Unit 8: Acid-Base Equilibrium

Definitions of acids and bases

Arrhenius Theory

An Arrhenius acid increases the concentration of $\text{H}^+$ (really $\text{H}_3\text{O}^+$ in water) in aqueous solution; an Arrhenius base increases $[\text{OH}^-]$. The model is useful for water-based chemistry but does not describe ammonia as a base in water without extra bookkeeping, and it does not address nonaqueous systems.

Example. Ammonia contains no hydroxide ion in its formula. Can it nevertheless increase aqueous hydroxide concentration? Explain the limitation of identifying bases only by an OH group in their formulas.

Ammonia accepts a proton from water: $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$. It therefore increases hydroxide concentration without dissociating into preexisting hydroxide ions. Inspecting the formula alone misses the reaction with solvent. The Brønsted-Lowry definition describes this proton-transfer behavior directly.

Brønsted–Lowry Theory

A Brønsted–Lowry acid is a proton donor; a Brønsted–Lowry base is a proton acceptor. When an acid $\text{HA}$ donates a proton to water,

$$\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^-(aq),$$

the species $\text{A}^-$ is the conjugate base of $\text{HA}$, and $\text{H}_3\text{O}^+$ is the conjugate acid of $\text{H}_2\text{O}$. Every Brønsted acid has a conjugate base, and every base has a conjugate acid, differing by one $\text{H}^+$ in the formula and one charge unit. For AP purposes, we will generally use this theory.

A key consequence is an inverse strength relationship: the stronger an acid, the weaker its conjugate base, and vice versa. A strong acid like $\text{HCl}$ ionizes almost completely precisely because its conjugate base $\text{Cl}^-$ has essentially no tendency to grab a proton back. A weak acid like $\text{HF}$ ionizes only slightly because its conjugate base $\text{F}^-$ is a reasonably good proton acceptor that pulls the equilibrium back toward the molecular form. This is the qualitative idea behind $K_a\times K_b=K_w$ (derived below).

Example. In $\mathrm{HCO_3^-+H_2O\rightleftharpoons CO_3^{2-}+H_3O^+}$, identify the acid and conjugate base. Can bicarbonate act as a base in a different reaction?

Bicarbonate donates a proton here, so it is the acid and carbonate is its conjugate base. With an acid, bicarbonate can instead accept a proton to form $\mathrm{H_2CO_3}$, making it a base in that reaction. Its negative charge does not fix its role: the actual direction of proton transfer determines the classification.

Lewis Theory

A Lewis acid accepts an electron pair; a Lewis base donates an electron pair. This picture includes reactions without proton transfer (e.g. $\text{BF}_3$ with $\text{NH}_3$) and matches how metal ions bind ligands in Unit 7 complex-ion formation. This is usually not covered on the AP exam.

Example. In $\mathrm{BF_3+NH_3\rightarrow F_3B-NH_3}$, identify the Lewis acid and base. Explain why this reaction shows that Lewis acid-base chemistry is broader than proton transfer.

Ammonia donates nitrogen’s lone pair to form the B-N bond, so it is the Lewis base. Boron trifluoride accepts the pair and is the Lewis acid. No proton changes partners in this reaction. Electron-pair donation and acceptance can form an acid-base adduct even when neither reactant supplies an acidic proton.

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Nomenclature (summary)

Binary acids (hydrogen + one other nonmetal): the anion name ending -ide becomes hydro-…-ic acid (e.g. $\text{HCl}$, hydrochloric acid). Oxyacids use the oxyanion stem: -ate → -ic acid ($\text{NO}_3^-$ → nitric acid), -ite → -ous acid ($\text{NO}_2^-$ → nitrous acid); prefixes such as hypo- and per- carry over.

Ionic hydroxides are named as cation + hydroxide. Molecular bases include ammonia ($\text{NH}_3$), amines (e.g. $\text{CH}_3\text{NH}_2$), and related nitrogen compounds that accept protons in water.

