point.

• Strong acid–strong base: many indicators work.
• Weak acid–strong base: choose an indicator that changes color above 7.
• Strong acid–weak base: choose an indicator that changes color below 7.
• Weak acid–weak base: the pH change near equivalence may be too gradual for a sharp endpoint (often not ideal with simple indicators).

Worked example 1: strong acid–strong base pH during titration

You titrate 25.00\text{ mL} of 0.100\text{ M} HCl with 0.100\text{ M} NaOH. Find the pH after adding 10.00\text{ mL} of base.

Stage 1 (stoichiometry):

Moles HCl initially:

n(HCl)=0.02500\times 0.100=2.50\times 10^{-3}

Moles OH^- added:

n(OH^-)=0.01000\times 0.100=1.00\times 10^{-3}

Neutralization consumes equal moles, so excess acid remains:

n(H_3O^+\text{ remaining})=2.50\times 10^{-3}-1.00\times 10^{-3}=1.50\times 10^{-3}

Total volume:

V_{tot}=0.02500+0.01000=0.03500\text{ L}

Hydronium concentration (strong acid, so direct):

[H_3O^+]=\frac{1.50\times 10^{-3}}{0.03500}=4.29\times 10^{-2}

Stage 2 (pH):

pH=-\log(4.29\times 10^{-2})

pH\approx 1.37

Worked example 2: weak acid–strong base at half-equivalence and equivalence

You titrate 50.00\text{ mL} of 0.100\text{ M} CH_3COOH with 0.100\text{ M} NaOH. Given K_a=1.8\times 10^{-5}, find (a) the pH at the half-equivalence point and (b) the pH at equivalence.

First compute initial moles of acid:

n(HA)=0.05000\times 0.100=5.00\times 10^{-3}

Equivalence requires the same moles of OH^-, so volume of base at equivalence:

V_{eq}=\frac{5.00\times 10^{-3}}{0.100}=0.05000\text{ L}

So half-equivalence occurs at 25.00\text{ mL} added.

(a) Half-equivalence pH

At half-equivalence, [A^-]=[HA], so:

pH=pK_a

Compute pK_a:

pK_a=-\log(1.8\times 10^{-5})\approx 4.74

So pH\approx 4.74.

(b) Equivalence-point pH

At equivalence, all HA has become A^-. Now pH comes from hydrolysis:

A^- + H_2O \rightleftharpoons HA + OH^-

We need K_b for A^-:

K_b=\frac{K_w}{K_a}

K_b=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}

Find [A^-] at equivalence. Moles A^- equals initial moles acid, 5.00\times 10^{-3}. Total volume at equivalence is 0.05000+0.05000=0.10000\text{ L}.

[A^-]=\frac{5.00\times 10^{-3}}{0.10000}=0.0500

Let x=[OH^-] produced by hydrolysis. Then:

K_b=\frac{x^2}{0.0500-x}

Assume 0.0500-x\approx 0.0500:

5.56\times 10^{-10}\approx \frac{x^2}{0.0500}

x^2\approx 2.78\times 10^{-11}

x\approx 5.27\times 10^{-6}

So pOH=-\log(5.27\times 10^{-6})\approx 5.28 and:

pH=14.00-5.28=8.72

This confirms the key idea: equivalence point is basic for a weak acid–strong base titration.

Reading titration curves conceptually (what AP questions often probe)

Titration curve questions aren’t always computational. You may be asked to label regions or justify trends:

• The buffer region is where both HA and A^- are present in significant amounts, so pH changes gradually.
• The steep rise near equivalence occurs because the limiting reactant switches; small volume additions cause big changes in excess H_3O^+ or OH^-.
• The half-equivalence point is identifiable because it ties directly to pK_a (a common way to experimentally determine pK_a).

Exam Focus

• Typical question patterns:
  • Calculate pH at specific volumes in a titration (initial, before equivalence, half-equivalence, equivalence, after equivalence) using the two-stage stoichiometry-then-equilibrium method.
  • Interpret or sketch titration curves, identifying buffer region, equivalence point pH (above/below 7), and best indicator range.
  • Use titration data to determine unknown concentration, or use the half-equivalence pH to determine pK_a.
• Common mistakes:
  • Treating the equivalence point of a weak acid–strong base titration as pH 7 (it is basic due to conjugate-base hydrolysis).
  • Using Henderson-Hasselbalch at equivalence (there is no HA left, so it’s not a buffer).
  • Forgetting dilution when converting remaining moles into concentrations after adding titrant.