Comprehensive Study Notes on Acid-Base Reactions, Titrations, Redox Processes, and Chemical Equilibrium

Fundamental Concepts of Acid-Base Reactions

• An acid is a substance that produces hydrogen ions ($H^+$) when dissolved in water ($H_2O$).

• A base is a substance that produces hydroxide ions ($OH^-$) when dissolved in water ($H_2O$).

• An acid-base reaction is also referred to as a neutralization reaction.

• General acid dissociation in water:      $HX\left(aq\right)\xrightarrow{H_2O}H^{+}(aq)+X^{-}(aq)$

• General base dissociation in water:      $MOH(aq)\xrightarrow{H_2O}M^{+}(aq)+OH^{-}(aq)$

• The $H^+$ Ion in Aqueous Solution:

  • A free $H^+$ ion is a bare proton and interacts strongly with water molecules.

  • In aqueous solutions, $H^+$ exists as a solvated hydronium ion ($H_3O^+$).

Classification of Acids, Bases, and Electrolytes

• Strong Acids and Bases:

  • Dissociate completely into ions in aqueous solution.

  • Act as strong electrolytes and conduct electricity extremely well.

• Weak Acids and Bases:

  • Dissociate very little into ions in aqueous solution.

  • Act as weak electrolytes and conduct electricity poorly.

• List of Strong and Weak Acids:

  • Strong Acids:

  • Hydrochloric acid ($HCl$)

  • Hydrobromic acid ($HBr$)

  • Hydriodic acid ($HI$)

  • Nitric acid ($HNO_3$)

  • Sulfuric acid ($H_2SO_4$)

  • Perchloric acid ($HClO_4$)

  • Weak Acids (Examples):

  • Hydrofluoric acid ($HF$)

  • Phosphoric acid ($H_3PO_4$)

  • Acetic acid ($HC_2H_3O_2$ or $CH_3COOH$)

• List of Strong and Weak Bases:

  • Strong Bases:

  • Group 1 Hydroxides: Lithium hydroxide ($LiOH$), Sodium hydroxide ($NaOH$), Potassium hydroxide ($KOH$), Rubidium hydroxide ($RbOH$), Cesium hydroxide ($CsOH$)

  • Heavy Group 2 Hydroxides: Calcium hydroxide ($Ca(OH)_2$), Strontium hydroxide ($Sr(OH)_2$), Barium hydroxide ($Ba(OH)_2$)

  • Weak Bases (Example):

  • Ammonia ($NH_3$)

• Classification Summary of Soluble Compounds:

  • Strong Electrolytes: Ionic compounds, Strong acids, Strong bases

  • Weak Electrolytes: Weak acids, Weak bases

  • Nonelectrolytes: Covalent (molecular) compounds

Acid-Base Calculations and Ion Stoichiometry

• Determining the number of $H^+$ (or $OH^-$) ions in solution:

  • Problem: Calculate the number of $H^+(aq)$ ions present in $25.3\,mL$ of $1.4\,M$ nitric acid ($HNO_3$).

  • Plan:

  1. Convert volume from $mL$ to $L$ using $10^3\,mL = 1\,L$.

  2. Multiply volume ($L$) by molarity ($M$) to find moles of $HNO_3$.

  3. Use the stoichiometric ratio ($1\,mol\,H^+ / 1\,mol\,HNO_3$) to determine moles of $H^+$ ions.

  4. Multiply moles of $H^+$ by Avogadro's number ($6.022 \times 10^{23}\,ions/mol$) to obtain total ions.

  • Chemical Equation:          $HNO_3(aq) \xrightarrow{H_2O} H^+(aq) + NO_3^-(aq)$

  • Step 1: Calculate moles of acid:          $25.3\,mL\text{ soln} \times \frac{1\,L}{10^3\,mL} \times \frac{1.4\,mol\,HNO_3}{1\,L\text{ soln}} = 0.035\,mol\,HNO_3$

  • Step 2: Calculate number of $H^+$ ions:          $0.035\,mol\,HNO_3 \times \frac{1\,mol\,H^+}{1\,mol\,HNO_3} \times \frac{6.022 \times 10^{23}\,ions}{1\,mol} = 2.1 \times 10^{22}\,H^+\text{ ions}$

Chemical Equations for Acid-Base Reactions

• Reaction Between Strong Acids and Strong Bases:

  • Molecular Equation:          $HX(aq) + MOH(aq) \rightarrow MX(aq) + H_2O(l)$

  • Total Ionic Equation (using $H_3O^+$):          $H_3O^+(aq) + X^-(aq) + M^+(aq) + OH^-(aq) \rightarrow H_2O(l) + X^-(aq) + M^+(aq) + H_2O(l)$

  • Net Ionic Equation:          $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$          or          $H_3O^+(aq) + OH^-(aq) \rightarrow 2\,H_2O(l)$

• Reaction Between Weak Acids and Strong Bases:

  • A weak acid is written as an undissociated intact molecule ($HX$).

