Unit 8: Acids and Bases

What Makes an Acid or Base?

Definitions that AP Chemistry Uses (and Why There Are Several)

Chemists use multiple acid–base definitions because “acid” and “base” describe patterns of behavior, and different situations highlight different patterns.

An Arrhenius acid produces hydronium in water (often written as hydrogen ion). An Arrhenius base produces hydroxide in water. This definition is straightforward, but it is limited to aqueous solutions.

A Brønsted–Lowry acid is a proton donor, and a Brønsted–Lowry base is a proton acceptor. This is the main framework used for Unit 8 equilibrium reasoning because it directly describes proton-transfer equilibria, conjugate pairs, and equilibrium constants.

A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. This is broader (it can describe reactions without protons), but in Unit 8 it is most helpful for recognizing that some bases do not contain hydroxide (for example, ammonia uses a lone pair to accept a proton).

Conjugate Acid–Base Pairs

When a Brønsted–Lowry acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

A key rule: conjugate pairs differ by exactly one proton.

Example reaction in water:

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

• $HA$ is the acid
• $A^-$ is the conjugate base
• $H_2O$ acts as the base
• $H_3O^+$ is the conjugate acid

Acid strength and base strength are linked: the stronger the acid, the weaker its conjugate base (and vice versa). A strong acid must have an extremely weak conjugate base.

Amphiprotic Species

An amphiprotic substance can act as either an acid or a base depending on what it reacts with. Water is the most common example:

• With a strong acid, $H_2O$ acts as a base (accepts a proton).
• With a base, $H_2O$ can act as an acid (donates a proton).

This idea becomes especially important for ions like bicarbonate or dihydrogen phosphate that can go in either direction.

Strong vs Weak: What “Strength” Actually Means

In Unit 8, strength refers to the extent of reaction with water (equilibrium position), not the amount of acid you have.

• A strong acid reacts essentially completely with water to form hydronium.
• A weak acid reacts only partially; a significant amount of molecular acid remains.

A common misconception is confusing strength with concentration. A strong acid can be dilute, and a weak acid can be concentrated.

Worked example: identify conjugate pairs

For:

$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

• Acid: $H_2O$
• Base: $NH_3$
• Conjugate acid: $NH_4^+$
• Conjugate base: $OH^-$

Exam Focus

• Typical question patterns:
  • Identify acids/bases and conjugate pairs in a reaction (often with water participating).
  • Decide whether a species is amphiprotic and predict its behavior in context.
  • Rank acid or base strengths using conjugate relationships.
• Common mistakes:
  • Thinking something is a base only if it contains $OH^-$ (bases like $NH_3$ generate $OH^-$ by reacting with water).
  • Pairing conjugates incorrectly (they must differ by exactly one proton).
  • Confusing “strong” with “concentrated.”

Quantifying Acidity: pH, pOH, pK, and the Ion Product of Water

The Hydronium Ion and What pH Measures

In water, free hydrogen ion is better represented as hydronium, $H_3O^+$. In AP Chemistry, $[H^+]$ is commonly used as shorthand for $[H_3O^+]$.

pH and pOH are logarithmic concentration measures (base-10 logs):

$pH = -\log[H_3O^+]$

$pOH = -\log[OH^-]$

Because the scale is logarithmic, each pH unit corresponds to a factor of 10 change in hydronium concentration. Increasing pH means decreasing $[H^+]$ and making a solution less acidic; decreasing pH means increasing $[H^+]$ and making a solution more acidic.

At $25^\circ C$:

• If $[H^+] = [OH^-]$, the solution is neutral and pH is 7.
• If $[H^+] > [OH^-]$, the solution is acidic and pH is less than 7.
• If $[H^+] < [OH^-]$, the solution is basic and pH is greater than 7.

(Also, pH values can be greater than 14 or less than 0; these are uncommon in typical aqueous lab situations but are possible in very concentrated solutions.)

Autoionization of Water and $K_w$

Pure water autoionizes slightly:

$2H_2O \rightleftharpoons H_3O^+ + OH^-$

The equilibrium constant is:

$K_w = [H_3O^+][OH^-]$

At $25^\circ C$:

$K_w = 1.0 \times 10^{-14}$

This relationship must hold in any aqueous solution: even in acidic or basic solutions, hydronium and hydroxide concentrations are linked by $K_w$.

