A 25.0 ML Sample Of 0.150 M Acetic Acid Is Titrated With A 0.150 M NaOH Solution. What Is The PH After
Understanding the titration process between acetic acid and sodium hydroxide (NaOH) is fundamental in analytical chemistry, especially when calculating the pH of a solution at various stages of titration. In this article, we will explore how to determine the pH after adding a certain volume of NaOH to a fixed volume of acetic acid, focusing on step-by-step calculations, chemical principles, and key concepts involved in acid-base titrations.
---
Introduction to Acetic Acid and Sodium Hydroxide Titration
Titration is a laboratory technique used to determine the concentration of an unknown solution by adding a titrant of known concentration until the reaction reaches its equivalence point. In this case, acetic acid (CH₃COOH), a weak acid, reacts with sodium hydroxide (NaOH), a strong base.
Key features of acetic acid:
- Weak acid with a typical dissociation constant (Ka) of approximately 1.8 × 10⁻⁵.
- Partial dissociation in aqueous solution, which influences the pH before the equivalence point.
Key features of NaOH:
- Strong base that dissociates completely in water.
- Used to neutralize acids in titrations.
---
Understanding the Titration Process
During titration, the pH of the solution changes depending on the amount of titrant added:
- Initial stage: Acidic pH, dominated by acetic acid.
- Before equivalence point: Mixture of acetic acid and acetate ions.
- At equivalence point: Complete neutralization, solution contains only acetate ions.
- Beyond equivalence point: Excess NaOH determines the pH.
In our specific problem, we are asked to find the pH after a certain volume of NaOH has been added to the acetic acid solution, but the exact volume isn't specified. To illustrate, we will analyze three key points:
- Before any NaOH is added.
- At the half-equivalence point.
- At the equivalence point.
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Calculating the Initial Conditions
Given data:
- Volume of acetic acid sample, V₁ = 25.0 mL = 0.025 L
- Concentration of acetic acid, C₁ = 0.150 M
- Concentration of NaOH, C₂ = 0.150 M
Initial moles of acetic acid:
\[ n{acid} = C1 \times V_1 = 0.150\, \text{mol/L} \times 0.025\, \text{L} = 3.75 \times 10^{-3}\, \text{mol} \]
---
Determining the Volume of NaOH Added
The moles of NaOH added after some volume V₂ (in mL or L):
\[ n{NaOH} = C2 \times V_2 \]
The reaction:
\[ \text{CH}3\text{COOH} + \text{NaOH} \rightarrow \text{CH}3\text{COO}^- + \text{H}_2\text{O} \]
- At any point, the moles of acetic acid remaining:
- Moles of acetate ions formed:
---
Calculating pH Before the Equivalence Point
When NaOH has been added but not enough to reach the equivalence point, the solution contains a mixture of unreacted acetic acid and acetate ions (its conjugate base). The pH can be calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \( pK_a \) of acetic acid ≈ 4.76
- \( [\text{A}^-] \) = concentration of acetate ions
- \( [\text{HA}] \) = concentration of acetic acid
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Case Study: Calculating pH at Different Points
2.1. Before Any NaOH Is Added
At zero volume of NaOH added:
- The solution contains only acetic acid.
- Using the initial molarity and volume:
\[
\text{Initial concentration of acetic acid} = 0.150\, \text{M}
\]
Since acetic acid is a weak acid, its dissociation equilibrium:
\[
\text{CH}3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}3\text{COO}^-
\]
Using the expression for Ka:
\[
Ka = \frac{[\text{H}^+][\text{CH}3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}
\]
Assuming initial dissociation:
\[
[\text{H}^+] = x
\]
\[
[\text{CH}_3\text{COOH}] \approx 0.150 - x \approx 0.150
\]
\[
[\text{CH}_3\text{COO}^-] \approx x
\]
Since Ka is small, \( x \ll 0.150 \), so:
\[
x = \sqrt{Ka \times C{initial}} = \sqrt{1.8 \times 10^{-5} \times 0.150} \approx 1.65 \times 10^{-3}\, \text{M}
\]
pH:
\[
pH = -\log x \approx -\log (1.65 \times 10^{-3}) \approx 2.78
\]
Therefore, initial pH ≈ 2.78.
---
2.2. At the Half-Equivalence Point
This occurs when half of the acetic acid has been neutralized:
\[
n{NaOH} = 0.5 \times n{acid,initial} = 1.88 \times 10^{-3}\, \text{mol}
\]
Corresponding volume of NaOH:
\[
V{half} = \frac{n{NaOH}}{C_2} = \frac{1.88 \times 10^{-3}}{0.150} \approx 0.0125\, \text{L} = 12.5\, \text{mL}
\]
At this point:
- [HA] = [A⁻], since half of the acid has been converted.
- pH = pKa (by the Henderson-Hasselbalch equation).
\[
pH = pK_a = 4.76
\]
This is a characteristic feature of weak acid-strong base titrations.
---
2.3. At the Equivalence Point
The equivalence point is reached when:
\[
n{NaOH} = n{acid,initial} = 3.75 \times 10^{-3}\, \text{mol}
\]
Corresponding volume:
\[
V_{equiv} = \frac{3.75 \times 10^{-3}}{0.150} = 0.025\, \text{L} = 25.0\, \text{mL}
\]
At this point:
- The solution contains only acetate ions, which form a basic solution.
- The pH is determined by the hydrolysis of acetate:
\[
\text{CH}3\text{COO}^- + \text{H}2\text{O} \leftrightarrow \text{CH}_3\text{COOH} + \text{OH}^-
\]
Using the \(K_b\) for acetate:
\[
Kb = \frac{Kw}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10}
\]
Concentration of acetate:
\[
[\text{A}^-] = \frac{n_{NaOH}}{\text{total volume}} = \frac{3.75 \times 10^{-3}}{0.025\, \text{L}} = 0.150\, \text{M}
\]
Hydrolysis of acetate:
\[
K_b = \frac{[\text{OH}^-]^2}{[\text{A}^-]}
\]
\[
[\text{OH}^-] = \sqrt{K_b \times [\text{A}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.150} \approx 9.13 \times 10^{-6}\, \text{M}
\]
pOH:
\[
pOH = -\log 9.13 \times 10^{-6} \approx 5.04
\]
pH