  • Molecular Equation:          $HX(aq) + MOH(aq) \rightarrow MX(aq) + H_2O(l)$

  • Total Ionic Equation:          $HX(aq) + M^+(aq) + OH^-(aq) \rightarrow X^-(aq) + M^+(aq) + H_2O(l)$

  • Net Ionic Equation:          $HX(aq) + OH^-(aq) \rightarrow X^-(aq) + H_2O(l)$

• Comparison Table of Reactions:

  • Strong Acid + Strong Base ($HCl + NaOH$):

  • Molecular Equation: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

  • Total Ionic Equation: $H_3O^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow H_2O(l) + Na^+(aq) + Cl^-(aq) + H_2O(l)$

  • Net Ionic Equation: $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

  • Weak Acid + Strong Base ($CH_3COOH + NaOH$):

  • Molecular Equation: $CH_3COOH(aq) + NaOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)$

  • Total Ionic Equation: $CH_3COOH(aq) + Na^+(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + Na^+(aq) + H_2O(l)$

  • Net Ionic Equation: $CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$

• Gas-Forming Reactions with Weak Acids:

  • Reaction of sodium bicarbonate with acetic acid produces carbon dioxide gas:

  • Molecular Equation: $NaHCO_3(aq) + HC_2H_3O_2(aq) \rightarrow NaC_2H_3O_2(aq) + CO_2(g) + H_2O(l)$

  • Total Ionic Equation: $Na^+(aq) + HCO_3^-(aq) + HC_2H_3O_2(aq) \rightarrow C_2H_3O_2^-(aq) + Na^+(aq) + CO_2(g) + H_2O(l)$

  • Net Ionic Equation: $HCO_3^-(aq) + HC_2H_3O_2(aq) \rightarrow C_2H_3O_2^-(aq) + CO_2(g) + H_2O(l)$

• Sample Problem: Writing Ionic and Net Ionic Equations:

  • System (a): Hydroiodic acid ($HI$) + Calcium hydroxide ($Ca(OH)_2$)

  • Molecular Equation: $2\,HI(aq) + Ca(OH)_2(aq) \rightarrow CaI_2(aq) + 2\,H_2O(l)$

  • Total Ionic Equation ($H^+$ representation): $2\,H^+(aq) + 2\,I^-(aq) + Ca^{2+}(aq) + 2\,OH^-(aq) \rightarrow 2\,I^-(aq) + Ca^{2+}(aq) + 2\,H_2O(l)$

  • Net Ionic Equation ($H^+$ representation): $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

  • Total Ionic Equation ($H_3O^+$ representation): $2\,H_3O^+(aq) + 2\,I^-(aq) + Ca^{2+}(aq) + 2\,OH^-(aq) \rightarrow 2\,I^-(aq) + Ca^{2+}(aq) + 4\,H_2O(l)$

  • Net Ionic Equation ($H_3O^+$ representation): $H_3O^+(aq) + OH^-(aq) \rightarrow 2\,H_2O(l)$

  • Spectator Ions: $Ca^{2+}(aq)$ and $I^-(aq)$; Salt formed: Calcium iodide ($CaI_2$)

  • System (b): Potassium hydroxide ($KOH$) + Propanoic acid ($CH_3CH_2COOH$)

  • Molecular Equation: $KOH(aq) + CH_3CH_2COOH(aq) \rightarrow CH_3CH_2COOK(aq) + H_2O(l)$

  • Total Ionic Equation: $K^+(aq) + OH^-(aq) + CH_3CH_2COOH(aq) \rightarrow K^+(aq) + H_2O(l) + CH_3CH_2COO^-(aq)$

  • Net Ionic Equation: $CH_3CH_2COOH(aq) + OH^-(aq) \rightarrow CH_3CH_2COO^-(aq) + H_2O(l)$

  • Spectator Ion: $K^+(aq)$; Salt formed: Potassium propanoate ($CH_3CH_2COOK$)

  • System (c): Nitric acid ($HNO_3$) + Potassium sulfite ($K_2SO_3$)

  • Molecular Equation: $2\,HNO_3(aq) + K_2SO_3(aq) \rightarrow 2\,KNO_3(aq) + H_2O(l) + SO_2(g)$