Relationship Between pH and pOH

Taking negative logs of the autoionization relationship gives:

$pH + pOH = pK_w$

At $25^\circ C$:

$pK_w = 14.00$

So under typical AP conditions at room temperature:

$pH + pOH = 14.00$

Be careful: $K_w$ (and therefore $pK_w$) changes with temperature.

pK Notation (Including $pK_a$ and $pK_b$)

The “p” notation means negative log:

$pK_a = -\log K_a$

A related and often-tested conjugate relationship at $25^\circ C$ is:

$K_a K_b = K_w$

Taking negative logs gives:

$pK_a + pK_b = 14.00$

pH and Solubility: Common Ion Effect with Hydroxides

pH can strongly affect the solubility of hydroxide salts because hydroxide is a common ion. For example:

$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$

• In high pH (significant $OH^-$ already present), the equilibrium shifts left, so **less** $Mg(OH)_2$ dissolves.
• In low pH (abundant $H^+$), added $H^+$ removes $OH^-$ by forming water, decreasing $[OH^-]$ and pulling dissolution to the right, so **more** $Mg(OH)_2$ dissolves.

Notation Reference Table

Worked example: convert between pH and concentrations

If $pH = 3.25$, find $[H_3O^+]$ and $[OH^-]$ at $25^\circ C$.

1) Use the definition:

$[H_3O^+] = 10^{-pH} = 10^{-3.25}$

So $[H_3O^+] \approx 5.6 \times 10^{-4}$.

2) Use $K_w$:

$[OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-4}}$

$[OH^-] \approx 1.8 \times 10^{-11}$

What Goes Wrong: Log and Significant-Figure Pitfalls

• pH is reported with decimal places; the number of decimal places corresponds to the significant figures in $[H_3O^+]$.
• Students sometimes compute pOH from $[H_3O^+]$ directly instead of finding $[OH^-]$ or using $pH + pOH = 14.00$.
• Using natural log instead of base-10 log causes major errors.

Exam Focus

• Typical question patterns:
  • Convert between $[H_3O^+]$, $[OH^-]$, pH, and pOH.
  • Use $K_w$ to link acid and base concentrations.
  • Identify whether a solution is acidic/basic/neutral based on relative $[H^+]$ and $[OH^-]$ (usually at $25^\circ C$).
  • Explain how pH affects hydroxide solubility using the common ion effect.
• Common mistakes:
  • Forgetting the temperature assumption behind $pH + pOH = 14.00$.
  • Treating pH changes as linear rather than logarithmic.
  • Ignoring that $[H^+][OH^-]$ must always match $K_w$ in aqueous solutions.

Strength vs Concentration: Strong Acids/Bases, Neutralization, and Their Calculations

What “Strong” Means Operationally

A strong acid dissociates essentially 100% in water for AP-level problems, so hydronium concentration is determined by stoichiometry rather than an ICE-table equilibrium calculation.

For a monoprotic strong acid $HX$:

$HX \rightarrow H_3O^+ + X^-$

If you start with $C$ mol/L of $HX$, then approximately:

$[H_3O^+] \approx C$

A strong base like $NaOH$ dissociates to give hydroxide directly:

$NaOH \rightarrow Na^+ + OH^-$

So for a strong base of concentration $C$:

$[OH^-] \approx C$

In practice, strong acids/bases are treated as going to completion (their equilibrium constants are effectively enormous), so AP problems typically do not require writing $K_a$ or $K_b$ expressions for them.

Common Strong Acids and Strong Bases (Lists)

Important strong acids:

• HCl, HBr, HI, HNO3, HClO4, H2SO4

Important strong bases:

• LiOH, NaOH, KOH, Ba(OH)2, Sr(OH)2

The Role of Stoichiometry in Mixing Problems

Many AP questions involve mixing acids and bases or adding acid/base to a solution. For strong acids/bases, a reliable workflow is:

1) Convert molarity and volume to moles.
2) Do neutralization stoichiometry.
3) Determine what remains (excess acid/base or neither).
4) Convert leftover moles to concentration using total volume.
5) Compute pH or pOH.

A common net ionic neutralization is:

$H_3O^+ + OH^- \rightarrow 2H_2O$

Neutralization Reaction Patterns (Common AP Classifications)

Neutralization reactions occur when an acid donates a proton to a base. Typical categories include:

• Strong acid + strong base: both dissociate completely; the key reacting ions are hydronium and hydroxide.

Example:

$HCl + NaOH \rightarrow NaCl + H_2O$

• Strong acid + weak base: the strong acid donates a proton to the weak base to form the conjugate acid.

