Chem 121 Chapter 4 Notes: Stoichiometry, Electrolytes, Acids, Bases, and Redox Reactions

Chem 121 Chapter 4: Stoichiometry of Chemical Reactions

Chemical Symbols on Different Levels and Balancing Chemical Equations

• Microscopic Level Chemistry: Deals with individual atoms and molecules.

• Macroscopic Level Chemistry: Deals with observable quantities of substances.

• Chemical Equation: A symbolic representation of a chemical reaction.

  • Example: C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H_2O(g)

• Components of a Chemical Reaction:

  • Correct chemical formula for each reactant and product.

  • Symbol to designate state of matter for each: (s) (solid), (l) (liquid), (g) (gas), (a⁢q) (aqueous solution).

  • Coefficient to balance the equation.

• Law of Conservation of Mass: Chemical equations must be balanced to ensure that no atoms are created or destroyed during a reaction.

  • Reactants (left side) yield products (right side).

  • A balanced chemical equation has the same number and types of atoms on both the reactants and products side.

  • Example balancing: C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H_2O(g)

    • Reactants:

      • C = 3

      • H = 8

      • O = (5×2)=10

    • Products:

      • C = 3

      • H = (4×2)=8

      • O = (3×2)+(4×1)=6+4=10

• Balancing Chemical Equations (Example 1):

  • Equation to balance: Na3PO4(aq) + CaCl2(aq) \rightarrow NaCl(aq) + Ca3(PO4)2(s)

  • Initial Atom Count (Reactants vs. Products):

    • Na: 3 vs. 1

    • P: 1 vs. 2

    • O: 4 vs. 8

    • Ca: 1 vs. 3

    • Cl: 2 vs. 1

  • Balanced Equation (step-by-step mental process/example shown in transcript):

    • Need 3 Ca on reactants: Na3PO4(aq) + 3CaCl_2(aq)

    • Need 2 P on reactants: 2Na3PO4(aq) + 3CaCl_2(aq)

    • Now balance Na and Cl on products side: 2Na3PO4(aq) + 3CaCl2(aq) \rightarrow 6NaCl(aq) + Ca3(PO4)2(s)

  • Final Atom Count (Balanced):

    • Na: (2×3)=6 vs. 6

    • P: (2×1)=2 vs. 2

    • O: (2×4)=8 vs. (2×4)=8

    • Ca: (3×1)=3 vs. 3

    • Cl: (3×2)=6 vs. 6

• Smallest Whole-Number Coefficients: Coefficients in a balanced equation should be the smallest possible whole numbers.

  • Incorrectly balanced example: 6Al(s) + 3Fe2O3(s) \rightarrow 3Al2O3(s) + 6Fe(l)

    • All coefficients are divisible by 3.

    • Al = 6 vs. (3×2)=6

    • Fe = (3×2)=6 vs. 6

    • O = (3×3)=9 vs. (3×3)=9

  • Correctly balanced (simplified): 2Al(s) + Fe2O3(s) \rightarrow Al2O3(s) + 2Fe(l)

    • Al = 2 vs. 2

    • Fe = 2 vs. 2

    • O = 3 vs. 3

Electrolytes and Nonelectrolytes

• What happens when a compound dissolves in water?

  • In aqueous solutions, water is always the solvent.

  • Soluble ionic compounds behave as strong electrolytes.

    • They dissolve to give solutions of hydrated ions, meaning the ions are surrounded by water molecules.

    • This process of ionic compounds dissolving to form ions in solution is called dissociation.

    • Example: N⁢a⁡C⁢l⁡(s)→H2⁢ON⁢a+⁡(a⁡q)+C⁢l−⁡(a⁡q).

  • Many molecular compounds behave as nonelectrolytes.

    • They dissolve to give hydrated molecules, meaning the molecules remain intact but are surrounded by water.

    • Example: C{12}H{22}O{11}(s) \xrightarrow{H2O} C{12}H{22}O_{11}(aq).

  • Acids are an exception among molecular compounds, as they can behave as electrolytes.

    • Example: H⁡C⁢l⁡(g)→H2⁡OH+⁡(a⁢q)+C⁢l−⁡(a⁢q).

• Definitions:

  • Electrolyte: A substance that dissolves in water, giving a solution that can conduct electricity.

  • Nonelectrolyte: A substance that also dissolves in water, but gives a solution that does not conduct electricity.

• Strong vs. Weak Electrolytes:

  • Strong electrolytes: Dissociate completely (approaching 100%) in aqueous solution.

    • Represented by a full forward arrow in the dissociation equation.

    • Example: K⁡C⁢l⁡(s)→H2⁢OK+⁡(a⁢q)+C⁢l−⁡(a⁢q).

