Reactivity 3. What are the mechanisms of chemical change?

Pair species that differ by exactly one H⁺ to identify conjugate acid–base pairs. Charge alone does not decide the role: in NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, NH₄⁺ is the proton donor and water is the acceptor.

NH₄⁺/NH₃ and H₂CO₃/HCO₃⁻ are conjugate pairs because each pair differs by one H⁺. Removing H⁺ lowers charge by one; adding H⁺ raises it by one. Do not pair species merely because they occur on opposite sides of an equation—trace the specific proton transfer.

For HCO₃⁻, donation gives CO₃²⁻ whereas acceptance gives H₂CO₃. Showing both reactions is the evidence for amphiprotic behaviour; one acid–base equation alone is insufficient.

For [H⁺] = 2.0 × 10⁻³ mol dm⁻³, pH = 2.70; the leading 2 makes the answer non-integer. A colour indicator estimates a range, whereas a calibrated probe supports a numerical measurement.

At 25 °C, Kw = 1.0 × 10⁻¹⁴, so a neutral solution has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³. Neutrality always means equal ion concentrations; neutral pH is not necessarily 7 when temperature changes.

At a fixed temperature, Kw is constant, so [OH-] = Kw/[H+]: a higher [H+] means a lower [OH-]. For example, at pH 9.3 and 25 C, [OH-] = 2.0 x 10^-5 mol dm^-3. Classify a solution from the ion comparison; do not assume neutral pH is 7 at every temperature.

Represent a strong acid with essentially complete ionization and a weak acid with an equilibrium containing substantial undissociated acid. Strength is an equilibrium property, while concentration is an initial amount per volume; pH depends on both, so strength alone cannot rank arbitrary solutions.

Balance proton capacity as well as atoms: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, while an acid–carbonate reaction also releases CO₂. To identify parents of Na₂SO₄, trace SO₄²⁻ to the acid and Na⁺ to the base rather than treating the salt name as a reaction equation.

For a strong acid titrated with strong base at 25 °C, calculate the initial pH from excess acid, locate equivalence from stoichiometric moles, and place the steep section around pH 7. Equivalence is a mole condition; it is not the same as equal solution volumes unless concentrations and stoichiometry make it so.

A reliable route is [OH⁻] → pOH → pH → [H⁺], or the reverse, with each logarithm shown. Use pH + pOH = pKw; replacing pKw by 14 is valid only at 25 °C.

Worked pOH example: for $0.025\,\mathrm{mol\,dm^{-3}}$ $\ce{KOH(aq)}$, complete dissociation gives $[\ce{OH^-}]=0.025\,\mathrm{mol\,dm^{-3}}$. Therefore $\mathrm{pOH}=-\log_{10}(0.025)=1.60$. The low pOH is consistent with a basic solution; at 298 K, $\mathrm{pH}=14.00-1.60=12.40$.

Large Ka and small pKa both indicate the stronger acid; large Kb and small pKb indicate the stronger base. Strength describes extent of ionization, whereas concentration describes amount per volume—dilute and weak are not synonyms.

Match the constants to one conjugate pair: Ka(HA) × Kb(A⁻) = Kw. A stronger acid therefore has a weaker conjugate base at the same temperature; do not multiply constants belonging to unrelated species.

Worked conjugate-constant example at 298 K: methylamine has $\mathrm{p}K_b=3.34$, so for its conjugate acid $\ce{CH3NH3+}$, $\mathrm{p}K_a=\mathrm{p}K_w-\mathrm{p}K_b=14.00-3.34=10.66$. Equivalently, $K_aK_b=K_w$. This relationship applies only to a conjugate acid–base pair at the same temperature.

Spectator ions from strong parents do not control pH. NH₄Cl is acidic because NH₄⁺ donates a proton to water, while a carbonate salt is basic because CO₃²⁻ accepts one; write the hydrolysing ion, not the intact salt, in the equilibrium.

A weak acid–strong base curve starts at a higher pH than an equally concentrated strong acid, contains a buffer region, has pH = pKa at half-equivalence, and has an alkaline equivalence point from conjugate-base hydrolysis. Weak–weak titrations often lack a sufficiently steep jump for a simple indicator endpoint.

Added acid shifts HInd ⇌ H⁺ + Ind⁻ toward the HInd colour; added base favours Ind⁻. The visible transition spans a range around pKa because both colours must change in relative abundance. Universal indicator combines several such equilibria and is not one substance with every colour.

Overlay the indicator transition range on the titration curve and require the whole visible change to fall inside the steep region. A weak acid–strong base equivalence is alkaline and favours an alkaline-range indicator; a strong acid–weak base equivalence is acidic. Endpoint proximity, not a memorized indicator name alone, is the criterion.