Example. Compare the names and chlorine oxidation numbers in $\mathrm{HClO}$ and $\mathrm{HClO_3}$. Explain why the naming difference is not a statement about how many protons each donates.

These are hypochlorous acid and chloric acid. With H at $+1$ and O at $-2$, chlorine is $+1$ and $+5$ respectively. Both formulas contain one ionizable proton; the oxyanion-derived names distinguish oxygen content, not proton count.

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Strength of acids and structural trends

Strong acids and strong bases are treated as complete ionization or dissociation in dilute aqueous solution for stoichiometry and pH estimates. Weak species reach equilibrium between the unionized form and ions.

The unifying principle behind every acid-strength trend is conjugate-base stability: anything that makes the conjugate base more stable (better able to hold the negative charge after the proton leaves) makes the acid stronger, because it pulls the ionization equilibrium toward products.

For binary acids $\text{HX}$, bond polarity and bond strength both matter: across a period, polarity toward $\text{X}$ can strengthen the acid; down a group, longer/weaker $\text{H–X}$ often dominates and acidity increases ($\text{HF}$ is a weak acid in water; $\text{HCl}$, $\text{HBr}$, $\text{HI}$ are strong). The down-a-group trend wins because the larger halogen forms a longer, weaker bond to hydrogen that breaks more easily, and the resulting larger anion spreads its charge over more volume.

For oxoacids with the same central atom, more electronegative atoms attached to that center or a higher oxidation state (more terminal oxygens) generally strengthens the acid: those extra electronegative oxygens pull electron density away from the O–H bond and spread out the negative charge of the conjugate base. This is why acid strength rises in the series $\text{HClO}<\text{HClO}_2<\text{HClO}_3<\text{HClO}_4$. For carboxylic acids, electron-withdrawing groups (such as the chlorines in chloroacetic acids) stabilize the conjugate base and increase $K_a$, while the resonance delocalization of the carboxylate anion is what makes carboxylic acids more acidic than alcohols in the first place.

Acid-base reactions favor formation of the weaker acid and weaker base. A quick way to predict direction is to compare acid strengths: the side with the larger $K_a$ acid tends to react toward the side with the smaller $K_a$ acid. In $\text{p}K_a$ language, reactions tend to go from lower $\text{p}K_a$ acid to higher $\text{p}K_a$ acid.

Example. Two equal-concentration acids are $\mathrm{CH_3COOH}$ and $\mathrm{ClCH_2COOH}$. Predict which has lower pH using the conjugate bases.

Chlorine withdraws electron density and stabilizes negative charge on the chloroacetate conjugate base. That favors acid ionization, so chloroacetic acid has larger Ka and lower pH at equal concentration. The comparison concerns stability after proton loss, not simply the number of H atoms.

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Strong acids and strong bases

Common strong acids (memorize for AP): $\text{HCl}$, $\text{HBr}$, $\text{HI}$ (hydrohalic acids), $\text{HNO}_3$, $\text{HClO}_4$, $\text{HClO}_3$, and $\text{H}_2\text{SO}_4$ (oxoacids)for the first proton only (the second proton is weak in the dilute-solution sense: $\text{HSO}_4^-$ is a weak acid). A notable exception to hydrohalic trend is that $\text{HF}$ is weak.

Strong bases are the group 1 hydroxides ($\text{LiOH}$, $\text{NaOH}$, $\text{KOH}$, …) and the heavier group 2 hydroxides commonly used in lab ($\text{Ca(OH)}_2$, $\text{Sr(OH)}_2$, $\text{Ba(OH)}_2$). $\text{Mg(OH)}_2$ is only slightly soluble but what dissolves is essentially fully dissociated.