  • Weak electrolytes: Dissociate incompletely (less than 100%) in aqueous solution.

    • Represented by forward and reverse arrows (equilibrium arrows) in the dissociation equation.

    • Example (acetic acid): CH3COOH(l) \xrightleftharpoons{H2O} CH_3COO^-(aq) + H^+(aq).

• Classification of Substances as Electrolytes or Nonelectrolytes:

  • Strong Electrolytes:

    • Strong Acids: HCl, HBr, HI, HClO4,HNO3,H2SO4

    • Soluble Ionic Compounds: KBr, NaCl, NaOH, KOH, and other soluble ionic compounds

  • Weak Electrolytes:

    • Weak Acids: CH3CO2H (acetic acid), HF, HCN

    • Weak Bases: N⁢H3 (ammonia)

  • Nonelectrolytes:

    • H2⁢O

    • C⁢H3⁢O⁢H (methyl alcohol)

    • C2H5OH (ethyl alcohol)

    • C{12}H{22}O_{11} (sucrose)

    • Most compounds of carbon (organic compounds) that are not acids.

Molecular, Ionic, and Net Ionic Equations

• Molecular Equation: A chemical equation where all compounds are written in their undissociated molecular form.

  • Example: Pb(NO3)2(aq) + 2KI(aq) \rightarrow 2KNO3(aq) + PbI2(s).

• Ionic Equation: Obtained by splitting aqueous ionic compounds into their constituent ions.

  • Solids, liquids, and gases CANNOT be split up into ions; they remain in their molecular or formula unit form.

  • Example for above reaction: Pb^{2+}(aq) + 2NO3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow 2K^+(aq) + 2NO3^-(aq) + PbI_2(s).

• Net Ionic Equation: An ionic equation written without spectator ions.

  • Spectator Ions: Ions that appear the same on both the reactants and products side of the ionic equation (they do not participate in the reaction).

  • Example (identifying spectator ions from above): The NO3−
    and K+
    are spectator ions.

  • Net Ionic Equation: P⁢b2+⁡(a⁢q)+2⁢I−⁡(a⁢q)→P⁢b⁡I2⁡(s).

Precipitation Reactions and Solubility Rules

• Precipitation Reactions: Reactions where two aqueous solutions combine to form a solid precipitate.

  • The precipitate is the insoluble product that forms.

  • Example: $Pb(NO<em>3)</em>2(aq) + 2KI(aq) \rightarrow 2KNO<em>3(aq) + PbI</em>2(s).H⁢e⁢r⁢e,\text{PbI}_2$ is the precipitate.

• Solubility Guidelines (Rules) to Predict Precipitate Formation:

  • Compounds containing these ions will ALWAYS be soluble (aq):

    • Ammonium (N⁢H4+)

    • Nitrate (N⁢O3−)

    • Acetate (C⁢H3⁢C⁢O⁢O−)

    • Perchlorate (C⁢l⁢O4−)

    • Group 1A Cations: L⁢i+,K+,N⁢a+,R⁢b+,C⁢s+

  • Halides (C⁢l−,B⁢r−,I−): Always soluble EXCEPT when with Ag+
    , Hg22+
    , or Pb2+
    .

  • Sulfate ($SO<em>4^{2-}$ ): Always soluble EXCEPT when with Sr2+
    , Ba2+
    , $\text{Hg}</em>2^{2+}$
    , or Pb2+
    .

  • Insoluble Compounds (except when with Group 1A cations or NH4+ ):

    • Carbonate ($CO<em>3^{2-}$ ): Always insoluble EXCEPT with Group 1A cations, $\text{NH}</em>4^+$
      .

    • Sulfide (S2− ): Always insoluble EXCEPT with Group 1A cations, NH4+
      , Sr2+
      , Ba2+
      .

    • Phosphate ($PO<em>4^{3-}$ ): Always insoluble EXCEPT with Group 1A cations, $\text{NH}</em>4^+$
      .

    • Hydroxide (O⁢H− ): Always insoluble EXCEPT with Group 1A cations, NH4+
      , Sr2+
      , Ba2+
      .

Acids and Bases: Identification and Naming

• The Arrhenius Definition of Acids and Bases:

  • Acid: A molecule or polyatomic ion that produces H+ (hydrogen ions) in aqueous solution.

    • H+
      is very reactive and typically associates with water to form the hydronium ion (H3⁢e⁢x⁢t⁢O+
      ).

    • Thus, when discussing aqueous acids, H+=H3⁢e⁢x⁢t⁢O+
      .

    • Example: H⁡C⁢l⁡(g)→H2⁡OH+⁡(a⁢q)+C⁢l−⁡(a⁢q)

    • Characteristics of Acids (which hydrogens are acidic?): To be acidic, a hydrogen must be bonded to a very electronegative atom or group that is stable as an ion.

