Acetic acid and propionic acid have `K_(a)` values `1.75xx10^(-5)` and `1.3xx10^(-5)` respectively at a cetrain temperature. An equimolar solution of a mixture of the two acids is partially neutralised by NaOH. How is the ratio of the contents of acetate and propionate ions related to the `K_(a)` values and the molarity ?

Options

A. `((alpha)/(1-alpha))=(1.75)/(1.3)xx((beta)/(1-beta))`, where `alpha`
and `beta` are ionized fractions of the acids
B. The ratio is unrelated to the `K_(a)` values
C. The ratio is unrelated to the molarity
D. The ratio is unrelated to the pH of the solution

Answer

To solve the problem, we need to analyze the dissociation of acetic acid and propionic acid in an equimolar mixture and how they relate to their dissociation constants (Ka values) after partial neutralization with NaOH.

### Step-by-Step Solution:

1. **Identify the Acids and Their Dissociation Constants**:
- Acetic acid (CH₃COOH) has a Ka of 1.75 x 10⁻⁵.
- Propionic acid (C₂H₅COOH) has a Ka of 1.3 x 10⁻⁵.

2. **Set Up the Equilibrium Expressions**:
- For acetic acid:
\[ CH₃COOH \rightleftharpoons CH₃COO^- + H^+ \]
The equilibrium expression is:
\[ K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]} \]

- For propionic acid:
\[ C₂H₅COOH \rightleftharpoons C₂H₅COO^- + H^+ \]
The equilibrium expression is:
\[ K_a = \frac{[C₂H₅COO^-][H^+]}{[C₂H₅COOH]} \]

3. **Define Molar Concentrations**:
- Let’s assume we start with 1 mole of each acid in a total volume of V liters.
- The initial concentrations of both acids are:
\[ [CH₃COOH] = [C₂H₅COOH] = \frac{1}{V} \]

4. **Introduce Degree of Dissociation**:
- Let α be the degree of dissociation of acetic acid and β be the degree of dissociation of propionic acid.
- After dissociation:
- Concentration of acetate ions (CH₃COO⁻) = α
- Concentration of propionate ions (C₂H₅COO⁻) = β
- Concentration of H⁺ ions = α + β
- Remaining acetic acid = 1 - α
- Remaining propionic acid = 1 - β

5. **Write the Equilibrium Expressions**:
- For acetic acid:
\[ K_a = \frac{α(α + β)}{1 - α} \]
- For propionic acid:
\[ K_a = \frac{β(α + β)}{1 - β} \]

6. **Set Up the Ratio of the Dissociation Constants**:
- The ratio of the dissociation constants can be expressed as:
\[ \frac{K_{a, \text{acetic}}}{K_{a, \text{propionic}}} = \frac{α(α + β)/(1 - α)}{β(α + β)/(1 - β)} \]
- Simplifying this gives:
\[ \frac{K_{a, \text{acetic}}}{K_{a, \text{propionic}}} = \frac{α(1 - β)}{β(1 - α)} \]

7. **Substituting the Values of Ka**:
- Substitute the given Ka values:
\[ \frac{1.75 \times 10^{-5}}{1.3 \times 10^{-5}} = \frac{α(1 - β)}{β(1 - α)} \]

8. **Final Expression**:
- Rearranging gives us the relationship between the concentrations of acetate and propionate ions in terms of their Ka values and degrees of dissociation.