In an ethanoic acid/ethanoate buffer, CH₃COO⁻ consumes added H⁺ and CH₃COOH consumes added OH⁻, so the conjugate ratio changes only slightly. A buffer resists small additions but has finite capacity; once one component is nearly exhausted, the pH can change sharply.

When a small amount of acid is added, A⁻ removes it to form HA; added base is removed by HA to form A⁻. Dilution leaves their ratio nearly unchanged but reduces buffer capacity, so resistance to a large addition is not unchanged.

Worked buffer example: an ethanoate buffer contains $0.100\,\mathrm{mol\,dm^{-3}}$ $\ce{CH3COOH}$ and $0.200\,\mathrm{mol\,dm^{-3}}$ $\ce{CH3COO^-}$, with $\mathrm{p}K_a=4.76$. Substitution gives $\mathrm{pH}=\mathrm{p}K_a+\log_{10}([\ce{A^-}]/[\ce{HA}])=4.76+\log_{10}(0.200/0.100)=5.06$. The pH is above $\mathrm{p}K_a$ because the conjugate base is more concentrated than the acid.

In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn rises from 0 to +2 and is oxidized, so it is the reducing agent; Cu²⁺ falls from +2 to 0 and is reduced, so it is the oxidizing agent. Name agents from what happens to them, not from the process they cause in the other species.

For MnO₄⁻ → Mn²⁺ in acid, balance O with 4H₂O, H with 8H⁺ and charge with 5e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. After combining halves, cancel electrons and any identical H⁺ or H₂O, then recheck both atoms and net charge.

Zinc displaces Cu²⁺ because Zn more readily oxidizes: Zn + Cu²⁺ → Zn²⁺ + Cu. Chlorine displaces Br⁻ because Cl₂ more readily reduces. Keep the metal and halogen trends in their correct electron directions instead of using one vague 'more reactive' rule.

Balance the electron transfer behind the molecular equation: metal atoms are oxidized and 2H⁺ + 2e⁻ → H₂ is the reduction. Use the metal charge and acid anion to construct the salt rather than assuming every metal forms a 2+ ion.

Name electrodes from the half-reactions before assigning signs. Electrons leave the anode and reach the cathode through the external circuit; a power supply reverses the polarities in an electrolytic cell but never changes where oxidation and reduction occur.

In a Zn|Zn²⁺ || Cu²⁺|Cu cell, Zn is oxidized at the negative anode and electrons travel through the wire to the positive Cu cathode, where Cu²⁺ is reduced. Salt-bridge anions migrate toward the anode compartment and cations toward the cathode compartment to prevent charge buildup; electrons do not flow through the bridge.

Write the discharge half-equations first. Charging a secondary cell requires an external potential to drive their reverse, whereas a primary cell is not designed for safe efficient reversal and a fuel cell continues only while reactants are supplied. Rechargeability is a reaction-design property, not simply the presence of a power socket.

Molten MgCl₂ contains only Mg²⁺ and Cl⁻: Mg²⁺ + 2e⁻ → Mg at the cathode and 2Cl⁻ → Cl₂ + 2e⁻ at the anode. The melt conducts by ion migration; do not introduce H₂, O₂ or water-based competition into a molten-salt question.

In a primary-alcohol experiment, distil the aldehyde as it forms to limit further oxidation; heat under reflux when the carboxylic acid is required. Tertiary alcohols lack the required hydrogen on the carbon bearing –OH and are not oxidized in the same way.

Track the carbon functional group rather than only the reagent: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol. Hydride supplies an electron-rich H unit to the carbonyl carbon; named reducing agents and detailed mechanisms are outside this objective.

Count π bonds to determine hydrogen demand: one mole of H₂ saturates one C=C, while full conversion of one C≡C to C–C needs two moles of H₂. Keep the carbon skeleton unchanged when drawing the product.

The standard hydrogen electrode uses H₂(g) at 100 kPa in contact with aqueous H⁺ of unit activity (commonly represented as 1 mol dm⁻³) at 298 K on an inert platinum surface, and is assigned E° = 0.00 V. Pair an unknown half-cell with this reference, use polarity to identify reduction, then interpret more positive reduction potential as stronger oxidizing tendency.

Select the more positive reduction potential as the cathode reaction, keep both tabulated values as reduction potentials, and calculate E°cell = E°cathode − E°anode. Do not multiply an electrode potential when a half-equation is scaled.

Worked $E^\circ_{cell}$ example: $E^\circ(\ce{Ag+/Ag})=+0.80\,\mathrm{V}$ and $E^\circ(\ce{Cu^{2+}/Cu})=+0.34\,\mathrm{V}$. Silver is the cathode, so $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-0.34=+0.46\,\mathrm{V}$. The positive result predicts the spontaneous reaction $\ce{2Ag+ + Cu -> 2Ag + Cu^{2+}}$ under standard conditions. Do not multiply $E^\circ$ when doubling the silver half-equation.