For a strong acid at moderate concentration, $[\text{H}_3\text{O}^+] \approx$ the analytical concentration of the acid (if one proton per formula unit). For a strong diprotic acid such as $\text{H}_2\text{SO}_4$, treat the first step as complete and the second with $K_{a2}$ if the problem requires it.

Example. A student calls $0.0010\ M$ HCl weaker than $0.10\ M$ acetic acid because the HCl is more dilute. Explain the distinction.

Strength describes the extent of ionization; concentration describes amount per volume. HCl is still the strong acid because it ionizes essentially completely. A concentrated weak acid can nevertheless produce more hydronium than a very dilute strong acid, so pH alone cannot label acid strength without concentration information.

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Weak acids: $K_a$ and ICE tables

For a weak monoprotic acid $\text{HA}$,

$$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]},$$

with the usual equilibrium concentrations. The same logic as Unit 7 ICE tables applies: define $x$ as the amount of $\text{HA}$ that ionizes per liter, then solve $K_a = x^2/(C - x)$ (or the quadratic if $x$ is not negligible). When $C \gg K_a$ and $x \ll C$, the approximation $K_a \approx x^2/C$ is common; check with a percent-ionization or “5%” rule if your course uses it.

$$\text{p}K_a = -\log K_a$$

Smaller $\text{p}K_a$ means a stronger acid (larger $K_a$).

Example. A hypothetical acid has $K_a=1.0\times10^{-3}$ and initial concentration $0.010\ M$. Test whether neglecting x is reasonable.

The shortcut gives $x=\sqrt{K_aC}=0.00316\ M$, or $31.6\%$ ionization, so it fails the 5% check. Solve $x^2/(0.010-x)=0.0010$ instead: $x=0.00270\ M$. Keeping the depleted denominator matters when a substantial fraction reacts.

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Weak bases: $K_b$

For a weak base $\text{B}$ (e.g. $\text{NH}_3$),

$$\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq),$$ $$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}, \qquad \text{p}K_b = -\log K_b.$$

ICE setup parallels weak acids, but you solve for $[\text{OH}^-]$ and then find pH from $K_w$ and pOH.

Example. A $0.10\ M$ weak base has $K_b=1.0\times10^{-5}$ at $25^\circ\mathrm{C}$. A student gets pH 3.00 from $-\log\sqrt{K_bC}$. Correct the result.

The square root estimates hydroxide, $[OH^-]=1.0\times10^{-3}\ M$. Its negative logarithm is pOH, not pH. Thus $\mathrm{pH}=14.00-3.00=11.00$. The estimate ionizes only $1\%$ of the base, consistent with neglecting depletion.

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Conjugate $K_a$ and $K_b$; $K_w$

For a conjugate pair $\text{HA}/\text{A}^-$ in water at a given temperature,

$$K_a \times K_b = K_w,$$

where $K_b$ refers to $\text{A}^-$ acting as a base toward water. Similarly $\text{p}K_a + \text{p}K_b = \text{p}K_w$ (at $25\,^\circ\text{C}$, $\text{p}K_w = 14.00$ when $K_w = 1.0 \times 10^{-14}$).

Autoionization of water:

$$2\,\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq), \qquad K_w = [\text{H}_3\text{O}^+][\text{OH}^-].$$

At $25\,^\circ\text{C}$, $K_w = 1.0 \times 10^{-14}$; $K_w$ depends on temperature, so $\text{pH} + \text{pOH} = 14$ is not universal outside standard conditions unless $K_w$ is updated.

Example. At a certain temperature, $K_w=4.0\times10^{-14}$. Find neutral pH and explain why a measured pH of 6.85 is not acidic at this temperature.

Neutrality requires equal hydronium and hydroxide: both are $\sqrt{K_w}=2.0\times10^{-7}\ M$. Neutral pH is $6.70$. At pH 6.85 hydronium is lower than its neutral value, so the solution is basic. The familiar boundary of 7.00 assumes $25^\circ\mathrm{C}$.

