Chapter 4 – Types of Chemical Reactions & Solution Stoichiometry

4.1 Water – the Common Solvent

• Ubiquitous, essential medium for life and chemistry.
• Polar molecule ➔ uneven charge distribution (δ⁺ on H, δ⁻ on O).
• Leads to strong ion–dipole attraction → excellent dissolving power.
• Capable of solvating (hydrating) ions – process called hydration:
$\text{ion}^{\pm}\; + \; nH<em>2O \;\longrightarrow \; [\text{ion}(H</em>2O)_n]^{\pm}$
• Water molecules orient so that the opposite charge faces the ion.

4.2 Nature of Aqueous Solutions – Strong & Weak Electrolytes

• Solution = homogeneous mixture of ≥ 2 substances.
• Solvent: component in greatest amount (often liquid water).
• Solute(s): component(s) present in smaller amount(s).
• Examples
• Soft drink (l): solvent = $H<em>2O$, solutes = sugar, $CO</em>2$.
• Air (g): solvent = $N<em>2$, solutes = $O</em>2, Ar, CH_4$.

Electrolyte vs. Nonelectrolyte
• Electrolyte: dissolves producing ions → conducts electricity.
• Nonelectrolyte: dissolves without ionization → no conductivity.
• Conductivity test: bright light = strong electrolyte, dim = weak, off = nonelectrolyte.

Strong Electrolytes (≈100 % dissociation)
• Ionic salts (e.g., $NaCl\;(s) \xrightarrow{H<em>2O} Na^{+}(aq)+Cl^{-}(aq)$). • Strong acids/bases (e.g., $HCl, HNO</em>3, NaOH$).

Weak Electrolytes (partial dissociation, reversible ⇌)
• Example: acetic acid
$CH<em>3COOH\,(aq) \rightleftharpoons CH</em>3COO^{-}(aq)+H^{+}(aq)$
• Establishes equilibrium; small fraction ionized → weak current.

Nonelectrolytes
• Molecular substances that remain intact (e.g., glucose)
$C<em>6H</em>{12}O<em>6(s) \xrightarrow{H</em>2O} C<em>6H</em>{12}O_6(aq)$

4.3 Composition of Solutions – Concentration, Molarity & Dilution

Molarity (M)
• Definition: $M = \dfrac{\text{moles of solute}}{\text{liters of solution}}$
• Example: $\dfrac{6\,\text{mol HCl}}{3\,\text{L}}=2.0\,M$.

Worked Exercise
• 500 g $K<em>3PO</em>4$ (Mₘ = 212 g mol⁻¹) in 1.50 L →
$n = \dfrac{500}{212}=2.36\,mol \;\;\Rightarrow\; M = \dfrac{2.36}{1.50}=1.57\,M$.

Preparation of Specific Ionic Concentrations
• For 0.25 M $CaCl<em>2$: $CaCl</em>2 \rightarrow Ca^{2+}+2Cl^{-}$
$[Ca^{2+}] = 1\times0.25=0.25\,M$ ; $[Cl^{-}]=2\times0.25=0.50\,M$.

Dilution
• Adding solvent lowers concentration while moles stay constant:
$M<em>1V</em>1 = M<em>2V</em>2$
• NaOH example: What volume of 2.00 M stock makes 150 mL of 0.800 M?
$V<em>c = \tfrac{M</em>dV<em>d}{M</em>c}= \tfrac{0.800\times150\,mL}{2.00}=60.0\,mL$.
• HNO₃ example: prepare 60 mL of 0.200 M from 4.00 M stock → 3 mL stock + 57 mL water.

4.4 / 4.5 Types of Chemical Reactions – Focus on Precipitation

Categories (this chapter):
• Precipitation • Acid–Base • Oxidation–Reduction.

Precipitation Reactions
• Double-displacement where an insoluble solid forms.
• Solubility governed by empirical rules (Table 4.2). Key rules:
• All alkali metal ions $\big(Li^{+},Na^{+},K^{+},Rb^{+},Cs^{+}\big)$ and $NH<em>4^{+}$ salts are soluble. • All nitrates $NO</em>3^{-}$, bicarbonates $HCO<em>3^{-}$ & chlorates $ClO</em>3^{-}$ soluble.
• Most chlorides, bromides, iodides soluble EXCEPT with $Ag^{+}, Hg<em>2^{2+}, Pb^{2+}.$ • Most sulfates soluble EXCEPT $Ag^{+}, Ca^{2+}, Sr^{2+}, Ba^{2+}, Hg</em>2^{2+}, Pb^{2+}$.
• Carbonates $CO<em>3^{2-}$, phosphates $PO</em>4^{3-}$, chromates $CrO<em>4^{2-}$, sulfides $S^{2-}$ → insoluble except with rule-1 cations. • Hydroxides insoluble except with alkali metals & $Ba^{2+}$. • Visual cues: colored precipitates (e.g., $CdS$ yellow, $Ni(OH)</em>2$ green).

