Two weak acids `HX` and `HY` have `K_(a)` values `1.75 xx 10^(-5)` and `1.3 xx 10^(-5)`, respectively, at a certain temperature. An equimolar solution of mixture of two acids is parially neutralised by `NaOH`. How is the ratio of the contents of `X^(Theta)` and `Y^(Theta)` ions related to the `K_(a)` values and molarity?

Options

A. `[(alpha)/(1-alpha)] = (1.75)/(1.3) xx [(alpha')/(1-alpha')],0`, where `alpha` and `alpha'` are ionised fractions of the acids `HX` and `HY` respectively.
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.

Explanation

To solve the problem, we need to analyze the dissociation of the two weak acids, HX and HY, and how their ionization degrees relate to their \( K_a \) values and the concentration of the acids in the solution.

### Step-by-Step Solution:

1. **Write the Dissociation Reactions**:
- For acid HX:
\[ HX \rightleftharpoons H^+ + X^- \]
- For acid HY:
\[ HY \rightleftharpoons H^+ + Y^- \]

2. **Define the Degree of Ionization**:
- Let \( \alpha \) be the degree of ionization of HX.
- Let \( \alpha' \) be the degree of ionization of HY.

3. **Concentration Relationships**:
- If the initial concentration of both acids is \( C \), after ionization:
- The concentration of \( H^+ \) from HX will be \( \alpha C \).
- The concentration of \( X^- \) will also be \( \alpha C \).
- The remaining concentration of HX will be \( C(1 - \alpha) \).
- Similarly, for HY:
- The concentration of \( H^+ \) from HY will be \( \alpha' C \).
- The concentration of \( Y^- \) will be \( \alpha' C \).
- The remaining concentration of HY will be \( C(1 - \alpha') \).

4. **Write the Expressions for \( K_a \)**:
- For HX:
\[ K_{a1} = \frac{[\text{H}^+][X^-]}{[HX]} = \frac{(\alpha C)(\alpha C)}{C(1 - \alpha)} = \frac{\alpha^2 C}{1 - \alpha} \]
- For HY:
\[ K_{a2} = \frac{[\text{H}^+][Y^-]}{[HY]} = \frac{(\alpha' C)(\alpha' C)}{C(1 - \alpha')} = \frac{\alpha'^2 C}{1 - \alpha'} \]

5. **Set Up the Ratio of \( K_a \) Values**:
- Taking the ratio of the two \( K_a \) expressions:
\[ \frac{K_{a1}}{K_{a2}} = \frac{\frac{\alpha^2 C}{1 - \alpha}}{\frac{\alpha'^2 C}{1 - \alpha'}} = \frac{\alpha^2 (1 - \alpha')}{\alpha'^2 (1 - \alpha)} \]

6. **Substitute the Given \( K_a \) Values**:
- Given \( K_{a1} = 1.75 \times 10^{-5} \) and \( K_{a2} = 1.3 \times 10^{-5} \):
\[ \frac{1.75 \times 10^{-5}}{1.3 \times 10^{-5}} = \frac{\alpha^2 (1 - \alpha')}{\alpha'^2 (1 - \alpha)} \]

7. **Conclusion**:
- The ratio of the concentrations of \( X^- \) to \( Y^- \) ions can be expressed in terms of their degrees of ionization and \( K_a \) values, showing that the ratio is dependent on the \( K_a \) values and the degrees of ionization but not directly on the molarity of the solution.