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pH and pOH

$$\text{pH} = -\log[\text{H}_3\text{O}^+], \qquad \text{pOH} = -\log[\text{OH}^-], \qquad \text{pH} + \text{pOH} = \text{p}K_w.$$

Neutral water at $25\,^\circ\text{C}$ has $\text{pH} = 7.00$ because $[\text{H}_3\text{O}^+] = [\text{OH}^-]$. $\text{pH} < 7$ is acidic and $\text{pH} > 7$ is basic at that temperature; at other temperatures, neutral pH shifts slightly because $K_w$ changes.

Because pH is logarithmic, a change of $1.00$ pH unit means a tenfold change in $[\text{H}_3\text{O}^+]$. A solution with pH $3$ has $100$ times the hydronium concentration of a solution with pH $5$.

Use inverse logarithms to move back from pH or pOH to concentration:

$$[\text{H}_3\text{O}^+] = 10^{-\text{pH}}, \qquad [\text{OH}^-] = 10^{-\text{pOH}}.$$

Problem-solving strategy

1. Strong acid or strong base only: use stoichiometric dissociation first, then pH/pOH.
2. Weak acid or weak base only: write $K_a$ or $K_b$ and use an ICE table.
3. Mixture with strong acid/base reaction: do limiting-reactant stoichiometry first; then decide what remains.
4. Weak acid + conjugate base or weak base + conjugate acid: use buffer logic.
5. At a titration equivalence point: identify the salt left behind and analyze its hydrolysis.

Example. Equal volumes of strong acid solutions at pH 2.00 and 4.00 are mixed. Find the final pH and explain why averaging pH values fails.

Average concentrations, not logarithms: $[H_3O^+]=(0.0100+0.000100)/2=0.00505\ M$. Therefore pH is $2.30$, not 3.00. The more concentrated acid supplies nearly all the hydronium.

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Percent ionization

Percent ionization (or percent dissociation for a weak acid) is

$$\%\ \text{ionization} = \frac{[\text{H}_3\text{O}^+]_{\text{eq}}}{[\text{HA}]_{\text{initial}}} \times 100\%,$$

using the initial analytical concentration of $\text{HA}$ in the denominator. For a weak base, an analogous expression uses $[\text{OH}^-]_{\text{eq}}/[\text{B}]_{\text{initial}}$. Adding common-ion $\text{A}^-$ or $\text{BH}^+$ suppresses ionization (Le Châtelier’s principle), lowering percent ionization.

Example. A weak acid is diluted by a factor of four while the small-x approximation remains valid. Predict the changes in hydronium concentration and percent ionization.

Since $[H_3O^+]\approx\sqrt{K_aC}$, hydronium halves. But percent ionization is proportional to $\sqrt{K_a/C}$ and doubles. A greater fraction of fewer acid molecules ionizes; higher percent ionization does not mean a higher hydronium concentration.

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Polyprotic acids

A polyprotic acid donates more than one proton. Successive $K_a$ values usually satisfy $K_{a1} > K_{a2} > K_{a3}$ because removing a positive proton from an increasingly negative anion is harder. Many calculations use only $K_{a1}$ if later steps are negligible contributors to $[\text{H}_3\text{O}^+]$; near the second equivalence point in a titration, the second dissociation matters.

Example. A diprotic acid has $K_{a1}=10^{-3}$ and $K_{a2}=10^{-8}$. Explain why treating a $0.10\ M$ solution as providing $0.20\ M$ hydronium fails.

Neither ionization is complete. The first step establishes hydronium, which further suppresses the much weaker second ionization. Two protons per formula unit specify neutralization capacity with sufficient base, not the free hydronium concentration before titration.

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Oxides and acid–base character

Nonmetal oxides tend to be acidic anhydrides (react with water to give acids). Metal oxides, especially ionic ones, tend to be basic anhydrides (give hydroxide or raise pH in water). Amphoteric oxides/hydroxides (e.g. $\text{Al}_2\text{O}_3$, $\text{Al(OH)}_3$) react with both strong acid and strong base.

