5.12 Equilibrium, Ksp Review, and Introduction to Redox Chemistry of Redox Reactions

Course Logistics and Important Dates

• Upcoming Schedule Exceptions:

  • There is no class scheduled for this coming Friday due to a Teacher In-Service Day.

  • There is no class on Memorial Day.

  • The final exam or a major due date is scheduled for the Friday after Memorial Day.

Quiz Review: Solubility and Ksp Calculations

• Problem 1: Precipitation of Lead (II) Chloride

  • The Scenario: Determine the concentration of sodium chloride ($NaCl$) required to precipitate lead (II) chloride ($PbCl_2$) from a solution with an initial lead concentration of $0.65\,M$.

  • Given Values:

  • $[Pb^{2+}] = 0.65\,M$

  • $K_{sp}(PbCl_2) = 1.7 \times 10^{-5}$

  • Waste solution volume = $125\,L$

  • Molar mass of $NaCl = 58.44\,g/mol$

  • Key Concept: Theory vs. Reality:

  • In theoretical "molar solubility" problems where no concentrations are known, we use variables like $s$ and $2s$, leading to "doubling and squaring" ($K_{sp} = [s][2s]^2$).

  • However, when "real" or initial concentrations are provided, we simply plug the known values into the equilibrium expression and solve for the unknown concentration.

  • Equilibrium Expression:

  • $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$

  • $K_{sp} = [Pb^{2+}][Cl^{-}]^2$

  • Calculation for Chloride Concentration:

  • $1.7 \times 10^{-5} = (0.65) \times (x^2)$

  • $x^2 = \frac{1.7 \times 10^{-5}}{0.65}$

  • $x = [Cl^{-}] = 5.11 \times 10^{-3}\,M$

  • Since chloride is added as sodium chloride ($NaCl$), the concentration of $NaCl$ needed is equal to the concentration of chloride ($5.11 \times 10^{-3}\,M$).

  • Calculation for Mass of NaCl:

  • To find the mass required for a $125\,L$ tank:

  • $(5.11 \times 10^{-3}\,mol/L) \times (125\,L) \times (58.44\,g/mol) = 37.336\,g$

  • Rounding to two significant figures (limited by the $K_{sp}$ value): $37\,g$.

  • Note: This calculation assumes the volume of the waste solution does not change significantly upon adding the solid salt.

• Problem 2: Enhancing Solubility of Lead Sulfate

  • The Scenario: A sample of lead sulfate ($PbSO_4$) is not dissolving well. How can it be made more soluble using acid-base chemistry?

  • Equilibrium Equation 1 (Solubility):

  • $PbSO_4(s) \rightleftharpoons Pb^{2+}(aq) + SO_4^{2-}(aq)$

  • Equilibrium Equation 2 (Conjugate Base Reactivity):

  • Sulfate ($SO_4^{2-}$) is a conjugate base. It reacts with water:

  • $SO_4^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_4^{-}(aq) + OH^{-}(aq)$

  • Solution: Adding acid ($H^{+}$) helps dissolve the salt because the acid neutralizes the hydroxide ($OH^{-}$) produced in the second equilibrium. By Le Chatelier's Principle, removing $OH^{-}$ pulls the second equilibrium to the right, which consumes $SO_4^{2-}$. Removing $SO_4^{2-}$ in turn pulls the first equilibrium to the right, causing more $PbSO_4$ to dissolve.

  • Important Caveat: Avoid using sulfuric acid ($H_2SO_4$) to achieve this. Sulfuric acid would add sulfate ions back into the solution, which could counteract the dissolution through the common ion effect. Nitric acid ($HNO_3$) is a better choice.

• Problem 3: Selective Precipitation (Magnesium vs. Strontium)

  • The Scenario: A waste tank contains $0.45\,M\,Mg^{2+}$ and $0.34\,M\,Sr^{2+}$. Carbonate is added to precipitate the metals.

  • Given Values:

  • $K_{sp}(MgCO_3) = 1.1 \times 10^{-5}$

  • $K_{sp}(SrCO_3) = 9.3 \times 10^{-10}$

  • Choosing a Reagent: Sodium carbonate ($Na_2CO_3$) or potassium carbonate ($K_2CO_3$) are preferred over ammonium carbonate ($(NH_4)_2CO_3$). Ammonium is a weak acid, which adds mathematical complexity to the modeling; using alkali metal spectators keeps the chemistry simpler.

  • Part A: Concentration for Strontium to start precipitating:

  • $K_{sp}(SrCO_3) = [Sr^{2+}][CO_3^{2-}]$

  • $9.3 \times 10^{-10} = (0.34) \times [CO_3^{2-}]$

  • $[CO_3^{2-}] = 2.7 \times 10^{-9}\,M$

  • Part B: Concentration of remaining Strontium when Magnesium begins to precipitate:

  • 1. Find the carbonate concentration required to start magnesium precipitation:

    • $1.1 \times 10^{-5} = (0.45) \times [CO_3^{2-}]$

    • $[CO_3^{2-}] = 2.44 \times 10^{-5}\,M$ (approx. $2.2 \times 10^{-5}\,M$ based on classroom discussion shorthand).

  • 2. Use this carbonate concentration to find the remaining strontium concentration:

    • $9.3 \times 10^{-10} = [Sr^{2+}]_{rem} \times (2.2 \times 10^{-5})$

    • $[Sr^{2+}]_{rem} = 4.2 \times 10^{-5}\,M$

Introduction to Redox Reactions

• The Planetary Context:

  • The Earth's atmosphere consists of four major gases: Nitrogen ($N_2$), Oxygen ($O_2$), Argon ($Ar$), and Water Vapor ($H_2O$).