Example:

$HCl + NH_3 \rightarrow NH_4^+ + Cl^-$

• Weak acid + strong base: the strong base removes a proton from the weak acid to form the conjugate base and water.

Example:

$HC_2H_3O_2 + NaOH \rightarrow NaC_2H_3O_2 + H_2O$

• Weak acid + weak base: primarily a proton-transfer equilibrium; extent depends on relative strengths.

Example:

$HC_2H_3O_3 + NH_3 \rightleftharpoons C_2H_3O_3^- + NH_4^+$

Worked example: strong acid–strong base mixing

Mix $25.0 \text{ mL}$ of $0.100 \text{ M}$ $HCl$ with $40.0 \text{ mL}$ of $0.0800 \text{ M}$ $NaOH$. Find the pH.

1) Moles of acid:

$n_{acid} = 0.100 \times 0.0250 = 2.50 \times 10^{-3}$

2) Moles of base:

$n_{base} = 0.0800 \times 0.0400 = 3.20 \times 10^{-3}$

3) Base is in excess. Leftover moles $OH^-$:

$n_{excess} = 3.20 \times 10^{-3} - 2.50 \times 10^{-3} = 7.0 \times 10^{-4}$

4) Total volume:

$V_{tot} = 0.0250 + 0.0400 = 0.0650 \text{ L}$

5) Concentration of leftover $OH^-$:

$[OH^-] = \frac{7.0 \times 10^{-4}}{0.0650} \approx 1.08 \times 10^{-2}$

6) pOH then pH:

$pOH = -\log(1.08 \times 10^{-2}) \approx 1.97$

$pH = 14.00 - 1.97 = 12.03$

Strong Polyprotic Acids (When It Matters)

Some acids can donate more than one proton (polyprotic acids). For stoichiometry, you must account for how many protons are released per formula unit only if the problem treats those dissociations as complete.

For example, if a problem explicitly states complete dissociation of $H_2SO_4$ for both protons, then 1 mole would yield 2 moles of hydronium. However, in many AP contexts, only the first dissociation of sulfuric acid is treated as strong and the second as weak; if a question expects that nuance, it will be framed accordingly.

Exam Focus

• Typical question patterns:
  • Compute pH/pOH from a strong acid/base molarity.
  • Neutralization and mixing problems with total-volume changes.
  • Recognize strong-acid/strong-base vs strong-acid/weak-base reaction products.
• Common mistakes:
  • Using initial molarity after mixing without recalculating using total volume.
  • Forgetting stoichiometric coefficients (especially for bases like $Ba(OH)_2$ or polyprotic acids).
  • Treating a weak acid as if it were strong unless the prompt clearly indicates complete dissociation.

Weak Acid and Weak Base Equilibria: $K_a$, $K_b$, ICE Tables, and Approximations

Why Weak Acids/Bases Require Equilibrium

A weak acid only partially ionizes in water, so you cannot assume hydronium equals the initial acid concentration.

For a weak acid $HA$:

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

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

For a weak base $B$, the base typically does not “dissociate into $OH^-$” by breaking apart; instead it accepts a proton from water, producing hydroxide:

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

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

Larger $K_a$ means a stronger acid (more dissociation). Larger $K_b$ means a stronger base (greater proton acceptance).

Using ICE Tables for Weak Acids

For a weak acid $HA$ with initial concentration $C$:

• Initial: $[HA] = C$, $[H_3O^+] \approx 0$, $[A^-] \approx 0$
• Change: $-x$, $+x$, $+x$
• Equilibrium: $[HA] = C - x$, $[H_3O^+] = x$, $[A^-] = x$

So:

$K_a = \frac{x^2}{C - x}$

If the acid is weak enough, $x$ is small and you can approximate:

$K_a \approx \frac{x^2}{C}$

$x \approx \sqrt{K_a C}$

Here, $x$ is $[H_3O^+]$.

The 5% Approximation Check

A common AP guideline to justify the approximation is:

$\frac{x}{C} \times 100\% < 5\%$

If not satisfied, solve the quadratic.

Worked example: pH of a weak acid

Find the pH of $0.100 \text{ M}$ acetic acid, $HC_2H_3O_2$, given:

$K_a = 1.8 \times 10^{-5}$

Approximate:

$K_a \approx \frac{x^2}{0.100}$

$x \approx \sqrt{(1.8 \times 10^{-5})(0.100)} = \sqrt{1.8 \times 10^{-6}}$

$x \approx 1.34 \times 10^{-3}$

Then:

$pH = -\log(1.34 \times 10^{-3}) \approx 2.87$

Check:

$\frac{1.34 \times 10^{-3}}{0.100} \times 100\% = 1.34\%$

So the approximation is valid.

