Titration and Buffer Solutions

Buffer Solutions

• When aiming for a specific pH, such as 3.20, you can choose between different acid/base pairs, like HNO2/NaNO2 (with Ka = 4.0 × 10^{-4}) or AcOH/AcONa (with Ka = 1.8 × 10^{-5}).
• The process involves generating 0.100 M buffer solutions (in acid concentration) and calculating the pH after adding 0.0010 M HCl.

AcOH/AcONa

• The Henderson-Hasselbalch equation is used:
  pH = pK_a + log \frac{[A^-]}{[AcOH]}

• To achieve a pH of 3.20:

  3.20 = -log(1.8 × 10^{-5}) + log(x)
  3.20 = 4.74 + log(x)
  log(x) = -1.54
  x = 0.0288

• Where x represents the ratio of [AcO^-] to [AcOH].
  \frac{[AcO^-]}{[AcOH]} = 0.0288
  [AcO^-] = 0.0288 × [AcOH]

• If [AcOH] = 0.100 M, then [AcO^-] = 0.0028 M.

Equilibrium Shift after Adding HCl

• The equilibrium is:
  AcOH = AcO^- + H^+

• Initial concentrations: 0.100 M AcOH and 0.0028 M AcO^-.

• Adding 0.0010 M HCl shifts the equilibrium:

  	AcOH	AcO⁻	H⁺
  Initial (I)	0.100 M	0.0028M	0.0010 M
  Change (C)	+0.0010	-0.0010	+0.0010
  Final (E)	0.101 M	0.0018M	0

• The new pH is:
  pH = 4.74 + log(\frac{0.0018}{0.101}) = 2.99

• The pH changes by one's place.

HNO2/NaNO2

• Using the Henderson-Hasselbalch equation:

  pH = pKa + log \frac{[NO2^-]}{[HNO_2]}

• Targeting a pH of 3.20:

  3.20 = -log(4.0 × 10^{-4}) + log(x)
  3.20 = 3.40 + log(x)
  log(x) = -0.20
  x = 0.631

• Where x is the ratio of [NO2^-] to [HNO2].
  \frac{[NO2^-]}{[HNO2]} = 0.631
  [NO2^-] = 0.631 × [HNO2]

• If [HNO2] = 0.100 M, then [NO2^-] = 0.0631 M.

Equilibrium Shift after Adding HCl

• The equilibrium is:
  HNO2 = NO2^- + H^+

• Initial concentrations: 0.100 M HNO2 and 0.0631 M NO2^-.

• Adding 0.0010 M HCl shifts the equilibrium:

  	HNO₂	NO₂⁻	H⁺
  Initial (I)	0.100 M	0.0631M	0.0010M
  Change (C)	+0.0010	-0.0010	+0.0010
  Final (E)	0.101 M	0.0621M	0

• The new pH is:
  pH = 3.40 + log(\frac{0.0621}{0.101}) = 3.19

• Resistant to pH change.

Titration

• Titration is an analytical technique used to determine the concentration of an unknown solution (analyte) by neutralizing it with a solution of known concentration (titrant).
• It involves a titrant against a solution.

Acid/Base Titrations

• Acid/base titrations are a subset of titrations.

Basic Analyte

• Basic analyte is titrated with an acidic titrant.

  • The graph of pH vs. volume of acid added shows a decrease in pH.
  • The equivalence point (Eq. Pt.) is when the moles of acid equal the moles of base.

Acidic Analyte

• Acidic analyte is titrated with a basic titrant.

  • The graph of pH vs. volume of base added shows an increase in pH.

• The endpoint of a titration is when the titration has ended. Ideally, the endpoint should be as close as possible to the equivalence point.
  Endpoint ≈ Eq. Pt.

Types of Titrations

• Titrations can involve strong acids vs. weak acids.
• Common types include:
  • Titrant = Strong Acid, Analyte = Strong Base
  • Titrant = Strong Acid, Analyte = Weak Base
  • Titrant = Strong Base, Analyte = Strong Acid
  • Titrant = Strong Base, Analyte = Weak Acid

Strong Base / Strong Acid

• When titrating a strong acid with a strong base, at the equivalence point, the pH is 7.0 because only A^- is in solution.

Strong Base with Weak Acid

• When titrating a weak acid (HA) with a strong base, a buffer region is formed.
  HA = H^+ + A^-
• At the 1/2 equivalence point, pH = pKa.
• At the equivalence point, the pH is greater than 7.