4.6 Describing Reactions in Solution – Three Equation Forms

• Molecular (formula) equation – compounds as neutral species.
Example: $AgNO<em>3(aq)+NaCl(aq) \rightarrow AgCl(s)+NaNO</em>3(aq)$
• Complete ionic equation – strong electrolytes written as ions:
$Ag^{+}+NO<em>3^{-}+Na^{+}+Cl^{-} \rightarrow AgCl(s)+Na^{+}+NO</em>3^{-}$
• Net ionic equation – spectators removed:
$Ag^{+}(aq)+Cl^{-}(aq) \rightarrow AgCl(s)$
• Spectator Ions: ions unchanged (e.g., $Na^{+}, NO_3^{-}$ above).
• Procedure for writing NIEs: write molecular → dissociate → cancel spectators.

Practice (CoCl₂ + NaOH)
• Net ionic: $Co^{2+}(aq)+2OH^{-}(aq) \rightarrow Co(OH)_2(s)$.

4.7 Stoichiometry of Precipitation Reactions

Problem-solving algorithm:

1. Identify species & possible reaction.

2. Write balanced net ionic equation.

3. Compute moles of reactants.

4. Identify limiting reactant.

5. Calculate moles (→ grams, concentration) of required product/ion.

Worked Example (PbSO₄ mass)
• Mixing 1.25 L 0.0500 M $Pb(NO<em>3)</em>2$ and 2.00 L 0.0250 M $Na<em>2SO</em>4$.
• Net ionic: $Pb^{2+}+SO<em>4^{2-} \rightarrow PbSO</em>4(s)$.
• $n<em>{Pb^{2+}}=0.0625\,mol$ ; $n</em>{SO<em>4^{2-}}=0.0500\,mol$ (limiting). • $n</em>{PbSO_4}=0.0500\,mol$ → mass$=0.0500\times303.3=15.2\,g$.

Mini-Concept Checks
• Lead(II) phosphate precipitation from 10 mL 0.30 M $Na<em>3PO</em>4$ + 20 mL 0.20 M $Pb(NO<em>3)</em>2$ → 1.1 g $Pb<em>3(PO</em>4)<em>2$. • [$NO</em>3^{-}$] post-reaction = 0.27 M ; [$PO_4^{3-}$] remaining = 0.011 M.

4.8 Acid–Base (Neutralization) Reactions

Properties of Acids
• Sour taste, change plant dye color, react with metals (→ $H<em>2$) & carbonates (→ $CO</em>2$), conduct electricity.
Properties of Bases
• Bitter, slippery, dye changes, conductive.

Acid–Base Definitions
• Arrhenius: Acid → $H^{+}$ in water, Base → $OH^{-}$.
• Brønsted–Lowry: Acid = proton donor, Base = proton acceptor.
• Requires acids to have at least one ionizable proton.

Polyprotic Acids
• Monoprotic: $HCl, HNO<em>3, CH</em>3COOH$.
• Diprotic: $H<em>2SO</em>4\;(1^{st}$ strong, $2^{nd}$ weak).
• Triprotic: $H<em>3PO</em>4$ (all steps weak).

Brønsted Examples
• $HI$ (acid), $CH<em>3COO^{-}$ (base), $H</em>2PO_4^{-}$ (amphiprotic).

Neutralization
• General: $\text{acid}+\text{base} \rightarrow \text{salt}+H<em>2O$. • Net ionic for strong acid–strong base: $H^{+}+OH^{-}\rightarrow H</em>2O$.

Titration Terminology
• Titrant: known $M$ solution delivered.
• Analyte: unknown concentration.
• Equivalence point: stoichiometric completion.
• Endpoint: indicator color change ≈ equivalence.

Titration Example (KHP vs. NaOH)
• Data: 23.48 mL NaOH required for 0.5468 g KHP.
$n<em>{KHP}=\frac{0.5468}{204.2}=2.678\times10^{-3}\,mol$ 1:1 ratio ⇒ $n</em>{NaOH}=2.678\times10^{-3}\,mol$
$M=\frac{2.678\times10^{-3}}{0.02348}=0.114\,M$.

Stoichiometric Question
• For 1.00 L 0.500 M $H<em>2SO</em>4$, moles $H<em>2SO</em>4=0.500$.
Reaction $H<em>2SO</em>4+2NaOH$ → need $1.00$ mol $NaOH$.

Acid/Base with Oxides
• Metal oxides (basic anhydrides) act like bases with acids: $CaO+2HCl\rightarrow CaCl<em>2+H</em>2O$.
• Non-metal oxides (acid anhydrides) act like acids with bases: $SO<em>2+2NaOH\rightarrow Na</em>2SO<em>3+H</em>2O$.

