Aqueous Equilibria

Aqueous Equilibria

Introduction to Chemistry

Buffers

• Definition: Buffers are solutions composed of a weak conjugate acid–base pair.

• Functionality: They are particularly resistant to pH changes, even when strong acids or bases are added.

Reaction Example of Buffers

• In an equimolar solution of Hydrofluoric Acid (HF) and Sodium Fluoride (NaF):
    - If a small amount of hydroxide (OH⁻) is added, HF reacts to form Fluoride ions (F⁻) and water:
      - $HF + OH⁻ <br>ightarrow H_2O + F⁻$

• pH Impact:
    - The presence of HF counteracts the addition of base, leading to a small increase in pH.
    - Conversely, if acid is added, F⁻ reacts to form HF and water, thus counteracting the addition of acid as well.

Buffer Calculations

• Dissociation Reaction of Generic Acid HA:
    - $HA + H_2O <br>ightleftharpoons H_3O^+ + A^-$

• Equilibrium Constant Expression (Ka):
    - $Ka = rac{[H_3O^+][A^-]}{[HA]}$

Deriving pKa from Ka

• Rearranging the expression leads us to:
    - $rac{[A^-]}{[HA]} = rac{Ka}{[H_3O^+]}$

• By taking the negative logarithm of both sides:
    - $rac{[A^-]}{[HA]} = - ext{log}(Ka) + ext{log}([H_3O^+])$
    - Which results in:
    - $pH = pKa + ext{log} rac{[base]}{[acid]}$

Henderson–Hasselbalch Equation

• Henderson–Hasselbalch Equation:
    - $pH = pKa + ext{log} rac{[base]}{[acid]}$

• Example Calculation:

• What is the pH of a buffer that is 0.12 M in lactic acid, CH₃CH(OH)COOH, and 0.10 M in sodium lactate?
    - Given:
      - $K_a ext{ for lactic acid} = 1.4 imes 10^{-4}$
    - Calculation:
      - $pKa = - ext{log}(1.4 imes 10^{-4}) = 3.85$
      - $pH = 3.85 + ext{log} rac{0.10}{0.12} = 3.77$

pH Range of Buffers

• Definition: The pH range is the range of pH values over which a buffer system operates effectively.

• Selection Criterion: It is best to choose an acid with a pKa value close to the desired pH for optimal buffer functionality.

Response to Strong Acids or Bases in Buffers

• When strong acids or bases are added to a buffer, it is generally assumed that all of the strong acid or base is consumed in the reaction.

• Steps for Analysis:
    1. Assess how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
    2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers

• Example Reaction Setup:
    - A buffer is prepared by adding 0.300 mol of acetic acid (HC₂H₃O₂) and 0.300 mol of sodium acetate (NaC₂H₃O₂) to make a total volume of 1.00 L solution.

• Initial pH Calculation:
    - $ext{pH} = 4.74$

• Adding NaOH to the System:
    - After adding 0.020 mol of NaOH, reaction with acetic acid follows:
      - $HC_2H_3O_2 + OH⁻ <br>ightarrow C_2H_3O_2⁻ + H_2O$
    - Before the reaction concentrations:
      - Acetic acid: 0.300 mol
      - Sodium acetate: 0.300 mol
      - Sodium hydroxide: 0.020 mol
    - After the reaction:
      - Acetic acid: 0.280 mol
      - Sodium acetate: 0.320 mol
      - Sodium hydroxide: 0.000 mol

New pH Calculation Using Henderson–Hasselbalch Equation

• Calculation:
    - $pH = 4.74 + ext{log} rac{0.320}{0.200}$
    - Result:
      - $pH = 4.80$

Titration Concept

• Definition: Titration is a technique where a known concentration of a base or acid is slowly added to a solution of acid or base.

• Measurement: Utilize a pH meter or indicators to ascertain when the solution reaches the equivalence point.

Indicator pH Ranges

• Different indicators will showcase color changes at specific pH values.

• Examples include:
    - Crystal Violet: pH 3
    - Thymol Blue: pH 8
    - Phenolphthalein: pH 10
    - Bromothymol Blue: pH range 6-7

Titration of Strong Acids with Strong Bases

• Behavior Overview:

• From the start to near the equivalence point, the pH increases slowly.

• Near the equivalence point, the pH experiences a rapid increase.