Example. Equal moles of $\mathrm{Na_2O}$ and $\mathrm{CO_2}$ are separately introduced into water. Predict opposite acid-base effects and support them with reactions.

Sodium oxide gives $\mathrm{Na_2O+H_2O\rightarrow2Na^++2OH^-}$, raising pH. Dissolved carbon dioxide participates in $\mathrm{CO_2+2H_2O\rightleftharpoons H_3O^++HCO_3^-}$, lowering pH. Oxygen in a formula does not by itself establish acid or base behavior.

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Amphoteric species

An amphoteric substance can act as acid or base. Water is the usual example: it donates a proton to $\text{NH}_3$ and accepts one from $\text{HCl}$. Polyprotic anions such as $\text{HCO}_3^-$ and $\text{HSO}_4^-$ can donate or accept a proton depending on what they meet.

Example. Show how bicarbonate can consume either added H+ or added OH-, and identify its role in each reaction.

With acid, $\mathrm{HCO_3^-+H^+\rightarrow H_2CO_3}$, followed by possible carbon dioxide loss; bicarbonate accepts a proton. With base, $\mathrm{HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O}$; bicarbonate donates a proton. It is amphiprotic because it can do both.

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Acid–base properties of salts

Salts dissociate into ions that may hydrolyze (react with water). A salt of strong acid + strong base (e.g. $\text{NaCl}$) gives neutral pH (neglecting tiny temperature effects). Weak acid + strong base (e.g. $\text{CH}_3\text{COONa}$) gives a basic solution because $\text{A}^-$ is a base. Strong acid + weak base (e.g. $\text{NH}_4\text{Cl}$) gives an acidic solution because $\text{NH}_4^+$ is an acid. Weak + weak salts require comparing $K_a$ of the cation acid and $K_b$ of the anion base.

Useful salt classification:

Salt source	pH prediction	Reason
Strong acid + strong base	Neutral	Neither ion hydrolyzes significantly
Weak acid + strong base	Basic	Conjugate base reacts with water to make $\text{OH}^-$
Strong acid + weak base	Acidic	Conjugate acid reacts with water to make $\text{H}_3\text{O}^+$
Weak acid + weak base	Compare $K_a$ and $K_b$	Larger constant dominates

For an anion from a weak acid,

$$\text{A}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HA}(aq) + \text{OH}^-(aq).$$

For a cation from a weak base,

$$\text{BH}^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{B}(aq) + \text{H}_3\text{O}^+(aq).$$

Example. A salt contains a cation with $K_a=10^{-9}$ and an anion with $K_b=10^{-5}$. Predict whether its dilute solution is acidic or basic and explain why ‘salts are neutral’ fails.

Both ions react with water, but the anion’s base reaction is much more favorable. The solution is basic. Electrical neutrality still holds: zero net bulk charge does not require equal hydronium and hydroxide concentrations.

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Buffers

A buffer resists pH change when modest amounts of strong acid or strong base are added. It contains a weak acid and its conjugate base in comparable amounts (or a weak base + conjugate acid).

The reason it works is that a buffer keeps a reservoir of both a proton donor and a proton acceptor on hand. When a small amount of strong acid is added, the conjugate base $\text{A}^-$ neutralizes it (soaking up the added $\text{H}_3\text{O}^+$ to form $\text{HA}$); when a small amount of strong base is added, the weak acid $\text{HA}$ neutralizes it (donating a proton to form $\text{A}^-$). Because the strong acid or base is converted into a weak conjugate rather than left free, the pH barely moves—only the ratio $[\text{A}^-]/[\text{HA}]$ shifts slightly. The Henderson–Hasselbalch equation (same assumptions as the small-change approximation from equilibrium) is

$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right),$$

with concentrations evaluated after any same-volume mixing (or use moles in the ratio if volume is common to both). The equation is most reliable when both species are present and neither concentration is extremely small.