    • Examples of Acids: HCl, $\text{HNO}<em>3,CH</em>3COOH(CH<em>3CO</em>2H),\text{H}<em>2 ext{PO}</em>4^-$
      , HCO3−
      .

    • Examples of Non-Acids: $\text{CH}<em>4,\text{C}</em>2 ext{H}<em>6,o⁢t⁢h⁢e⁢r⁢h⁢y⁢d⁢r⁢o⁢c⁢a⁢r⁢b⁢o⁢n⁢s,a⁢n⁢d⁢a⁢l⁢c⁢o⁢h⁢o⁢l⁢s(CH</em>3CH_2OH$
      ).

  • Base: A molecule or ion that produces OH− (hydroxide ions) in water.

    • Strong Base Examples: NaOH, KOH, Ba(OH)2
      .

    • Weak Base Example: NH3
      (ammonia) is a common weak base.

Polyprotic Acids

• Polyprotic Acids: Acids that can donate more than one proton (H+).

  • Monoprotic Acids: Have one dissociable H+ .

    • HCl →H++C⁢l−
      (strong electrolyte / acid)

    • $\text{HNO}<em>3 \rightarrow H^+ + NO</em>3^-$
      (strong electrolyte / acid)

    • $CH<em>3COOH \xrightleftharpoons H^+ + CH</em>3COO^-$
      (weak electrolyte / acid)

  • Diprotic Acids: Have two dissociable H+ .

    • $\text{H}<em>2 ext{SO}</em>4 \rightarrow 2H^+ + SO_4^{2-}$
      (strong electrolyte / acid for the first proton).

  • Triprotic Acids: Have three dissociable H+ .

    • $\text{H}<em>3 ext{PO}</em>4 \xrightleftharpoons 3H^+ + PO_4^{3-}$
      (weak electrolyte / acid).

Strong Acids and Strong Bases

• Strong Acid or Base: A molecule or ion that dissociates completely in water.

• Weak Acids and Bases: Dissociate very little in water.

• Memorize the 6 Strong Acids: HCl, HBr, HI, $\text{HClO}<em>4,\text{HNO}</em>3,\text{H}<em>2 ext{SO}</em>4$ (sulfuric acid).

  • Any other acid encountered will generally be weak.

• Strong Bases: All soluble metal hydroxides.

  • Group 1 metal hydroxides (e.g., NaOH, KOH).

  • Hydroxides of Ca2+
    , Ba2+
    , and Sr2+
    .

• Weak Bases: Ammonia (NH3 ) is the most common example.

  • Compounds with $\text{NH}<em>2g⁢r⁢o⁢u⁢p⁢s(e.g.,CH</em>3NH_2$
    , methylamine) are also weak bases.

Neutralization Reactions

• Definition: Acid-base reactions are called neutralization reactions.

• Reaction between Strong Acid and Strong Base: Produces a salt and water.

  • Molecular Equation: K⁢O⁡H⁡(a⁢q)+H⁡C⁢l⁡(a⁢q)→K⁢C⁢l⁡(a⁢q)+H2⁡O⁡(l)

    • Base + Acid →
      A Salt + Water

  • Ionic Equation: K+⁡(a⁢q)+O⁡H−⁡(a⁢q)+H+⁡(a⁢q)+C⁢l−⁡(a⁢q)→K+⁡(a⁢q)+C⁢l−⁡(a⁢q)+H2⁡O⁡(l)

  • Net Ionic Equation (for strong acid + strong base): H+⁡(a⁢q)+O⁡H−⁡(a⁢q)→H2⁡O⁡(l)

• Neutralization with a Carbonate Base:

  • Example: $Na<em>2CO</em>3(aq) + HNO<em>3(aq) \rightarrow NaNO</em>3(aq) + H<em>2O(l) + CO</em>2(g)$

• Neutralization with a Bicarbonate Base:

  • Example: $NaHCO<em>3(aq) + CH</em>3COOH(aq) \rightarrow CH<em>3COONa(aq) + H</em>2O(l) + CO_2(g)$

Oxidation-Reduction (Redox) Reactions

• Definition: Redox reactions involve the transfer of electrons between species.

• Identification: Redox reactions are identified by changes in oxidation numbers or charges of atoms and ions.

• Example (Ionic Reaction): S⁢n2+⁡(a⁢q)+F⁢e3+⁡(a⁢q)→S⁢n4+⁡(a⁢q)+F⁢e2+⁡(a⁢q)

  • Tin (S⁢n): Goes from a +2 charge to a +4 charge.

    • This is a loss of electrons (2 electrons lost).

  • Iron (F⁢e): Goes from a +3 charge to a +2 charge.

    • This is a gain of electrons (1 electron gained).

