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.

	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

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.

	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

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

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

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}