Buffer capacity increases with total concentration of buffer components. When $[\text{HA}] = [\text{A}^-]$, $\text{pH} = \text{p}K_a$ and the system can absorb equal challenge from added acid or base in a symmetric sense (maximum buffering range is often quoted near $\text{p}K_a \pm 1$).

Buffer stoichiometry before equilibrium

When a strong acid or strong base is added to a buffer, do the neutralization reaction before using Henderson-Hasselbalch.

Added strong acid consumes conjugate base:

$$\text{A}^- + \text{H}_3\text{O}^+ \longrightarrow \text{HA} + \text{H}_2\text{O}.$$

Added strong base consumes weak acid:

$$\text{HA} + \text{OH}^- \longrightarrow \text{A}^- + \text{H}_2\text{O}.$$

After the stoichiometry step, use the new moles of $\text{HA}$ and $\text{A}^-$ in the Henderson-Hasselbalch ratio. If either buffer component is used up, the solution is no longer a buffer and the excess strong acid/base controls pH. Note that all pH-pKa pairs can be substituted for pOH-pKb pairs.

Example. Two equal-volume buffers have the same acid/base ratio, but one contains ten times as many moles of each component. Compare initial pH and response to an equal small acid addition.

Henderson-Hasselbalch predicts the same initial pH because the ratios match. The acid addition converts the same number of conjugate-base moles to acid in each buffer, causing a smaller fractional ratio change in the more concentrated buffer. Equal pH does not imply equal capacity.

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Titrations

In a titration, a solution of known concentration (titrant) is added from a buret to the analyte until reaction is complete. For acid–base work, the equivalence point is the stoichiometric point: moles of $\text{H}^+$ supplied equal moles of $\text{OH}^-$ accepted (account for diprotic acids and stoichiometry).

Titration curve shape:

• Strong acid / strong base: equivalence near $\text{pH} = 7$ at $25\,^\circ\text{C}$, steep vertical jump.
• Weak acid / strong base: equivalence $\text{pH} > 7$ (conjugate base hydrolysis).
• Weak base / strong acid: equivalence $\text{pH} < 7$ (conjugate acid).

At the half-equivalence point of a weak acid titrated with strong base, $[\text{HA}] \approx [\text{A}^-]$ and $\text{pH} \approx \text{p}K_a$ (buffer maximum in that sense). Polyprotic acids show multiple equivalence steps and multiple near-plateau regions corresponding to each $\text{p}K_a$.

Titration calculation stages

For a weak acid $\text{HA}$ titrated with strong base:

Region	What controls pH?	Usual method
Before base is added	Weak acid equilibrium	$K_a$ ICE table
Before equivalence	Buffer mixture of $\text{HA}$ and $\text{A}^-$	Stoichiometry, then Henderson-Hasselbalch
Half-equivalence	$[\text{HA}] = [\text{A}^-]$	$\text{pH} = \text{p}K_a$
Equivalence	Conjugate base $\text{A}^-$	$K_b = K_w/K_a$ ICE table
After equivalence	Excess strong base	Stoichiometry for leftover $\text{OH}^-$

For a weak base titrated with strong acid, swap the acid/base roles: the buffer contains $\text{B}$ and $\text{BH}^+$, the half-equivalence point gives $\text{pOH} = \text{p}K_b$ or $\text{pH} = \text{p}K_a$ for $\text{BH}^+$, and the equivalence point is acidic.

If an acid can dissociate more than once, it’s titration curve follows a polyprotic titration curve:

Example. Titrate $25.0\ \mathrm{mL}$ of $0.100\ M$ weak monoprotic acid with $0.100\ M$ NaOH. Why must the pH method change between $12.5$, $25.0$, and $30.0\ \mathrm{mL}$ of added base?