4.9 Oxidation–Reduction (Redox) Reactions

Fundamentals
• Electron transfer → oxidation (loss, ↑oxidation state) & reduction (gain, ↓oxidation state).
• Oxidizing agent gains e⁻ (is reduced); reducing agent loses e⁻ (is oxidized).

Half-Reaction Illustration (Mg + O₂)
• $2Mg \rightarrow 2Mg^{2+}+4e^{-}$ (oxidation).
• $O<em>2+4e^{-} \rightarrow 2O^{2-}$ (reduction). • Overall: $2Mg+O</em>2 \rightarrow 2MgO$.

Zn & CuSO₄ Reaction
• $Zn\;\text{oxidized}$, $Cu^{2+}\;\text{reduced}$.
• Cu²⁺ is oxidizing agent; Zn metal reducing agent.

Oxidation Numbers – Rules Recap

1. Elemental form = 0.

2. Monatomic ion = its charge.

3. O = –2 (peroxides –1).

4. H = +1 (with metals –1).

5. F = –1 always; group 1 = +1; group 2 = +2.

6. Sum in molecule = 0; in ion = charge.
   Examples
   • $HCO<em>3^{-}$ → O = –2, H = +1, find C: $3(-2)+1+x=-1⇒x=+4$. • $NaIO</em>3$ → I = +5; $IF<em>7$ → I = +7; $K</em>2Cr<em>2O</em>7$ → Cr = +6.

Redox Reaction Types
• Combination: $2Al+3Br<em>2\rightarrow2AlBr</em>3$.
• Decomposition: $2KClO<em>3\rightarrow2KCl+3O</em>2$.
• Combustion: $S+O<em>2\rightarrow SO</em>2$.
• Displacement
• Hydrogen displacement: $Sr+2H<em>2O\rightarrow Sr(OH)</em>2+H<em>2$. • Metal displacement: $Ti+2MgCl</em>2\rightarrow TiCl<em>4+2Mg$. • Halogen displacement: $Cl</em>2+2KBr\rightarrow2KCl+Br<em>2$. • Disproportionation (same element up & down): $2H</em>2O<em>2\rightarrow2H</em>2O+O<em>2$ ; $Cl</em>2+2OH^{-}\rightarrow ClO^{-}+Cl^{-}+H_2O$.

Balancing Redox by Oxidation-State Method (Zn + HCl example)

1. Write skeleton: $Zn+HCl\rightarrow Zn^{2+}+Cl^{-}+H_2$.

2. Assign oxidation states. 3. Show e⁻ transfer. 4. Equalize electrons (×2 for Cl). 5. Balance rest → $Zn+2HCl\rightarrow ZnCl<em>2+H</em>2$ (then split ions if aqueous).

Classification Practice
• $Ca^{2+}+CO<em>3^{2-}\rightarrow CaCO</em>3$ – Precipitation.
• $NH<em>3+H^{+}\rightarrow NH</em>4^{+}$ – Acid–Base.
• $Zn+2HCl\rightarrow ZnCl<em>2+H</em>2$ – Redox (H₂ displacement).
• $Ca+F<em>2\rightarrow CaF</em>2$ – Redox (combination).

Oxidation-State Exercises (answers)
• $K<em>2Cr</em>2O<em>7$: K = +1, O = –2, Cr = +6. • $CO</em>3^{2-}$: O = –2, thus C = +4.
• $MnO<em>2$: O = –2, Mn = +4. • $PCl</em>5$: Cl = –1, P = +5.
• $SF_4$: F = –1, S = +4.

Identifying Redox
• a) $Zn+2HCl\rightarrow ZnCl<em>2+H</em>2$ ⇒ Redox; Zn reducing agent, H⁺ oxidizing agent.
• b) $Cr<em>2O</em>7^{2-}+2OH^{-}\rightarrow2CrO<em>4^{2-}+H</em>2O$ ⇒ Redox; $Cr<em>2O</em>7^{2-}$ oxidizing/ reducing self? (Cr reduces from +6 to +6… actually not redox; oxidation state unchanged) – NOT redox.
• c) $2CuCl\rightarrow CuCl_2+Cu$ ⇒ Disproportionation; Cu⁺ both oxidized & reduced.

Balancing Steps (general)

1. Write unbalanced skeleton. 2. Assign oxidation numbers. 3. Tie-line electrons. 4. Scale coefficients to equalize e⁻. 5. Balance remaining atoms/charges. 6. Indicate physical states.

These bullet-point notes consolidate every significant fact, rule, example, equation, procedure, and numerical illustration from Chapter 4, providing a stand-alone, step-by-step study resource for solution stoichiometry and reaction chemistry in aqueous media.