  • Total Ionic Equation: $2\,H^+(aq) + 2\,NO_3^-(aq) + 2\,K^+(aq) + SO_3^{2-}(aq) \rightarrow 2\,K^+(aq) + 2\,NO_3^-(aq) + H_2O(l) + SO_2(g)$

  • Net Ionic Equation: $2\,H^+(aq) + SO_3^{2-}(aq) \rightarrow H_2O(l) + SO_2(g)$

  • Spectator Ions: $K^+(aq)$ and $NO_3^-(aq)$; Salt formed: Potassium nitrate ($KNO_3$)

Acid-Base Titrations

• Core Principles of Titration:

  • A titration uses the concentration of a standard solution to find the unknown concentration of another solution.

  • A standard solution of base is incrementally added to an acid solution of unknown molarity.

  • An acid-base indicator displays different colors in acidic and basic solutions to track reaction progress.

  • Equivalence Point: The exact stage where moles of $H^+$ from acid equal moles of $OH^-$ from base.

  • End Point: The stage where a permanent color change occurs due to a slight excess of base ($OH^-$).

• Sample Problem: Stoichiometry of Stomach Acid Neutralization:

  • Problem: Calculate the volume (in liters) of $0.10\,M\,HCl$ stomach acid required to react completely with an antacid tablet containing $0.10\,g$ of magnesium hydroxide [$Mg(OH)_2$].

  • Chemical Equation:          $Mg(OH)_2(s) + 2\,HCl(aq) \rightarrow MgCl_2(aq) + 2\,H_2O(l)$

  • Step 1: Moles of $Mg(OH)_2$ (Molar mass = $58.32\,g/mol$):          $0.10\,g\,Mg(OH)_2 \times \frac{1\,mol\,Mg(OH)_2}{58.32\,g\,Mg(OH)_2} = 1.7 \times 10^{-3}\,mol\,Mg(OH)_2$

  • Step 2: Moles of $HCl$ needed:          $1.7 \times 10^{-3}\,mol\,Mg(OH)_2 \times \frac{2\,mol\,HCl}{1\,mol\,Mg(OH)_2} = 3.4 \times 10^{-3}\,mol\,HCl$

  • Step 3: Volume of $HCl$ solution:          $3.4 \times 10^{-3}\,mol\,HCl \times \frac{1\,L\text{ soln}}{0.10\,mol\,HCl} = 3.4 \times 10^{-2}\,L\,HCl$

• Sample Problem: Concentration of Acid from Titration Data:

  • Problem: A $50.00\,mL$ sample of $H_2SO_4$ is titrated with $0.1524\,M\,NaOH$. The buret reading is $0.55\,mL$ initially and $33.87\,mL$ at the end point. Calculate the molarity of $H_2SO_4$

  • Chemical Equation:          $H_2SO_4(aq) + 2\,NaOH(aq) \rightarrow Na_2SO_4(aq) + 2\,H_2O(l)$

  • Step 1: Calculate volume of $NaOH$ added:          $\text{Volume of } NaOH = 33.87\,mL - 0.55\,mL = 33.32\,mL$

  • Step 2: Calculate moles of $NaOH$ used:          $33.32\,mL\text{ soln} \times \frac{1\,L}{10^3\,mL} \times \frac{0.1524\,mol\,NaOH}{1\,L\text{ soln}} = 5.078 \times 10^{-3}\,mol\,NaOH$

  • Step 3: Calculate moles of $H_2SO_4$ reacted:          $5.078 \times 10^{-3}\,mol\,NaOH \times \frac{1\,mol\,H_2SO_4}{2\,mol\,NaOH} = 2.539 \times 10^{-3}\,mol\,H_2SO_4$

  • Step 4: Calculate molarity of $H_2SO_4$ solution:          $\frac{2.539 \times 10^{-3}\,mol\,H_2SO_4}{50.00\,mL\text{ soln}} \times \frac{10^3\,mL}{1\,L} = 0.05078\,M\,H_2SO_4$

Oxidation-Reduction (Redox) Reactions

• Definitions:

  • Oxidation: The loss of electrons. Oxidation number increases.

  • Reduction: The gain of electrons. Oxidation number decreases.

  • Oxidizing Agent: The species that gains electrons and becomes reduced.

  • Reducing Agent: The species that loses electrons and becomes oxidized.

  • Oxidation and reduction always take place simultaneously.

• Redox Process in Compound Formation:

  • Ionic Compounds: Direct transfer of electrons (e.g., $Mg(s) + O_2(g) \rightarrow MgO(s)$).

