Titrations, Equilibrium & Solubility

Titrations, Equilibrium & Solubility

Part 1 of 2 — Acid-Base Titrations

Where this fits. The Acids, Bases & Buffers lesson covered pH, $K_a$ and $K_b$, the Henderson–Hasselbalch equation and how buffers resist pH change. This lesson puts those tools to work on titration curves, then widens to equilibrium in general: $K_c$ and $K_p$, Le Chatelier's principle and solubility ($K_{sp}$).

Titration Fundamentals

A titration adds a standard solution (titrant) of known concentration to an unknown solution until the reaction is complete (equivalence point).

$\text{moles acid} = \text{moles base at equivalence point}$

$M_a V_a = M_b V_b \quad (\text{for 1:1 reactions})$

For polyprotic acids, account for the proton-to-hydroxide ratio: $n_a M_a V_a = n_b M_b V_b$

Titration Curve Shapes

Strong Acid + Strong Base

• Before equivalence: pH calculated from excess strong acid
• At equivalence: pH = 7.00 (neutral salt, no hydrolysis)
• After equivalence: pH calculated from excess strong base
• Sharp endpoint: any common indicator works, because the near-vertical jump spans about pH 4–10

Weak Acid + Strong Base

• Initially: pH is higher than for a strong acid of the same concentration (the weak acid only partially dissociates)
• Buffer region: HA and $\text{A}^-$ coexist, so the Henderson–Hasselbalch equation gives the pH: $\text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}])$
• At half-equivalence: $[\text{HA}] = [\text{A}^-]$ → pH = $pK_{a}$ ← MCAT key fact
• At equivalence: pH > 7 (conjugate base hydrolyzes: $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$)
• After equivalence: pH from excess strong base

Weak Base + Strong Acid

• At equivalence: pH < 7 (conjugate acid hydrolyzes)
• At half-equivalence: pOH = pKb, or equivalently pH = 14 − pKb = $pK_{a}$ of conjugate acid

Key Features of Titration Curves

Acid-Base Indicators

Indicators are weak acids (HIn) where HIn and $In^{-}$ are different colors:

$\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-$

Color changes at pH ≈ $pK_{a}$ (indicator). The indicator transitions over ~pH = $pK_{a}$ ± 1.

Common indicators:

Choosing the right indicator: Pick one whose transition range includes the equivalence point pH.

• Strong acid + strong base: any common indicator, because the near-vertical jump spans about pH 4–10
• Weak acid + strong base: equivalence point pH > 7 → use phenolphthalein
• Weak base + strong acid: equivalence point pH < 7 (about 5 for $NH_3$ + HCl) → an indicator changing on the acidic side, such as methyl red (4.4–6.2); methyl orange suits only a very acidic endpoint

Worked Example — Titrating Acetic Acid with NaOH

Problem 1. $50.0$ mL of $0.100$ M acetic acid ($\text{p}K_a = 4.74$) is titrated with $0.100$ M NaOH.

• (a) What volume of NaOH reaches the equivalence point?
• (b) What is the pH after $10.0$ mL of NaOH?
• (c) What is the pH after $25.0$ mL of NaOH?

(a) Equivalence volume. Moles of acid $= (0.0500\text{ L})(0.100\text{ M}) = 5.00$ mmol, so $V_b = 5.00\text{ mmol}/0.100\text{ M} = 50.0$ mL.

(b) Buffer region. $10.0$ mL of NaOH is $1.00$ mmol $\text{OH}^-$, which converts $1.00$ mmol HA to $\text{A}^-$ and leaves $4.00$ mmol HA: $\text{pH} = 4.74 + \log\frac{1.00}{4.00} = 4.74 - 0.60 = 4.14$.

(c) Half-equivalence. $25.0$ mL is half of the $50.0$ mL from (a), leaving $2.50$ mmol each of HA and $\text{A}^-$, so $\text{pH} = \text{p}K_a = 4.74$.

Problem 2. The same titration reaches equivalence: all $5.00$ mmol of acetic acid ($K_a = 1.8 \times 10^{-5}$) is now acetate, in $100.0$ mL of solution.

• (a) What is $[\text{A}^-]$ at the equivalence point?
• (b) What is $[\text{OH}^-]$ there?
• (c) What is the pH at the equivalence point?
• (d) Phenolphthalein (8.2–10.0) or methyl orange (3.1–4.4): which indicator suits this endpoint?

(a) Acetate concentration. $5.00\text{ mmol}/100.0\text{ mL} = 0.0500$ M: the volume doubled, so the concentration halved.