  • Oxygen is the only highly reactive gas among these. Because of its prevalence and reactivity, we refer to Earth as having an oxidizing atmosphere.

  • Common evidence includes the corrosion or rusting of metals over time.

• Definitions of Oxidation and Reduction:

  • Ancient Definition: Based on the presence of oxygen.

  • Oxidation: The gain of, or reaction with, oxygen (e.g., forming metal oxides from free metals).

  • Reduction: The removal of oxygen (e.g., refining metal ores back into free metals).

  • Modern Subatomic Definition: Focuses on the transfer of electrons.

  • Oxidation: The loss of electrons.

  • Reduction: The gain of electrons.

  • Comparison to Acid-Base Chemistry: Just as Bronsted-Lowry acid-base chemistry is defined by the transfer of a proton ($H^{+}$), Redox (Reduction-Oxidation) chemistry is defined by the transfer of an electron ($e^{-}$).

• Mnemonics for Redox:

  • LEO the lion says GRRR: Loss of Electrons is Oxidation; Gain of Electrons is Reduction.

  • OIL RIG: Oxidation Is Loss; Reduction Is Gain.

Oxidation States (Oxidation Numbers)

• Purpose: A bookkeeping system to track the transfer of electrons in reactions, especially in covalent compounds where "charges" are not literal.

• Rules for Assigning Oxidation States:

  • Free Elements: The oxidation state of any element in its pure form (e.g., $C$, $Fe$, $O_2$) is $0$.

  • Monatomic Ions: The oxidation state is equal to the charge of the ion (e.g., $Pb^{2+}$ is $+2$).

  • Oxygen: Usually $-2$.

  • Exception: In peroxides (like $H_2O_2$ or $CaO_2$), oxygen is $-1$.

  • Hydrogen: Usually $+1$.

  • Exception: In metal hydrides (like $LiAlH_4$), hydrogen is $-1$.

  • Sum of States:

  • In a neutral compound, the sum of all oxidation states must equal $0$.

  • In a polyatomic ion, the sum must equal the charge of the ion.

• Examples of Oxidation State Assignments:

  • Carbon Compounds:

  • Carbon Monoxide ($CO$): $O = -2$, therefore $C = +2$.

  • Carbon Dioxide ($CO_2$): $O = -2 \times 2 = -4$, therefore $C = +4$.

  • Carbonate ($CO_3^{2-}$): $O = -2 \times 3 = -6$; net charge is $-2$, therefore $C = +4$.

  • Methane ($CH_4$): $H = +1 \times 4 = +4$, therefore $C = -4$.

  • Nitrogen Compounds:

  • Ammonia ($NH_3$): $H = +1 \times 3 = +3$, therefore $N = -3$.

  • Ammonium ($NH_4^{+}$): $H = +1 \times 4 = +4$; net charge is $+1$, therefore $N = -3$.

  • Nitrogen Dioxide ($NO_2$): $O = -2 \times 2 = -4$, therefore $N = +4$.

  • Nitrite ($NO_2^{-}$): $O = -2 \times 2 = -4$; net charge is $-1$, therefore $N = +3$.

Organic Redox: Hydrogenation and Fatty Acids

• Fatty Acid Structure:

  • Defined as a carboxylic acid with a long hydrocarbon chain.

  • Textbooks differ on the size, but usually, it starts around $10$ to $12$ carbon atoms.

  • Natural fatty acids typically have an even number of carbons because they are biosynthesized from two-carbon fragments called Acetyl-CoA.

• Saturation States:

  • Saturated Fatty Acids: Contain only single bonds between carbons. They are typically solid at room temperature and derived from animal products (e.g., lard, tallow, schmaltz).

  • Unsaturated Fatty Acids: Contain at least one double bond ($C=C$). They are typically liquid at room temperature and derived from plant products (e.g., olive oil, canola oil).

• Hydrogenation (Industrial Reduction):

  • This is the process of adding hydrogen ($H_2$) to unsaturated oils to make them saturated (solid).

  • Hydrogenation is a reduction reaction (addition of hydrogen).

  • De-hydrogenation is an oxidation reaction (removal of hydrogen).

• Trans-Fats and the Problem of Equilibrium:

  • Natural unsaturated fats are almost exclusively in the cis configuration (hydrogens on the same side of the double bond).

  • In industrial hydrogenation, a metal catalyst is used. The reaction is an equilibrium. As oils are partially hydrogenated, some fats re-oxidize back to double bonds.

  • Unlike biological enzymes, this industrial process is not selective, resulting in a 50/50 mix of cis and trans double bonds.

  • Trans-fatty acids are the result of this partial hydrogenation, creating molecular structures that do not exist significantly in nature.

Chemical Agents

• Oxidizing Agent: The reactant that causes oxidation in another substance by accepting its electrons. The oxidizing agent itself is reduced.

  • Common examples: Oxygen ($O_2$), bleach, peroxides, and metal perborates.

• Reducing Agent: The reactant that causes reduction in another substance by donating electrons. The reducing agent itself is oxidized.

  • Common examples: Hydrogen ($H_2$), carbon (coke/soot), and active metals like Magnesium ($Mg$) or Zinc ($Zn$).

• Sacrificial Anode: In engineering, active metals like Zinc are attached to steel structures (like bridges). The Zinc acts as a sacrificial reducing agent, oxidizing itself to prevent the iron in the steel from corroding. These are also conceptually related to antioxidants in biology.