Weak Bases and Finding pH

For weak bases, you often find hydroxide first, then convert.

Worked example: pH of a weak base

Find pH of $0.200 \text{ M}$ ammonia, $NH_3$, given:

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

Let $x = [OH^-]$:

$K_b \approx \frac{x^2}{0.200}$

$x \approx \sqrt{(1.8 \times 10^{-5})(0.200)} = \sqrt{3.6 \times 10^{-6}}$

$x \approx 1.90 \times 10^{-3}$

Then:

$pOH = -\log(1.90 \times 10^{-3}) \approx 2.72$

$pH = 14.00 - 2.72 = 11.28$

Connecting $K_a$ and $K_b$ for Conjugate Pairs

For a conjugate acid–base pair $HA$ and $A^-$:

$K_a K_b = K_w$

So:

$K_b = \frac{K_w}{K_a}$

and similarly:

$K_a = \frac{K_w}{K_b}$

This connection is central in titrations and salt hydrolysis.

Percent Ionization (Percent Dissociation)

Percent ionization tells you what fraction of a weak acid dissociates:

$\%\text{ ionization} = \frac{[H_3O^+]_{eq}}{[HA]_{initial}} \times 100\%$

For weak acids, percent ionization increases as the solution becomes more dilute. A helpful way to say this conceptually is that with more water present (lower initial acid concentration), there is more opportunity for proton transfer to water, so a larger fraction of the acid is ionized.

Exam Focus

• Typical question patterns:
  • Calculate pH of a weak acid/base using $K_a$ or $K_b$ with an ICE table.
  • Convert between $K_a$ and $K_b$ for conjugate pairs using $K_w$.
  • Compare percent ionization at different initial concentrations.
• Common mistakes:
  • Using $[H_3O^+] = C$ for weak acids.
  • Forgetting to check whether the approximation is valid.
  • Writing incorrect equilibrium expressions (water is omitted from the denominator).

Comparing Acid Strength from Structure

The Big Idea: Acid Strength Depends on Conjugate Base Stability

Whether proton donation is favorable depends heavily on how stable the conjugate base is after losing the proton. A more stable conjugate base corresponds to a stronger acid.

This structural viewpoint is powerful because it lets you justify rankings instead of memorizing lists.

Factors That Stabilize (or Destabilize) the Conjugate Base

1) Electronegativity (Across a Period)

For binary acids $H-X$ across a period, acid strength generally increases as $X$ becomes more electronegative because the conjugate base is stabilized.

A common comparison set is $CH_4$, $NH_3$, $H_2O$, $HF$.

2) Atomic Size and Bond Strength (Down a Group)

Down a group, acid strength of $H-X$ increases because the larger atom stabilizes negative charge better and the $H-X$ bond is weaker.

Example trend: $HI$ is a stronger acid than $HF$. Even though fluorine is highly electronegative, HF is weak primarily because the H–F bond is very strong, so it resists proton loss.

3) Resonance Stabilization

If the conjugate base can delocalize negative charge through resonance, it is stabilized and the acid is stronger.

Carboxylic acids are much stronger than alcohols because carboxylate conjugate bases are resonance-stabilized.

4) Inductive Effects

Electron-withdrawing groups pull electron density through sigma bonds, stabilizing negative charge and strengthening the acid.

Example: $CF_3COOH$ is stronger than $CH_3COOH$.

Oxyacids: The “Number of Oxygens” Pattern

For oxyacids (containing H, O, and a central atom), two common ranking patterns are:

1) Same central atom, different number of oxygens: more oxygens usually means stronger acid because of resonance and inductive stabilization.

2) Same number of oxygens, different central atom electronegativity: more electronegative central atom tends to give a stronger acid.

A useful conceptual example is comparing oxyacids such as $HOF$ and $HOBr$: oxygen sits between H and the halogen, and a more electronegative halogen withdraws electron density through oxygen more strongly, which can make the O–H bond more polar and influence acidity trends. On AP-style questions, the safest justification is still in terms of conjugate base stabilization (induction/resonance) rather than memorizing one-off cases.

Worked example: ranking oxyacids

Rank $HClO$, $HClO_2$, $HClO_3$ by acid strength.