(b) Hydroxide. Acetate is a weak base, so use the weak-base shortcut $[\text{OH}^-] \approx \sqrt{K_b C}$ (valid because so little of the base reacts that $C$ barely changes), with $K_b = K_w/K_a \approx 5.6 \times 10^{-10}$ and $C = 0.0500$ M from (a): $[\text{OH}^-] \approx \sqrt{2.8 \times 10^{-11}} = \sqrt{28 \times 10^{-12}} \approx 5.3 \times 10^{-6}$ M.

(c) pOH, then pH. With $[\text{OH}^-] \approx 5.3 \times 10^{-6}$ M from (b): $\text{pOH} = 6 - \log 5.3 \approx 6 - 0.72 = 5.28$, so $\text{pH} = 14 - \text{pOH} \approx 8.72$.

(d) Indicator. pH 8.72 lies inside phenolphthalein's 8.2–10.0 range; methyl orange (3.1–4.4) would change colour long before equivalence.

Polyprotic Titrations

Titrating $\text{H}_2\text{CO}_3$ or $\text{H}_3\text{PO}_4$ with NaOH gives multiple equivalence points. The number of equivalence points = number of ionizable protons.

For $\text{H}_3\text{PO}_4$ + NaOH:

• 1st equiv. point: $\text{H}_3\text{PO}_4 \to \text{NaH}_2\text{PO}_4$
• 2nd equiv. point: $\text{NaH}_2\text{PO}_4 \to \text{Na}_2\text{HPO}_4$
• 3rd equiv. point: $\text{Na}_2\text{HPO}_4 \to \text{Na}_3\text{PO}_4$

Each stage has its own buffer region. Halfway to the first equivalence point, pH = $\text{p}K_{a1}$; halfway between the first and second, pH = $\text{p}K_{a2}$. At the first equivalence point the solution is the amphiprotic $\text{H}_2\text{PO}_4^-$ (it can both donate and accept a proton), and its pH sits midway between the two: $\text{pH} \approx (\text{p}K_{a1} + \text{p}K_{a2})/2$.

• In practice the third equivalence point of $\text{H}_3\text{PO}_4$ is barely visible in water: $\text{p}K_{a3} \approx 12.3$ is so close to the pH of the NaOH titrant that the curve shows almost no jump there.

Titrations & Indicators 🎯

Key Takeaways — Part 1

• Equivalence: moles match. At equivalence, moles of acid = moles of base; $M_aV_a = M_bV_b$ for a 1:1 reaction.
• Half-equivalence: pH = pKa. Halfway, $[\text{HA}] = [\text{A}^-]$, so pH = $\text{p}K_a$: the most tested titration fact.
• Equivalence pH follows the salt. Strong + strong gives 7.00; a weak acid gives > 7; a weak base gives < 7.
• Indicator brackets equivalence. Pick one whose range (about its $\text{p}K_a \pm 1$) contains the equivalence pH.

Common traps

• Equivalence is always neutral. Only strong + strong lands at 7; acetate's hydrolysis puts a weak acid's above 7.
• Flat region as equivalence. The flat buffer region centres on half-equivalence; equivalence is the steepest point.
• Methyl orange for a weak acid. Its 3.1–4.4 range misses a basic equivalence point; phenolphthalein fits.
• One equivalence point per acid. A polyprotic acid has one equivalence point for each ionizable proton.

Titrations, Equilibrium & Solubility

Part 2 of 2 — Equilibrium: $K_c$, $K_p$, Le Chatelier's Principle & $K_{sp}$

Equilibrium Constant Expressions

For a reaction: $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$

$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$

Rules: only gases and aqueous species appear in the expression; pure solids and liquids are NOT included.

$K > 1$: products favored at equilibrium; $K < 1$: reactants favored.

$K_p$ vs $K_c$

For reactions involving gases: $K_p = K_c (RT)^{\Delta n}$

where $\Delta n = \text{moles gaseous products} - \text{moles gaseous reactants}$.

$K_p = K_c$ whenever $\Delta n = 0$, at any temperature.

Reaction Quotient Q

$Q = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \quad (\text{at any point, not just equilibrium})$

Le Chatelier's Principle

If a system at equilibrium is disturbed, it shifts to partially counteract the disturbance.

Key distinction: Adding or removing species and changing pressure shift the position of equilibrium but do not change $K$. Only temperature changes $K$.

Solubility Product ($K_{sp}$)

For a sparingly soluble salt: $\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)$

$K_{sp} = [\text{M}^+][\text{X}^-]$

For $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$:

$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$

Molar solubility: if $s$ = moles dissolved per liter, then for $\text{CaF}_2$ $[\text{Ca}^{2+}] = s$ and $[\text{F}^-] = 2s$, so $K_{sp} = (s)(2s)^2 = 4s^3$.

Common Ion Effect

Adding a common ion (e.g., adding NaF to a $\text{CaF}_2$ solution) shifts equilibrium left, decreasing solubility.