Find n from the balanced overall redox equation, not from a single unscaled half-equation. With E° in volts and F in C mol⁻¹, ΔG° is obtained in J mol⁻¹; convert to kJ mol⁻¹ only at the end.

Worked $\Delta G^\circ$ example: for $\ce{2H+ + Zn -> Zn^{2+} + H2}$, $n=2$ and $E^\circ_{cell}=+0.76\,\mathrm{V}$. Using $F=9.65\times10^4\,\mathrm{C\,mol^{-1}}$, $\Delta G^\circ=-(2)(9.65\times10^4)(0.76)=-1.47\times10^5\,\mathrm{J\,mol^{-1}}=-147\,\mathrm{kJ\,mol^{-1}}$. Its negative sign agrees with a spontaneous standard-cell reaction.

List the solute ion and water as competing possibilities at each electrode, then use electrode-potential data together with stated concentration conditions to select products. In aqueous sulfate, water commonly supplies the anode gas; in concentrated halide, halogen formation may compete. Do not transfer molten-salt products automatically to solution.

Trace metal atoms through the circuit: M atoms may oxidize at a soluble anode to maintain Mⁿ⁺, while Mⁿ⁺ gains electrons and deposits on the object. Reversing the object to the anode would remove metal rather than coat it.

The dot in Cl· or CH₃· represents one unpaired electron, not a positive or negative charge. Radicals react readily because pairing that electron can form a bond; track the dot through every equation so electron and atom accounting remain explicit.

Draw two single-barbed arrows from the breaking X–X bond, one toward each atom, to account for both electrons. The UV or heat step creates radicals and is initiation; a step that consumes one radical and forms another belongs to propagation.

For methane chlorination, initiation forms 2Cl· from Cl₂ under UV. Propagation uses Cl· + CH₄ → HCl + CH₃· and CH₃· + Cl₂ → CH₃Cl + Cl·; termination combines two radicals. Further substitution creates a mixture, so the mechanism does not guarantee only CH₃Cl.

Look for an available electron pair: OH⁻ and CN⁻ use a negative charge and lone pair, while NH₃ uses a lone pair without being an anion. A curly arrow must start at that pair and point toward the atom where the new bond forms.

For CH₃CH₂Br + OH⁻, the C–O bond forms as the C–Br bond breaks, giving CH₃CH₂OH + Br⁻. Account for charge and every atom in the product; the leaving group takes the bonding pair rather than departing as a neutral bromine atom.

Place the curly-arrow tail on the bond being broken and its head on the fragment receiving both electrons. Then assign charges from electron ownership: heterolysis creates ions, unlike homolysis, which gives one electron to each radical.

Identify the electron-poor atom, not merely a positive-looking formula. H⁺ and carbocations are electrophiles, and the δ⁺ carbon in a polar C–X bond can also accept a pair. The incoming curly arrow ends at this acceptor.

Treat the C=C as the reactive site and place the two added groups on its two carbon atoms. Bromine addition removes the double bond and forms a dibromoalkane; hydration forms an alcohol. At SL, deducing these products does not require a mechanism.

In BF₃ + NH₃ → F₃B←NH₃, NH₃ donates the pair and is the Lewis base; BF₃ accepts it and is the Lewis acid. Classify the roles from electron-pair movement rather than from whether H⁺ appears.

Show a coordination bond with an arrow from a ligand lone pair to the metal ion. The arrow records the origin of the shared pair; after formation the bond is not a different electrostatic species from other covalent bonds.

Read [Cu(NH₃)₄]²⁺ as one Cu centre with four NH₃ ligands and coordination number 4. Use ligand charges and the overall bracket charge to deduce the metal oxidation state; do not confuse coordination number with oxidation state.

SN2 is a concerted backside attack: the nucleophile bonds as the carbon–halogen bond breaks. SN1 first forms a carbocation; the nucleophile then attacks the planar intermediate. Primary substrates generally favour SN2 and tertiary substrates generally favour SN1 because steric access and carbocation stability differ. Secondary substrates can support either mechanism, so use the conditions as well as the substrate class.

For comparable halogenoalkanes, the weaker C–I bond usually makes iodide a better leaving group and substitution faster than for chloride. Explain the trend with bond enthalpy and the actual rate-determining bond change, not polarity alone.

Begin with an arrow from the alkene π bond to the electrophile. HX or acid-catalysed hydration can form a carbocation, whereas halogen addition uses a bridged halonium ion; the next arrow must start from the nucleophile's pair.

For an unsymmetrical alkene, draw both protonation routes and compare the carbocations: tertiary is generally more stable than secondary, then primary. Choose the product from the more stable intermediate; do not apply this shortcut to a halonium pathway.

The first step temporarily disrupts aromatic delocalization; loss of H⁺ then restores the ring and replaces H by E. An addition product would leave aromaticity broken, which explains why the assessed outcome is substitution.