Basic Titration pH Strategy for Weak Acid/Weak Bases

• For strong acid/base titrations:
  1. Convert the concentration from molarity (M) to moles.
  2. Perform a neutralization reaction.
  3. Convert the new moles back to concentration.
  4. Calculate the pH using: pH = -log[H+] or pOH = -log[OH-].
• For weak acid/base titrations:
  1. Determine which region of the titration you are in: initial, before equivalence point, at equivalence point, or after equivalence point.
  2. Initial: pH is only based on the weak acid.
     M1V1 = M2V2
  3. Before Equivalence Point: a buffer region exists; use the Henderson-Hasselbalch equation:
     pH = pK_a + log(\frac{[A^-]}{[HA]})
  4. At 1/2 Equivalence Point: pH = pKa.

Titration Calculations

• Problem: Calculate the molarity of an acetic acid solution if 34.57 mL of this solution are needed to neutralize 25.19 mL of 0.1025 M sodium hydroxide.
  CH3COOH + NaOH = NaCH3COO + H_2O
• Moles of CH_3COOH = Moles of NaOH
• Calculate moles of NaOH:
  Moles_{NaOH} = 25.19 mL × \frac{1 L}{1000 mL} × 0.1025 \frac{mols NaOH}{L} = 2.582 × 10^{-3} mols NaOH
• [CH3COOH] = \frac{2.582 × 10^{-3} mols CH3COOH}{0.03457 L} = 0.07469 M

Titration pH Calculation

• Problem: Calculate the pH of the solution if 30.00 mL of a 50.0 mM AcOH (pKa = 4.75) is added to 10.0 mL of a 20.0 mM sodium hydroxide solution.

• Moles of NaOH:

  0.0100 L × \frac{0.0200 mol}{L} = 0.000200 mols

• Moles of CH_3COOH:

  0.0300 L × \frac{0.0500 mol}{L} = 0.00150 mols

• Reaction:

  	CH₃COOH	NaOH	NaCH₃COO	H₂O
  Initial	0.00150	0.000200	0	-
  Change	-0.000200	-0.000200	+0.000200	-
  End	0.00130	0	0.000200	-

• Final concentrations:

  [CH3COOH] = \frac{0.00130 mols}{0.0400 L} = 0.0325 M [NaCH3COO] = \frac{0.000200 mols}{0.0400 L} = 0.00500 M

• pH Calculation:

  pH = 4.75 + log(\frac{0.00500}{0.0325}) = 3.94

Titrations of Polyprotic Acids

• Polyprotic acids have multiple dissociable protons, leading to multiple buffer regions and equivalence points.
• Example: H_2A
• Dissociation Steps:
  • H2A = HA^- + H^+ , Equilibrium constant = K{a1}
  • HA^- = A^{2-} + H^+ , Equilibrium constant = K_{a2}
• Titration Curve:
  • First Buffer Region: Dominated by H_2A and HA^-
    • First 1/2 Equivalence Point: pH = pK_{a1}
  • First Equivalence Point: [H^-] = [OH^-] Halfway between pKa1 and pKa2 , Use the average: pH = \frac{pKa1 + pKa2}{2}
  • Second Buffer Region: Dominated by HA^- and A^{2-}
    • Second 1/2 Equivalence Point: pH = pK_{a2}
  • Second Equivalence Point: corresponds to complete deprotonation to A^{2-}

Solubility Product

• Solubility Rules:
  • Soluble Anions (with exceptions):
    • NO_3^-: No exceptions
    • CH_3COO^-: No exceptions
    • Cl^-: Except with Ag^+, Hg_2^{2+}, Pb^{2+}
    • Br^-: Except with Ag^+, Hg_2^{2+}, Pb^{2+}
    • I^-: Except with Ag^+, Hg_2^{2+}, Pb^{2+}
    • SO4^{2-}: Except with Sr^{2+}, Ba^{2+}, Hg2^{2+}, Pb^{2+}
  • Insoluble Anions (with exceptions):
    • S^{2-}: Except with NH_4^+, alkali metals, Ca^{2+}, Sr^{2+}, Ba^{2+}
    • CO3^{2-}: Except with NH4^+, alkali metals
    • PO4^{3-}: Except with NH4^+, alkali metals
    • OH^-: Except with NH_4^+, alkali metals, Ca^{2+}, Sr^{2+}, Ba^{2+}
• Example:
  • AgNO3(s) = Ag^+(aq) + NO3^-(aq)
    • Equilibrium lies far to the right
  • AgCl(s) = Ag^+(aq) + Cl^-(aq)
    • Equilibrium lies far to the left
• Solubility Product (K_{sp}):
  • For AgCl(s), K_{sp} = [Ag^+][Cl^-] = 1.6 × 10^{-10}