Answer :
Sure, I'll provide a step-by-step solution to determine how to prepare 1 liter of a 0.1 M acetate buffer solution with a pH of 4.0, starting from a solution of 0.1 M acetic acid (CH₃COOH) and 5 M NaOH.
### Step 1: Understanding the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of the solution to the pKa (acid dissociation constant) and the ratio of the concentrations of the conjugate base (acetate ion, CH₃COO⁻) and the acid (acetic acid, CH₃COOH):
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Here:
- [tex]\(\text{pH} = 4.0\)[/tex]
- [tex]\(\text{pKa} = 4.76\)[/tex]
- [tex]\([\text{A}^-]\)[/tex] is the concentration of acetate ion (CH₃COO⁻)
- [tex]\([\text{HA}]\)[/tex] is the concentration of acetic acid (CH₃COOH)
### Step 2: Calculate the Ratio [A-]/[HA]
Using the Henderson-Hasselbalch equation:
[tex]\[ 4.0 = 4.76 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Rearranging the equation to solve for the ratio [tex]\(\frac{[\text{A}^-]}{[\text{HA}]}\)[/tex]:
[tex]\[ \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 4.0 - 4.76 = -0.76 \][/tex]
[tex]\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.76} \approx 0.1738 \][/tex]
So, the ratio [tex]\(\frac{[\text{A}^-]}{[\text{HA}]}\)[/tex] is approximately 0.1738.
### Step 3: Use the Ratio to Find Concentrations
Let [tex]\([\text{HA}] = 0.1 - [\text{A}^-]\)[/tex] (since the initial concentration of acetic acid is 0.1 M):
[tex]\[ \frac{[A^-]}{0.1 - [A^-]} = 0.1738 \][/tex]
Solving for [tex]\([A^-]\)[/tex]:
[tex]\[ [A^-] = 0.1738 (0.1 - [A^-]) \][/tex]
[tex]\[ [A^-] + 0.1738 [A^-] = 0.01738 \][/tex]
[tex]\[ [A^-](1 + 0.1738) = 0.01738 \][/tex]
[tex]\[ [A^-] = \frac{0.01738}{1.1738} \approx 0.01481 \][/tex]
Therefore:
[tex]\[ [\text{A}^-] \approx 0.01481 \, \text{M} \][/tex]
[tex]\[ [\text{HA}] = 0.1 - 0.01481 = 0.08519 \, \text{M} \][/tex]
### Step 4: Calculate Moles of Acetate (A-) and Acetic Acid (HA)
For a 1-liter solution:
[tex]\[ \text{Moles of acetate} = 0.01481 \, \text{M} \times 1 \, \text{L} = 0.01481 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of acetic acid} = 0.08519 \, \text{M} \times 1 \, \text{L} = 0.08519 \, \text{moles} \][/tex]
### Step 5: Determine Moles of NaOH Required
NaOH reacts with acetic acid to form acetate:
[tex]\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \][/tex]
Thus, moles of NaOH needed equals moles of acetate formed:
[tex]\[ \text{Moles of NaOH required} = 0.01481 \, \text{moles} \][/tex]
### Step 6: Determine Volume of 5 M NaOH Solution Needed
[tex]\[ \text{Volume of 5 M NaOH solution} = \frac{0.01481 \, \text{moles}}{5 \, \text{M}} \approx 0.00296 \, \text{L} = 2.96 \, \text{mL} \][/tex]
### Summary
To prepare 1 liter of 0.1 M acetate buffer solution with pH 4.0:
1. Calculate the required ratio of acetate to acetic acid.
2. The ratio is approximately 0.1738.
3. The concentration of acetate is approximately 0.01481 M, while that of acetic acid is 0.08519 M.
4. The moles of acetate needed are 0.01481.
5. The moles of acetic acid required are 0.08519.
6. The moles of NaOH needed to convert acetic acid to acetate are 0.01481.
7. The volume of 5 M NaOH solution required is approximately 2.96 mL.
### Step 1: Understanding the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of the solution to the pKa (acid dissociation constant) and the ratio of the concentrations of the conjugate base (acetate ion, CH₃COO⁻) and the acid (acetic acid, CH₃COOH):
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Here:
- [tex]\(\text{pH} = 4.0\)[/tex]
- [tex]\(\text{pKa} = 4.76\)[/tex]
- [tex]\([\text{A}^-]\)[/tex] is the concentration of acetate ion (CH₃COO⁻)
- [tex]\([\text{HA}]\)[/tex] is the concentration of acetic acid (CH₃COOH)
### Step 2: Calculate the Ratio [A-]/[HA]
Using the Henderson-Hasselbalch equation:
[tex]\[ 4.0 = 4.76 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Rearranging the equation to solve for the ratio [tex]\(\frac{[\text{A}^-]}{[\text{HA}]}\)[/tex]:
[tex]\[ \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 4.0 - 4.76 = -0.76 \][/tex]
[tex]\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.76} \approx 0.1738 \][/tex]
So, the ratio [tex]\(\frac{[\text{A}^-]}{[\text{HA}]}\)[/tex] is approximately 0.1738.
### Step 3: Use the Ratio to Find Concentrations
Let [tex]\([\text{HA}] = 0.1 - [\text{A}^-]\)[/tex] (since the initial concentration of acetic acid is 0.1 M):
[tex]\[ \frac{[A^-]}{0.1 - [A^-]} = 0.1738 \][/tex]
Solving for [tex]\([A^-]\)[/tex]:
[tex]\[ [A^-] = 0.1738 (0.1 - [A^-]) \][/tex]
[tex]\[ [A^-] + 0.1738 [A^-] = 0.01738 \][/tex]
[tex]\[ [A^-](1 + 0.1738) = 0.01738 \][/tex]
[tex]\[ [A^-] = \frac{0.01738}{1.1738} \approx 0.01481 \][/tex]
Therefore:
[tex]\[ [\text{A}^-] \approx 0.01481 \, \text{M} \][/tex]
[tex]\[ [\text{HA}] = 0.1 - 0.01481 = 0.08519 \, \text{M} \][/tex]
### Step 4: Calculate Moles of Acetate (A-) and Acetic Acid (HA)
For a 1-liter solution:
[tex]\[ \text{Moles of acetate} = 0.01481 \, \text{M} \times 1 \, \text{L} = 0.01481 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of acetic acid} = 0.08519 \, \text{M} \times 1 \, \text{L} = 0.08519 \, \text{moles} \][/tex]
### Step 5: Determine Moles of NaOH Required
NaOH reacts with acetic acid to form acetate:
[tex]\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \][/tex]
Thus, moles of NaOH needed equals moles of acetate formed:
[tex]\[ \text{Moles of NaOH required} = 0.01481 \, \text{moles} \][/tex]
### Step 6: Determine Volume of 5 M NaOH Solution Needed
[tex]\[ \text{Volume of 5 M NaOH solution} = \frac{0.01481 \, \text{moles}}{5 \, \text{M}} \approx 0.00296 \, \text{L} = 2.96 \, \text{mL} \][/tex]
### Summary
To prepare 1 liter of 0.1 M acetate buffer solution with pH 4.0:
1. Calculate the required ratio of acetate to acetic acid.
2. The ratio is approximately 0.1738.
3. The concentration of acetate is approximately 0.01481 M, while that of acetic acid is 0.08519 M.
4. The moles of acetate needed are 0.01481.
5. The moles of acetic acid required are 0.08519.
6. The moles of NaOH needed to convert acetic acid to acetate are 0.01481.
7. The volume of 5 M NaOH solution required is approximately 2.96 mL.