  • Covalent Compounds: Shift in electron density due to polar covalent bonding (e.g., $H_2(g) + Cl_2(g) \rightarrow 2\,HCl(g)$).

• Rules for Assigning Oxidation Numbers (O.N.):

  1. Elemental form: O.N. = $0$ for any atom in its pure element ($Na$, $C$, $O_2$, $Cl_2$, $P_4$).

  2. Monatomic ion: O.N. = ion charge (with sign written before numeral).

  3. Sum of O.N.s:

  • Equals $0$ for a neutral molecule or formula unit.

  • Equals the ion's charge for a polyatomic ion.

  1. Rules for Specific Groups/Elements:

  • Group 1: O.N. = $+1$ in all compounds.

  • Group 2: O.N. = $+2$ in all compounds.

  • Hydrogen: O.N. = $+1$ with nonmetals; O.N. = $-1$ with metals and boron ($NaH$).

  • Fluorine: O.N. = $-1$ in all compounds.

  • Oxygen: O.N. = $-1$ in peroxides ($H_2O_2$); O.N. = $-2$ in all other compounds (except combined with F).

  • Group 17: O.N. = $-1$ in combination with metals, nonmetals (except O), and lower halogens.

• Sample Problem: Assigning Oxidation Numbers:

  • (a) $ZnCl_2$: Each $Cl = -1$ (total $-2$); therefore $Zn = +2$.

  • (b) $SO_3$: Each $O = -2$ (total $-6$); therefore $S = +6$.

  • (c) $HNO_3$: $H = +1$, each $O = -2$ (total $-6$); therefore $N = +5$.

  • (d) $Cr_2O_7^{2-}$: Each $O = -2$ (total $-14$); overall charge is $-2$, so two $Cr$ atoms total $+12$; each $Cr = +6$

• Sample Problem: Identifying Redox Reactions and Agents:

  • Reaction (a): $2\,Al(s) + 3\,H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3\,H_2(g)$

  • O.N. changes: $Al$ changes from $0 \rightarrow +3$ (oxidized); $H$ changes from $+1 \rightarrow 0$ (reduced).

  • Classification: Redox reaction. $Al$ is the reducing agent; $H_2SO_4$ is the oxidizing agent.

  • Reaction (b): $H_2SO_4(aq) + 2\,NaOH(aq) \rightarrow Na_2SO_4(aq) + 2\,H_2O(l)$

  • O.N. changes: No atom changes O.N. ($H = +1$, $S = +6$, $O = -2$, $Na = +1$).

  • Classification: Not a redox reaction.

  • Reaction (c): $PbO(s) + CO(g) \rightarrow Pb(s) + CO_2(g)$

  • O.N. changes: $C$ changes from $+2 \rightarrow +4$ (oxidized); $Pb$ changes from $+2 \rightarrow 0$ (reduced).

  • Classification: Redox reaction. $CO$ is the reducing agent; $PbO$ is the oxidizing agent.

Redox Titrations

• Principles of Redox Titrations:

  • Similar to acid-base titrations, but based on an electron-transfer reaction.

  • Potassium permanganate ($KMnO_4$) acts as its own indicator: $MnO_4^-$ (purple) is reduced to $Mn^{2+}$ (faint pink/colorless).

• Sample Problem: Blood Calcium Determination by Redox Titration:

  • Problem: Calcium ions in $1.00\,mL$ of blood are precipitated as $CaC_2O_4$, redissolved in $H_2SO_4$, and titrated with $2.05\,mL$ of $4.88 \times 10^{-4}\,M\,KMnO_4$. Calculate the moles of $Ca^{2+}$ in the blood sample.

  • Balanced Reaction Equation:          $2\,KMnO_4(aq) + 5\,CaC_2O_4(s) + 8\,H_2SO_4(aq) \rightarrow 2\,MnSO_4(aq) + K_2SO_4(aq) + 5\,CaSO_4(s) + 10\,CO_2(g) + 8\,H_2O(l)$

  • Step 1: Calculate moles of $KMnO_4$:          $2.05\,mL\text{ soln} \times \frac{1\,L}{10^3\,mL} \times \frac{4.88 \times 10^{-4}\,mol\,KMnO_4}{1\,L\text{ soln}} = 1.00 \times 10^{-6}\,mol\,KMnO_4$

  • Step 2: Calculate moles of $CaC_2O_4$:          $1.00 \times 10^{-6}\,mol\,KMnO_4 \times \frac{5\,mol\,CaC_2O_4}{2\,mol\,KMnO_4} = 2.50 \times 10^{-6}\,mol\,CaC_2O_4$

  • Step 3: Calculate moles of $Ca^{2+}$:          $2.50 \times 10^{-6}\,mol\,CaC_2O_4 \times \frac{1\,mol\,Ca^{2+}}{1\,mol\,CaC_2O_4} = 2.50 \times 10^{-6}\,mol\,Ca^{2+}$

Types of Redox Reactions and Activity Series

• Classification of Redox Reactions:

  • Combination Reactions: Two or more reactants combine to form a single product ($X + Y \rightarrow Z$).