Same central atom (Cl). More oxygens stabilizes the conjugate base more:

$HClO_3 > HClO_2 > HClO$

Which Hydrogens Can Dissociate (Acid Structure and Formula Writing)

Only some hydrogens in a molecular formula are meaningfully acidic. For example, acetic acid is written as $HC_2H_3O_2$ rather than $C_2H_4O_2$ to highlight the acidic hydrogen.

General structural guideline: the acidic hydrogen is typically the one attached to a highly electronegative atom (often O, sometimes N or a halogen in binary acids). Hydrogens bonded to carbon almost never dissociate in typical aqueous acid–base chemistry because C–H bonds are relatively nonpolar (similar electronegativities).

A common practical clue: in many formulas, hydrogens written at the beginning are the acidic ones, while hydrogens embedded in the middle of an organic formula are often not acidic.

What Goes Wrong: “More H Means More Acidic”

Having more hydrogens in the formula does not automatically make something “more acidic.” What matters is conjugate base stability and the relevant dissociation constants. For polyprotic acids, each successive proton usually has a smaller $K_a$ (is less acidic) than the previous one.

Exam Focus

• Typical question patterns:
  • Rank acids by strength using electronegativity, size, resonance, and inductive effects.
  • Explain rankings in terms of conjugate base stability.
  • Compare oxyacids by number of oxygens and/or central-atom electronegativity.
  • Identify which hydrogen(s) in a structure are actually acidic.
• Common mistakes:
  • Ranking only by bond polarity and ignoring conjugate base stabilization.
  • Assuming more hydrogens automatically means stronger acid.
  • Forgetting that HF is weak because the H–F bond is unusually strong.

Polyprotic Acids and Acid–Base Species in Water

Polyprotic Acids Dissociate Stepwise

A polyprotic acid can donate more than one proton, but it does so in steps, each with its own equilibrium constant.

For a diprotic acid $H_2A$:

$H_2A + H_2O \rightleftharpoons H_3O^+ + HA^-$

$HA^- + H_2O \rightleftharpoons H_3O^+ + A^{2-}$

$K_{a1} = \frac{[H_3O^+][HA^-]}{[H_2A]}$

$K_{a2} = \frac{[H_3O^+][A^{2-}]}{[HA^-]}$

Usually:

$K_{a1} > K_{a2}$

After the first proton leaves, the species is negatively charged and holds onto additional protons more strongly.

Examples of Polyprotic Acids and Real $K_a$ Patterns

Common polyprotic acids include $H_2SO_4$ and $H_3PO_4$.

For phosphoric acid, typical values illustrate the stepwise decrease:

$K_{a1} = 7.1 \times 10^{-3}$

$K_{a2} = 6.3 \times 10^{-8}$

$K_{a3} = 4.5 \times 10^{-13}$

This supports the idea that $H_3PO_4$ donates its first proton more readily than species like $H_2PO_4^-$ donate their next proton.

When You Can Ignore Later Dissociations

In many AP pH calculations, $K_{a1}$ dominates hydronium production, and later steps contribute negligibly. You justify this by comparing magnitudes of the dissociation constants and/or checking whether later-produced hydronium is small relative to the first step.

Amphiprotic Intermediate Species

Intermediate ions like $HA^-$ can act as either acids or bases. These amphiprotic species appear frequently in buffers and titration curves.

Worked example: conceptual pH reasoning for a diprotic acid

Suppose $H_2A$ has:

$K_{a1} = 1.0 \times 10^{-3}$

$K_{a2} = 1.0 \times 10^{-8}$

Because $K_{a1}$ is much larger, the first dissociation produces far more hydronium than the second. For typical concentrations, the second step is negligible, so pH is primarily determined by the first equilibrium.

What Goes Wrong: Treating Polyprotic Acids Like Strong Acids

A frequent error is to multiply the acid concentration by the number of hydrogens (for example, assuming $0.10 \text{ M}$ $H_2A$ gives $0.20 \text{ M}$ $H^+$). That only makes sense if all protons dissociate completely, which is not true for weak polyprotic acids and may not even be true when only the first step is strong.

Exam Focus

• Typical question patterns:
  • Write stepwise dissociation reactions and $K_a$ expressions.
  • Decide whether later dissociation steps can be neglected.
  • Identify amphiprotic species and predict their behavior.
• Common mistakes:
  • Adding all possible protons as if they all dissociate completely.
  • Mixing up $K_{a1}$ and $K_{a2}$ expressions.
  • Forgetting that intermediate species can act as both acids and bases.

