Question
What is the calculated concentration of C2H3O2- at the equivalence point using the given average molarity of acetic acid, initial volume, and volume of NaOH added, and does it match the provided answer of 0.04916?
Answers
GPT-4o mini
To calculate the concentration of the acetate ion (C₂H₃O₂⁻) at the equivalence point of a titration between acetic acid (CH₃COOH) and sodium hydroxide (NaOH), we need to use the provided information:
1. **Average molarity of acetic acid (CH₃COOH)**: Let's denote this as M₁.
2. **Initial volume of acetic acid**: Let's denote this as V₁ (in liters).
3. **Volume of NaOH added at equivalence point**: Let's denote this as V₂ (in liters).
At the equivalence point, all of the acetic acid has reacted with the added NaOH, producing acetate ions (C₂H₃O₂⁻) in solution.
### Step-by-Step Calculation:
1. **Determine moles of acetic acid**:
\[
\text{Moles of CH₃COOH} = \text{Molarity (M₁)} \times \text{Volume (V₁)}
\]
2. **Determine moles of NaOH**:
Since the reaction between acetic acid and sodium hydroxide is a 1:1 stoichiometry:
\[
\text{Moles of NaOH} = \text{Volume (V₂)} \times \text{Molarity (M₂)}
\]
At the equivalence point, these two quantities are equal:
\[
\text{Moles of CH₃COOH} = \text{Moles of NaOH}
\]
3. **Calculate total volume at equivalence point**:
\[
\text{Total Volume} = V₁ + V₂
\]
4. **Calculate concentration of C₂H₃O₂⁻ at equivalence point**:
The concentration of acetate ions will be equal to the moles of acetic acid divided by the total volume at the equivalence point:
\[
[\text{C₂H₃O₂⁻}] = \frac{\text{Moles of CH₃COOH}}{\text{Total Volume}} = \frac{(\text{Molarity (M₁)} \times \text{Volume (V₁)})}{V₁ + V₂}
\]
### Example Calculation:
To check if it matches the provided answer of 0.04916 M, you will need to input the specific values for M₁, V₁, and V₂ into this formula.
Without those specific values, we cannot confirm whether it matches 0.04916 M. Please provide the molarity of acetic acid, initial volume of acetic acid, and volume of NaOH added for a more precise calculation.
1. **Average molarity of acetic acid (CH₃COOH)**: Let's denote this as M₁.
2. **Initial volume of acetic acid**: Let's denote this as V₁ (in liters).
3. **Volume of NaOH added at equivalence point**: Let's denote this as V₂ (in liters).
At the equivalence point, all of the acetic acid has reacted with the added NaOH, producing acetate ions (C₂H₃O₂⁻) in solution.
### Step-by-Step Calculation:
1. **Determine moles of acetic acid**:
\[
\text{Moles of CH₃COOH} = \text{Molarity (M₁)} \times \text{Volume (V₁)}
\]
2. **Determine moles of NaOH**:
Since the reaction between acetic acid and sodium hydroxide is a 1:1 stoichiometry:
\[
\text{Moles of NaOH} = \text{Volume (V₂)} \times \text{Molarity (M₂)}
\]
At the equivalence point, these two quantities are equal:
\[
\text{Moles of CH₃COOH} = \text{Moles of NaOH}
\]
3. **Calculate total volume at equivalence point**:
\[
\text{Total Volume} = V₁ + V₂
\]
4. **Calculate concentration of C₂H₃O₂⁻ at equivalence point**:
The concentration of acetate ions will be equal to the moles of acetic acid divided by the total volume at the equivalence point:
\[
[\text{C₂H₃O₂⁻}] = \frac{\text{Moles of CH₃COOH}}{\text{Total Volume}} = \frac{(\text{Molarity (M₁)} \times \text{Volume (V₁)})}{V₁ + V₂}
\]
### Example Calculation:
To check if it matches the provided answer of 0.04916 M, you will need to input the specific values for M₁, V₁, and V₂ into this formula.
Without those specific values, we cannot confirm whether it matches 0.04916 M. Please provide the molarity of acetic acid, initial volume of acetic acid, and volume of NaOH added for a more precise calculation.