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 $HNO<em>2/NaNO</em>2$ (with $K<em>a = 4.0 × 10^{-4}$) or $AcOH/AcONa$ (with $K</em>a = 1.8 × 10^{-5}$).
• 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.

$HNO<em>2/NaNO</em>2$

• Using the Henderson-Hasselbalch equation:

  $pH = pK<em>a + log \frac{[NO</em>2^-]}{[HNO_2]}$

• 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 $[NO<em>2^-]$ to $[HNO</em>2]$.
  $\frac{[NO<em>2^-]}{[HNO</em>2]} = 0.631$
  $[NO<em>2^-] = 0.631 × [HNO</em>2]$

• If $[HNO<em>2] = 0.100 M$, then $[NO</em>2^-] = 0.0631 M$.

Equilibrium Shift after Adding HCl

• The equilibrium is:
  $HNO<em>2 = NO</em>2^- + H^+$

• Initial concentrations: 0.100 M $HNO<em>2$ and 0.0631 M $NO</em>2^-$.

• 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.
     $M<em>1V</em>1 = M<em>2V</em>2$
  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.
  $CH<em>3COOH + NaOH = NaCH</em>3COO + H_2O$
• Moles of $CH_3COOH$ = Moles of NaOH
• Calculate moles of NaOH:
  $Moles_{NaOH} = 25.19 mL × \frac{1 L}{1000 mL} × 0.1025 \frac{mols NaOH}{L} = 2.582 × 10^{-3} mols NaOH$
• $[CH<em>3COOH] = \frac{2.582 × 10^{-3} mols CH</em>3COOH}{0.03457 L} = 0.07469 M$

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:

  $[CH<em>3COOH] = \frac{0.00130 mols}{0.0400 L} = 0.0325 M$$[NaCH</em>3COO] = \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:
  • $H<em>2A = HA^- + H^+$ , Equilibrium constant = $K</em>{a1}$
  • $HA^- = A^{2-} + H^+$ , 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: $pH = \frac{pKa<em>1 + pKa</em>2}{2}$
  • 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 $Ag^+, Hg_2^{2+}, Pb^{2+}$
    • $SO<em>4^{2-}$: Except with $Sr^{2+}, Ba^{2+}, Hg</em>2^{2+}, Pb^{2+}$
  • Insoluble Anions (with exceptions):
    • $S^{2-}$: Except with $NH_4^+,$ alkali metals, $Ca^{2+}, Sr^{2+}, Ba^{2+}$
    • $CO<em>3^{2-}$: Except with $NH</em>4^+,$ alkali metals
    • $PO<em>4^{3-}$: Except with $NH</em>4^+,$ alkali metals
    • $OH^-$: Except with $NH_4^+,$ alkali metals, $Ca^{2+}, Sr^{2+}, Ba^{2+}$
• Example:
  • $AgNO<em>3(s) = Ag^+(aq) + NO</em>3^-(aq)$
    • Equilibrium lies far to the right
  • $AgCl(s) = Ag^+(aq) + Cl^-(aq)$
    • Equilibrium lies far to the left
• Solubility Product ($K_{sp}$):
  • For $AgCl(s)$, $K_{sp} = [Ag^+][Cl^-] = 1.6 × 10^{-10}$