  • Example: $2\,K(s) + Cl_2(g) \rightarrow 2\,KCl(s)$

  • Decomposition Reactions: A single compound breaks down into two or more products ($Z \rightarrow X + Y$).

  • Example: $2\,HgO(s) \xrightarrow{\Delta} 2\,Hg(l) + O_2(g)$

  • Displacement Reactions:

  • Single Displacement: An active element displaces another from a compound ($X + YZ \rightarrow XZ + Y$).

    • Hydrogen displacement from water: $2\,Li(s) + 2\,H_2O(l) \rightarrow 2\,LiOH(aq) + H_2(g)$

    • Hydrogen displacement from acid: $Ni(s) + 2\,H^+(aq) \rightarrow Ni^{2+}(aq) + H_2(g)$

    • Metal displacement: $Cu(s) + 2\,AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2\,Ag(s)$

  • Double Displacement: Exchange of ions between two compounds ($AB + CD \rightarrow AD + CB$) (non-redox in precipitation/acid-base).

  • Combustion Reactions: Process of combining a substance with oxygen gas ($O_2$).

• The Activity Series of Metals:

  • Ranks metals by their strength as reducing agents (ability to lose electrons).

  • Can displace $H_2$ from liquid water: $Li$, $K$, $Ba$, $Ca$, $Na$

  • Can displace $H_2$ from steam: $Mg$, $Al$, $Mn$, $Zn$, $Cr$, $Fe$, $Cd$

  • Can displace $H_2$ from acids: $Co$, $Ni$, $Sn$, $Pb$

  • Cannot displace $H_2$ from any source: $Cu$, $Hg$, $Ag$, $Au$

• The Activity Series of Halogens:

  • Reactivity decreases down Group 17 ($F_2 > Cl_2 > Br_2 > I_2$).

  • Halogens higher in the group can displace (oxidize) halide ions below them:

  • $F_2$ oxidizes $Cl^-$, Br^-$, and I^-.\n - Cl_2$oxidizes$Br^-$and$I^-.\n - Br_2$oxidizes only$I^-.\n - I_2 cannot oxidize any halide ion above it.\n\n\n\n- Sample Problem: Classifying Redox Reactions:\n - **Reaction (a)**: 3\,Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)\n - Type: **Combination Reaction**.\n - Reducing Agent: Mg$; Oxidizing Agent:$N_2\n - **Reaction (b)**: 2\,H_2O_2(l) \rightarrow 2\,H_2O(l) + O_2(g)\n - Type: **Decomposition Reaction**.\n - Reducing Agent and Oxidizing Agent: H_2O_2\n - **Reaction (c)**: 2\,Al(s) + 3\,Pb(NO_3)_2(aq) \rightarrow 2\,Al(NO_3)_3(aq) + 3\,Pb(s)\n - Type: **Displacement Reaction**.\n - Total Ionic Equation: 2\,Al(s) + 3\,Pb^{2+}(aq) + 6\,NO_3^-(aq) \rightarrow 2\,Al^{3+}(aq) + 6\,NO_3^-(aq) + 3\,Pb(s)\n - Net Ionic Equation: 2\,Al(s) + 3\,Pb^{2+}(aq) \rightarrow 2\,Al^{3+}(aq) + 3\,Pb(s)\n - Reducing Agent: Al$; Oxidizing Agent:$Pb(NO_3)_2\n\n\n# Dynamic Reversibility and Chemical Equilibrium\n\n- Nonequilibrium Systems vs. Equilibrium Systems:\n - **Nonequilibrium System**: Reactions that proceed to completion because products escape or leave the system.\n - Example: Heating calcium carbonate (CaCO_3$) in an open container allows$CO_2 gas to escape:\n      \n      CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)\n\n - **Equilibrium System**: Reactions in a closed system reach a state of dynamic chemical equilibrium where forward and reverse reaction rates are equal.\n - Example: Heating CaCO_3$in a closed container traps$CO_2, establishing equilibrium:\n      \n      CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$$