Acid–Base Properties of Salts (Hydrolysis)

Why Salts Can Make Solutions Acidic or Basic

When an ionic compound dissolves, it produces ions. Some ions are spectators, but others react with water in hydrolysis reactions that produce hydronium or hydroxide.

Core idea: ions that are conjugates of weak acids/bases tend to react with water.

Classifying Ions

Cations

• Cations from strong bases (like $Na^+$, $K^+$) generally do not affect pH.
• Cations that are conjugate acids of weak bases (like $NH_4^+$) can make solution acidic.

Example:

$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$

Anions

• Anions that are conjugate bases of strong acids (like $Cl^-$, $NO_3^-$) generally do not affect pH.
• Anions that are conjugate bases of weak acids (like $F^-$, $C_2H_3O_2^-$) can make solution basic.

Example:

$F^- + H_2O \rightleftharpoons HF + OH^-$

Using $K_a$ and $K_b$ to Quantify Hydrolysis

For the conjugate base $A^-$ of a weak acid $HA$:

$K_b(A^-) = \frac{K_w}{K_a(HA)}$

For the conjugate acid $BH^+$ of a weak base $B$:

$K_a(BH^+) = \frac{K_w}{K_b(B)}$

Worked example: pH of sodium acetate

Find the pH of $0.100 \text{ M}$ $NaC_2H_3O_2$. Given for acetic acid:

$K_a = 1.8 \times 10^{-5}$

1) Basic ion is acetate, $C_2H_3O_2^-$.

2) Convert to $K_b$:

$K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.6 \times 10^{-10}$

3) Let $x = [OH^-]$ with initial $[C_2H_3O_2^-] = 0.100$:

$K_b \approx \frac{x^2}{0.100}$

$x \approx \sqrt{(5.6 \times 10^{-10})(0.100)} = \sqrt{5.6 \times 10^{-11}}$

$x \approx 7.5 \times 10^{-6}$

4) Convert:

$pOH = -\log(7.5 \times 10^{-6}) \approx 5.12$

$pH = 14.00 - 5.12 = 8.88$

Salts of Weak Acid and Weak Base

If both ions hydrolyze (for example, $NH_4F$), predicting pH requires comparing their tendencies using $K_a$ and $K_b$. Conceptually, the ion with the larger equilibrium tendency for reaction with water dominates the pH.

What Goes Wrong: Assuming “Salt = Neutral”

Many salts are neutral (like $NaCl$) because they come from a strong acid and a strong base, not because “salts are neutral.” Always trace ions back to their parent acid/base.

Exam Focus

• Typical question patterns:
  • Predict whether a salt solution is acidic, basic, or neutral.
  • Calculate pH of a salt solution using conjugate relationships.
  • Compare salts by relative acidity/basicity.
• Common mistakes:
  • Treating negative ions like $Cl^-$ as basic just because they are negative.
  • Forgetting to convert between $K_a$ and $K_b$.
  • Using salt concentration directly as $[H_3O^+]$ or $[OH^-]$.

Buffers: How They Work and How to Design Them

What a Buffer Is

A buffer resists changes in pH when small amounts of acid or base are added. Typical AP buffers are made from a weak acid with its conjugate base (often provided as a soluble salt), or a weak base with its conjugate acid.

Example: $NaC_2H_3O_2$ and $HC_2H_3O_2$ together form an acetic acid/acetate buffer because both components are present in significant amounts and do not fully neutralize each other.

Buffers cannot be effectively made from a strong acid and its conjugate base (or a strong base and its conjugate acid) because strong species dissociate essentially completely, leaving no meaningful equilibrium pair to resist pH changes.

How Buffers Resist pH Change (Le Châtelier’s Principle)

In a weak acid buffer with $HA$ and $A^-$:

If acid is added:

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

If base is added:

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

So the added strong acid/base is converted into the weak conjugate partner, limiting pH change. A buffer does not keep pH perfectly constant; it has limits, and once one component is consumed, buffering fails.

Buffer Capacity and Concentration Choice

Buffer capacity increases when the total concentration (total moles) of buffer components increases. Capacity is also best when $[HA]$ and $[A^-]$ are comparable.

Adding water to a buffer generally does not greatly change pH (the ratio can stay similar), but it can reduce buffer capacity because there are fewer moles per liter to neutralize added acid/base.

Henderson–Hasselbalch Equation (Weak Acid Buffers)

For a weak acid buffer:

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

with:

$pK_a = -\log K_a$

When $[A^-] = [HA]$, the log term is zero and:

$pH = pK_a$

Henderson–Hasselbalch works best when both buffer components are present in appreciable amounts.