At $12.5\ \mathrm{mL}$, equal amounts of acid and conjugate base form a buffer, so $\mathrm{pH}=\mathrm{p}K_a$. At $25.0\ \mathrm{mL}$, stoichiometric neutralization leaves conjugate base; use its hydrolysis equilibrium. At $30.0\ \mathrm{mL}$, excess hydroxide dominates: $[\mathrm{OH^-}]=(0.00300-0.00250)/0.0550=0.00909\ M$, giving pH about $11.96$ at $25^\circ\mathrm{C}$. An equilibrium expression is chosen only after identifying what remains from neutralization.

pH Indicators

Acid–base indicators are weak acids or bases whose conjugate forms have different colors. The endpoint is where the color change is observed; it should lie near the equivalence point of a titration.

Indicator	Approximate transition range	Acid color	Base color
Methyl orange	$3.1-4.4$	red	yellow
Bromothymol blue	$6.0-7.6$	yellow	blue
Phenolphthalein	$8.2-10.0$	colorless	pink
Universal indicator	broad range	red/orange	green/blue/purple

Choose an indicator whose transition range lies within the steep vertical region of the titration curve. A strong acid-strong base titration has a steep jump around pH $7$, so many indicators can work. A weak acid-strong base titration has an equivalence point above $7$, so phenolphthalein is often better than methyl orange. A weak base-strong acid titration has an equivalence point below $7$, so an acidic-range indicator is usually better.

Example. A weak-acid/strong-base titration has a steep pH jump from about 7 to 10 near equivalence. Indicator X changes color from pH 3 to 4; indicator Y changes from 8 to 9. Which is suitable, and what concentration error would an early endpoint cause?

Y changes within the steep region, so a small added volume carries it through its transition near equivalence. X changes too early, while acid remains unneutralized. Using that too-small base volume as the equivalence volume underestimates the initial acid amount and concentration. The best choice matches the curve’s steep interval, not a rule that every indicator must change at pH 7.

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Common ion effect

The common ion effect is the suppression of ionization of a weak electrolyte when a solution already contains one of its ions (from a salt). It is the same Le Châtelier’s principle logic as in Unit 7: added $\text{A}^-$ shifts $\text{HA}$ ionization left, lowering $[\text{H}_3\text{O}^+]$.

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Working checklist

1. Identify strong vs weak; write the correct net ionic chemistry.
2. Use $K_w$, $K_a$, and $K_b$ at a consistent temperature; link conjugates with $K_a K_b = K_w$.
3. Use ICE tables for weak acids/bases; watch dilution and the common ion effect.
4. Buffers: Henderson–Hasselbalch equation or full equilibrium when assumptions fail.
5. Titrations: stoichiometry first, then equilibrium at the equivalence point or half-equivalence point for $\text{p}K_a$.

Example. Adding sodium acetate to acetic acid raises pH. Does the acid’s Ka decrease? Explain using its equilibrium expression.

Ka stays fixed at a fixed temperature. Added acetate raises the numerator of $Q_a=[H_3O^+][A^-]/[HA]$ before adjustment, so some hydronium and acetate combine to form HA. The resulting lower hydronium concentration restores the same Ka, rather than creating a new constant.

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Reference: common strong acids and bases

Strong acids (typical list)	Strong bases (typical list)
$\text{HCl}$, $\text{HBr}$, $\text{HI}$	$\text{LiOH}$, $\text{NaOH}$, $\text{KOH}$, …
$\text{HNO}_3$, $\text{HClO}_4$, $\text{HClO}_3$	$\text{Ca(OH)}_2$, $\text{Sr(OH)}_2$, $\text{Ba(OH)}_2$
$\text{H}_2\text{SO}_4$ (first $\text{H}^+$ only)

$\text{HF}$ is weak; $\text{HSO}_4^-$ is a weak acid.