Precipitation Criterion

• If $Q_{sp} > K_{sp}$: precipitation occurs (solution is supersaturated)
• If $Q_{sp} < K_{sp}$: no precipitate (solution is unsaturated)
• If $Q_{sp} = K_{sp}$: saturated, at equilibrium

Worked Example — Q, Le Chatelier and Molar Solubility

Problem 1. The reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ is endothermic, with $K_c = 5.0 \times 10^{-3}$. A flask holds $0.20$ M $\text{N}_2\text{O}_4$ and $0.020$ M $\text{NO}_2$.

• (a) What is $Q$?
• (b) Which way does the reaction shift to reach equilibrium?
• (c) Once at equilibrium, which way does compressing the flask shift it, and does $K$ change?
• (d) Which way does heating shift it, and does $K$ change?

(a) Reaction quotient. $Q = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.020)^2}{0.20} = \dfrac{4.0 \times 10^{-4}}{0.20} = 2.0 \times 10^{-3}$.

(b) Direction. With $Q = 2.0 \times 10^{-3}$ from (a) below $K = 5.0 \times 10^{-3}$, there are too few products: the reaction shifts right, making $\text{NO}_2$.

(c) Compression. The left side has 1 mole of gas and the right side 2, so compression shifts the equilibrium left, toward fewer gas moles. $K$ is unchanged.

(d) Heating. The forward reaction is endothermic, so heat acts like a reactant: the equilibrium shifts right and $K$ increases. Temperature is the only change here that alters $K$.

Problem 2. $\text{CaF}_2$ has $K_{sp} = 4.0 \times 10^{-11}$, where $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$.

• (a) What is its molar solubility $s$ in pure water?
• (b) What is $s$ in $0.10$ M NaF, and how does it compare?
• (c) Would $1.0 \times 10^{-3}$ M $\text{Ca}^{2+}$ in this $0.10$ M NaF form a precipitate?

(a) In pure water. $[\text{Ca}^{2+}] = s$ and $[\text{F}^-] = 2s$, so $K_{sp} = (s)(2s)^2 = 4s^3 = 4.0 \times 10^{-11}$. Then $s^3 = 1.0 \times 10^{-11} = 10 \times 10^{-12}$, so $s = \sqrt[3]{10} \times 10^{-4} \approx 2.2 \times 10^{-4}$ M.

(b) Common ion. $[\text{F}^-] \approx 0.10$ M, since the $2s$ from $\text{CaF}_2$ is tiny. So $s = (4.0 \times 10^{-11})/(0.10)^2 = 4.0 \times 10^{-9}$ M, against $2.2 \times 10^{-4}$ M from (a): the common ion cuts solubility about 50,000-fold.

(c) Precipitation test. $Q_{sp} = (1.0 \times 10^{-3})(0.10)^2 = 1.0 \times 10^{-5}$, far above $K_{sp} = 4.0 \times 10^{-11}$, so $\text{CaF}_2$ precipitates.

Equilibrium, Le Chatelier & $K_{sp}$ 🎯

Key Takeaways — Part 2

• Compression favours fewer gas moles. Squeezing a gas mixture shifts it toward the side with fewer moles of gas.
• Q moves toward K. $Q < K$ shifts right and $Q > K$ left; for a salt, $Q_{sp} > K_{sp}$ means a precipitate forms.
• Only temperature changes K. Concentration and pressure move the position; heating favors the endothermic side.
• Ksp follows stoichiometry. Raise each ion to its coefficient: $\text{CaF}_2$ gives $K_{sp} = (s)(2s)^2 = 4s^3$.

Common traps

• Catalysts shift equilibrium. A catalyst speeds both directions equally; neither the position nor $K$ changes.
• Inert gas shifts equilibrium. Adding an inert gas at constant volume changes no partial pressure, so nothing shifts.
• Common ion, more dissolving. A shared ion pushes dissolution LEFT, so the salt becomes LESS soluble.
• Kp always equals Kc. Only when $\Delta n = 0$; otherwise $K_p = K_c(RT)^{\Delta n}$.

Titrations, Equilibrium & Solubility — Complete Topic Summary

Part 1: Titration curves, equivalence vs half-equivalence point (pH = pKa), indicators, strong/weak acid titration differences, polyprotic titrations.

Part 2: Kc, Kp, Δn, reaction quotient Q, Le Chatelier's principle (only T changes K), Ksp, molar solubility, common ion effect, precipitation criterion (Q vs Ksp).

Most Tested MCAT Concepts

1. Half-equivalence point: pH = pKa (titrations)
2. Le Chatelier shifts vs. K changes (only T changes K)
3. Common ion effect reduces solubility
4. Weak acid + strong base: pH > 7 at equivalence (the conjugate base hydrolyzes)
5. Ksp → molar solubility (account for stoichiometric coefficients)