Henderson–Hasselbalch Form for Weak Base Buffers

For a buffer made from a weak base $B$ and its conjugate acid $BH^+$, an analogous form is:

$pOH = pK_b + \log\left(\frac{[BH^+]}{[B]}\right)$

This is often convenient if $K_b$ is given directly.

Worked example: designing a buffer

You want a buffer of pH $4.50$ using acetic acid with:

$K_a = 1.8 \times 10^{-5}$

1) Compute:

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

2) Use Henderson–Hasselbalch:

$4.50 = 4.74 + \log\left(\frac{[A^-]}{[HA]}\right)$

$\log\left(\frac{[A^-]}{[HA]}\right) = -0.24$

$\frac{[A^-]}{[HA]} = 10^{-0.24} \approx 0.58$

So you need more acetic acid than acetate, with acetate about 0.58 times the acid concentration. In general, when choosing an acid for a buffer, pick one with a $pK_a$ close to the desired pH.

Buffer Calculations by Stoichiometry First

When strong acid or base is added to a buffer, the most reliable method is:

1) Do stoichiometry (neutralization) to update moles of $HA$ and $A^-$ (or $B$ and $BH^+$).
2) Use Henderson–Hasselbalch with the new ratio.

Worked example: adding strong acid to a buffer

A buffer contains $0.200$ mol $HA$ and $0.300$ mol $A^-$ in $1.00 \text{ L}$ with:

$K_a = 1.0 \times 10^{-5}$

Add $0.0500$ mol $HCl$. Find the new pH.

1) Compute:

$pK_a = -\log(1.0 \times 10^{-5}) = 5.00$

2) Stoichiometry: $A^-$ consumes added acid.

New moles:

• $A^-$: $0.300 - 0.0500 = 0.250$
• $HA$: $0.200 + 0.0500 = 0.250$

3) Henderson–Hasselbalch (ratios of moles equal ratios of concentrations in 1.00 L):

$pH = 5.00 + \log\left(\frac{0.250}{0.250}\right) = 5.00$

Exam Focus

• Typical question patterns:
  • Predict qualitatively how pH changes when acid/base is added to a buffer.
  • Use Henderson–Hasselbalch to compute pH or required ratio.
  • “Stoichiometry first, then Henderson–Hasselbalch” after adding strong acid/base.
• Common mistakes:
  • Using Henderson–Hasselbalch when the solution is not actually a buffer.
  • Updating concentrations without tracking moles (especially when volume changes).
  • Forgetting the direction of reaction: adding strong acid consumes $A^-$ and forms $HA$; adding strong base consumes $HA$ and forms $A^-$.

Acid–Base Titrations and Titration Curves

What a Titration Measures

A titration determines an unknown concentration by reacting it with a solution of known concentration (the titrant) while monitoring pH.

Key points on a titration curve:

• Initial pH
• Buffer region (for weak acid/base titrations)
• Half-equivalence point
• Equivalence point
• After equivalence

A common qualitative feature is that pH changes gradually at first and then sharply in the steep region near equivalence.

Strong Acid–Strong Base Titrations

For strong acid–strong base titrations, the equivalence point is approximately neutral at $25^\circ C$ (pH about 7) because the resulting salt does not significantly hydrolyze.

Before equivalence, pH is controlled by excess strong acid; after equivalence, by excess strong base.

Weak Acid–Strong Base Titrations

A very common AP case is titrating weak acid $HA$ with strong base:

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

As base is added:

• $HA$ converts to $A^-$, forming a buffer.
• pH rises gradually in the buffer region.
• At equivalence, the solution contains mostly $A^-$, a weak base, so pH is greater than 7.

The Half-Equivalence Point

At the half-equivalence point:

$[HA] = [A^-]$

So:

$pH = pK_a$

This allows you to read $pK_a$ directly from a titration curve.

Calculating pH at Key Points (Weak Acid Titrated with Strong Base)

Treat each point as a different type of problem:

1) Initial pH: weak acid equilibrium (ICE table with $K_a$).
2) Before equivalence: buffer calculation (stoichiometry to get moles, then Henderson–Hasselbalch).
3) Half-equivalence: $pH = pK_a$.
4) Equivalence: weak base solution of $A^-$ (use $K_b = K_w/K_a$ and ICE).
5) After equivalence: excess strong base stoichiometry.