Example. A student uses the strong-acid list to assign $[H_3O^+]=2C$ for every sulfuric acid solution. Explain the needed qualification.

The first ionization is treated as complete, but the second is governed by the bisulfate equilibrium. Its contribution depends on concentration and the hydronium already present. Two equivalents of strong base are needed per mole for complete neutralization, but that stoichiometric fact does not mean both ionizations are initially complete.

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Practice

MCQ

1. Mix $20.0\ \mathrm{mL}$ of $0.100\ M$ HCl with $30.0\ \mathrm{mL}$ of $0.100\ M$ NaOH at $25^\circ C$. Find pH, assuming additive volumes.

   (A) $1.70$
   (B) $7.00$
   (C) $12.30$
   (D) $13.00$

2. A buffer initially has $0.100\ \mathrm{mol}$ HA and $0.100\ \mathrm{mol}$ A-. Add $0.0200\ \mathrm{mol}$ HCl with negligible volume change. What is $\mathrm{pH}-\mathrm{p}K_a$ afterward?

   (A) $+0.176$
   (B) $-0.176$
   (C) $0$
   (D) $-0.699$

3. A weak acid is diluted 100-fold while its small-x approximation remains valid. What happens approximately to hydronium concentration and percent ionization?

   (A) Both decrease tenfold
   (B) Hydronium decreases 100-fold and percent is fixed
   (C) Both increase tenfold
   (D) Hydronium decreases tenfold and percent increases tenfold

4. At a temperature where $K_w=1.0\times10^{-12}$, which solution is neutral?

   (A) pH 6.00
   (B) pH 7.00
   (C) pH 12.00
   (D) pH 0.00

5. A $25.0\ \mathrm{mL}$ weak monoprotic acid sample reaches equivalence after $40.0\ \mathrm{mL}$ strong base. At $20.0\ \mathrm{mL}$ base its pH is 5.00. What is Ka?

   (A) $5.0$
   (B) $1.0\times10^{-9}$
   (C) $1.0\times10^{-5}$
   (D) It cannot be inferred because the original acid concentration is unknown

6. A salt contains an acidic cation with $K_a=2.0\times10^{-9}$ and a basic anion with $K_b=5.0\times10^{-6}$. Which prediction is best?

   (A) Acidic because the cation has positive charge
   (B) Basic because anion hydrolysis is stronger
   (C) Neutral because salt has zero net charge
   (D) Neutral because both ions react with water

FRQ

7. A $0.100\ M$ solution of acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$, has $K_a=1.8\times10^{-5}$.

   $(A)$ Write the acid-ionization equation.

   $(B)$ Calculate $[\text{H}_3\text{O}^+]$ using the small-$x$ approximation.

   $(C)$ Calculate the $\text{pH}$.

   $(D)$ Explain what happens to the percent ionization if sodium acetate is added.

   $(E)$ Original extension. A separate sample contains $0.0500\ \text{mol}$ acetic acid. Add $0.0200\ \text{mol}$ NaOH and dilute to $0.500\ \text{L}$. Calculate the pH, identifying the reaction that must be completed before using an equilibrium expression.

8. The 2026 AP Chemistry exam included a nitrous acid titration and indicator question. (Adapted from College Board, 2026 AP Chemistry FRQ 3.)

   $(A)$ Explain why the equivalence point of a weak acid-strong base titration has $\text{pH}>7$.

   $(B)$ A $35.0\ \text{mL}$ sample of $\text{HNO}_2$ is titrated to equivalence with $21.0\ \text{mL}$ of $0.160\ M\ \text{NaOH}$. Calculate the molarity of $\text{HNO}_2$.

   $(C)$ Explain why an indicator should change color near the steep part of the titration curve.

   $(D)$ Original extension. At $25^\circ\text{C}$, take $K_a(\text{HNO}_2)=4.0\times10^{-4}$. Calculate the equivalence-point pH for part B, assuming additive volumes, and check the small-change approximation.