• At the equivalence point, the moles of acid equal the moles of base, resulting in only water and the salt remaining in the solution.

Titration of Weak Acids with Strong Bases

• Key Differences:

• The conjugate base formed from the weak acid affects the pH.

• At the equivalence point, the pH is always greater than 7.

• Indicator Example: Phenolphthalein is commonly used as an indicator for these titrations.

Titration of Weak Bases with Strong Acids

• Traits: At the equivalence point, the pH is less than 7.

• Preferred Indicator: Methyl red is suitable in these titrations.

Titrations of Polyprotic Acids

• Overview: For polyprotic acids, there is an equivalence point for each dissociation throughout the titration process.

Common Ion Effect

• Definition: The common-ion effect describes the shift in equilibrium due to the addition of an ion already present in the equilibrium reaction.
    - Example:
      - In the equilibrium $AgCl(s) <br>ightleftharpoons Ag^+(aq) + Cl^−(aq)$, the addition of NaCl shifts the equilibrium position.

Equilibrium Shifts in Acetic Acid

• For a solution of acetic acid, if acetate ion is added, according to Le Châtelier’s principle, the equilibrium will shift to the left:
    - $CH_3COOH(aq) + H_2O(l) <br>ightleftharpoons H_3O^+(aq) + CH_3COO^−(aq)$

Calculating pH with Common-Ion Effect

• Example Calculation:

• For a solution 0.20 M in HF and 0.10 M in HCl with $K_a$ for HF as 6.8 x 10^-4:

• Initial concentrations before the reaction would be 0.20 M [HF], 0.10 M [H₃O⁺], and 0 [F⁻].

Change in Concentrations During Equilibrium

• The equilibrium calculation must take into account the changes to the concentrations based on the common ion:
    - Initial Conditions:
      
  
    - At equilibrium:
    - The change indicates that as the HF ionizes, the concentration of the existing ions changes according to the equilibrium constants.

Final pH Calculation After Changes

• Upon solving the equilibrium expressions, we arrive at the fluoride ion concentration and the respective pH:
    - $[F⁻] = x = 1.4 imes 10^{-3}$
    - $pH = - ext{log}(0.10) = 1.00$

Solubility Products

• Definition: The equilibrium in a saturated solution of a salt involves the salt dissolving and producing its respective ions in solution.
    - Example:
    - For Barium Sulfate:
    - $BaSO_4(s) <br>ightleftharpoons Ba^{2+}(aq) + SO_4^{2−}(aq)$

• Equilibrium Constant (Ksp):
    - $K_{sp} = [Ba^{2+}][SO_4^{2−}]$

• Note: Ksp is distinct from solubility.

Factors Affecting Solubility

• Common-Ion Effect: The presence of a common ion (like barium in BaSO₄) decreases solubility by shifting equilibrium:
    - $BaSO_4(s) <br>ightleftharpoons Ba^{2+}(aq) + SO_4^{2−}(aq)$

Solubility Product for Dissolved Solids

• For the dissolution process of Bismuth Sulfide:
    - Packaging in context of solubility product constants:
    - $Bi_2S_3(s) <br>ightleftharpoons 2Bi^{3+}(aq) + 3S^{2−}(aq)$
    - Ksp is calculated accordingly:
    - $K_{sp} = [Bi^{3+}]^2[S^{2−}]^3$

Dissolution Calculations

• 1/2 [Bi^{3+}] = solubility (s), so solubility is consistent across concentrations even when common ions are present, affecting the overall solubility product constant Ksp.

Selective Precipitation of Ions

• Overview: Using differences in solubility to separate ions from a mixture.
    - Generalized approach:
    - 1. Group 1: Insoluble Chlorides (like AgCl, PbCl₂, Hg₂Cl₂) are precipitated using 6 M HCl.
    - 2. Group 2: Acid-insoluble sulfides (such as CuS, CdS, Bi₂S₃, PbS) can be precipitated using H₂S in 0.2 M HCl.
    - 3. Group 3: Base-insoluble sulfides and hydroxides (like Al(OH)₃, Fe(OH)₃, Cr(OH)₃) can be separated using phosphates and ammonia.
    - 4. Final separations using phosphates for alkaline earth metals (like Ca₃(PO₄)₂) can help isolate specific metal cations without affecting alkali metals.