Worked example: pH at half-equivalence and equivalence

Titrate $50.0 \text{ mL}$ of $0.100 \text{ M}$ weak acid $HA$ with:

$K_a = 1.0 \times 10^{-5}$

using $0.100 \text{ M}$ $NaOH$.

1) Initial moles:

$n_{HA} = 0.100 \times 0.0500 = 5.00 \times 10^{-3}$

2) Equivalence volume:

$V_{eq} = \frac{5.00 \times 10^{-3}}{0.100} = 0.0500 \text{ L} = 50.0 \text{ mL}$

3) Half-equivalence volume: $25.0 \text{ mL}$.

At half-equivalence:

$pK_a = -\log(1.0 \times 10^{-5}) = 5.00$

So:

$pH = 5.00$

At equivalence: total volume is $0.1000 \text{ L}$, and all $HA$ becomes $A^-$.

$[A^-] = \frac{5.00 \times 10^{-3}}{0.1000} = 0.0500$

Compute:

$K_b = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9}$

Let $x = [OH^-]$:

$K_b \approx \frac{x^2}{0.0500}$

$x \approx \sqrt{(1.0 \times 10^{-9})(0.0500)} = \sqrt{5.0 \times 10^{-11}}$

$x \approx 7.1 \times 10^{-6}$

Then:

$pOH = -\log(7.1 \times 10^{-6}) \approx 5.15$

$pH = 14.00 - 5.15 = 8.85$

Weak Base–Strong Acid Titrations (Parallel Logic)

For a weak base titrated with a strong acid, you form a buffer of $B$ and $BH^+$. At equivalence, the solution contains mainly $BH^+$, which hydrolyzes to produce hydronium, so the equivalence point is acidic (pH less than 7).

What Goes Wrong: Using the Wrong Tool at the Wrong Point

Students often use Henderson–Hasselbalch at equivalence (when one buffer component is missing) or use a weak-acid ICE table in the buffer region (where stoichiometry dominates first).

Exam Focus

• Typical question patterns:
  • Compute pH at initial, half-equivalence, equivalence, or after-equivalence points.
  • Identify buffer region and connect half-equivalence pH to $pK_a$.
  • Interpret titration curves (steepness, equivalence-point pH, indicator choice).
• Common mistakes:
  • Forgetting stoichiometry before equilibrium.
  • Assuming equivalence point is always pH 7.
  • Using Henderson–Hasselbalch when one component has been essentially consumed.

Indicators, Endpoints, and Choosing an Indicator

Equivalence Point vs Endpoint

The equivalence point is the theoretical stoichiometric point where moles of acid and base match according to the balanced reaction.

The endpoint is when an indicator changes color (the experimental signal). A good indicator changes color very close to the equivalence point.

How Acid–Base Indicators Work

Indicators are typically weak acids (or weak bases) whose conjugate forms have different colors. For a weak-acid indicator $HIn$:

$HIn + H_2O \rightleftharpoons H_3O^+ + In^-$

As pH changes, the equilibrium shifts, changing the dominant form and therefore the color. The indicator has its own $K_a$ and $pK_a$. The visible color transition usually happens over a range near the indicator’s $pK_a$ (often approximated as about $pK_a \pm 1$). A commonly used idea is that at the midpoint of the color change:

$pH = pK_a$

Choosing an Indicator Based on the Titration Curve

Choose an indicator whose transition range lies within the steep vertical region of the titration curve.

• Strong acid–strong base: steep jump near pH 7, many indicators work.
• Weak acid–strong base: equivalence pH above 7, choose a basic-range indicator.
• Weak base–strong acid: equivalence pH below 7, choose an acidic-range indicator.

Worked example: indicator choice (conceptual)

If a weak acid–strong base titration has an equivalence point around pH 8.8, an indicator changing around pH 8–10 is appropriate. An indicator changing around pH 4–6 would flip too early, giving an endpoint far from equivalence.

What Goes Wrong: Picking Based Only on “Acid vs Base”

It is not enough to know whether the titrant is an acid or base. The correct criterion is the equivalence-point pH and the steep portion of the curve.

Exam Focus

• Typical question patterns:
  • Distinguish equivalence point from endpoint.
  • Choose an indicator given the titration type or curve.
  • Explain indicator behavior using equilibrium shifting and $pK_a$.
• Common mistakes:
  • Assuming all indicators work for all titrations.
  • Choosing an indicator whose transition range is far from equivalence.
  • Confusing endpoint (observed) with equivalence point (stoichiometric).