  • Balanced Ionic Equation (balancing electron transfer): S⁢n2+⁡(a⁢q)+2⁢F⁢e3+⁡(a⁢q)→S⁢n4+⁡(a⁢q)+2⁢F⁢e2+⁡(a⁢q)

• OIL RIG Mnemonic:

  • Oxidation Is Loss (of electrons).

  • Reduction Is Gain (of electrons).

• Species Oxidized and Reduced:

  • An element or ion that loses electrons is OXIDIZED (S⁢n2+
    is oxidized).

  • An element or ion that gains electrons is REDUCED (F⁢e3+
    is reduced).

• Recognizing Redox Reactions in Complex Compounds: If a reaction is not a precipitation or acid-base reaction, it is likely a redox reaction.

  • Example: $2Fe<em>2O</em>3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)$

• Assigning Oxidation Numbers (Rules):

  1. An atom in its elemental state has an oxidation number of 0.

     • Examples: C, $\text{H}<em>2,\text{Cl}</em>2,\text{S}<em>8,\text{Br}</em>2$
       , Mg.

  2. An atom in a monatomic ion has an oxidation number equal to its charge.

     • Examples: N⁢a+=+1
       , F⁢e3+=+3
       , O2−=−2
       .

  3. If a compound is ionic, the oxidation numbers of the elements are equal to their charge.

     • Example in $\text{Fe}<em>2 ext{O}</em>3$
       : Fe is +3
       , O is −2
       .

  4. For elements in a molecule or polyatomic ion (common assignments):

     • Oxygen (O): Usually −2
       .

     • Hydrogen (H): Usually +1
       .

     • Fluorine (F): Always −1
       .

  5. For other elements in a molecule or polyatomic ion: Use algebra.

     • The sum of the oxidation numbers must equal the charge on the molecule or polyatomic ion.

     • Example calculation for CO2
       : x+2⁢(−2)=0⇒x−4=0⇒x=+4
       for Carbon.

• Applying Oxidation Number Rules (Example: $2Fe<em>2O</em>3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)$):

  • Reactants:

    • Fe in $\text{Fe}<em>2 ext{O}</em>3$
      : +3

    • O in $\text{Fe}<em>2 ext{O}</em>3$
      : −2

    • C: 0

  • Products:

    • Fe: 0

    • C in CO2
      : +4

    • O in CO2
      : −2

  • Analysis of Changes:

    • Oxygen: No change ($-2 \rightarrow -2$).</p></li><li><p>Iron: +3→0 (oxidation number decreases, so <strong>Iron in$\mathbf{Fe2O3}$is reduced</strong>).</p></li><li><p>Carbon: 0→+4 (oxidation number increases, so <strong>Carbon in C is oxidized</strong>).</p></li></ul></li></ul></li><li><p><strong>Reducing Agents and Oxidizing Agents:</strong></p><ul><li><p><strong>Reducing Agent:</strong> The species that <strong>loses electrons</strong> and <strong>causes reduction</strong> in another species. It itself <strong>becomes oxidized</strong>.</p><ul><li><p>In the example: Carbon (C) is the reducing agent.</p></li></ul></li><li><p><strong>Oxidizing Agent:</strong> The species that <strong>gains electrons</strong> and <strong>causes oxidation</strong> in another species. It itself <strong>becomes reduced</strong>.</p><ul><li><p>In the example: Iron in$\text{Fe}2 ext{O}3$$
      is the oxidizing agent.

Balancing Simple Redox Reactions

• Steps for Balancing Redox Reactions in Acidic and Basic Solution:

  1. Write the two half-reactions representing the oxidation and reduction processes.

  2. Balance all elements in each half-reaction except oxygen and hydrogen.

  3. Balance oxygen atoms by adding H2⁢e⁢x⁢t⁢O
     molecules to the side that needs more oxygen.

  4. Balance hydrogen atoms by adding H+
     ions to the side that needs more hydrogen.

  5. Balance charge by adding electrons (e−
     ) to the more positive side of each half-reaction.

  6. Equalize electrons: If necessary, multiply each half-reaction's coefficients by the smallest possible integers to yield equal numbers of electrons transferred in both half-reactions.

  7. Add the balanced half-reactions together and simplify by removing species (like H2⁢e⁢x⁢t⁢O or H+) that appear on both sides of the equation.

  8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
     a. Add OH−
     ions to both sides of the equation in numbers equal to the number of H+
     ions.
     b. On the side of the equation containing both H+
     and OH−
     ions, combine these ions to yield water molecules (H++OH−→H2⁢e⁢x⁢t⁢O).
     c. Simplify the equation by removing any redundant water molecules from both sides.

  9. Final Check: Verify that both the number of atoms and the total charges are balanced on both sides of the final equation.