Calculate the ratio of `[HXOO^(-)]` and `[F^(-)]` in a mixture of 0.2 M HCOOH `(K_(a)=2xx10^(-4))` and 0.1 M HF `(K_(a)=6.6xx10^(-4)) :`
(a)`1:6.6`
(b)`1:3.3`
(c)`2:3.3`
(d)`3.3:2`

Options

A. `1:6.6`
B. `1:3.3`
C. `2:3.3`
D. `3.3:2`

Answer

To solve the problem, we need to calculate the ratio of the concentrations of the formate ion \([HCOO^-]\) and the fluoride ion \([F^-]\) in a mixture of 0.2 M formic acid (HCOOH) and 0.1 M hydrofluoric acid (HF). We will use the dissociation constants (\(K_a\)) provided for both acids.

### Step-by-Step Solution:

1. **Write the dissociation equations:**
- For formic acid (HCOOH):
\[ HCOOH \rightleftharpoons H^+ + HCOO^- \]
- For hydrofluoric acid (HF):
\[ HF \rightleftharpoons H^+ + F^- \]

2. **Define initial concentrations:**
- Initial concentration of HCOOH, \(C_1 = 0.2 \, \text{M}\)
- Initial concentration of HF, \(C_2 = 0.1 \, \text{M}\)

3. **Set up the equilibrium expressions:**
- For formic acid:
\[ K_{a1} = \frac{[H^+][HCOO^-]}{[HCOOH]} = \frac{x(x+y)}{C_1} \quad (1) \]
- For hydrofluoric acid:
\[ K_{a2} = \frac{[H^+][F^-]}{[HF]} = \frac{(x+y)y}{C_2} \quad (2) \]

Here, \(x\) is the amount of dissociation of formic acid, and \(y\) is the amount of dissociation of hydrofluoric acid.

4. **Assume that the degree of dissociation is small:**
- We can ignore \(x\) in the denominator for formic acid since \(C_1\) is much larger than \(x\).
- Similarly, ignore \(y\) in the denominator for hydrofluoric acid.

5. **Rewrite the equilibrium expressions:**
- For formic acid:
\[ K_{a1} = \frac{x(x+y)}{0.2} \quad (1) \]
- For hydrofluoric acid:
\[ K_{a2} = \frac{(x+y)y}{0.1} \quad (2) \]

6. **Divide equation (1) by equation (2):**
\[ \frac{x}{y} = \frac{K_{a1} \cdot C_2}{K_{a2} \cdot C_1} \]

7. **Substitute the given values:**
- \(K_{a1} = 2 \times 10^{-4}\)
- \(K_{a2} = 6.6 \times 10^{-4}\)
- \(C_1 = 0.2\)
- \(C_2 = 0.1\)

Thus,
\[ \frac{x}{y} = \frac{(2 \times 10^{-4}) \cdot (0.1)}{(6.6 \times 10^{-4}) \cdot (0.2)} \]

8. **Calculate the ratio:**
\[ \frac{x}{y} = \frac{2 \times 10^{-5}}{1.32 \times 10^{-4}} = \frac{2}{13.2} = \frac{1}{6.6} \]

9. **Final Ratio:**
The ratio of \([HCOO^-]\) to \([F^-]\) is:
\[ \frac{[HCOO^-]}{[F^-]} = \frac{x}{y} = \frac{1}{6.6} \]

### Conclusion:
The ratio of \([HCOO^-]\) to \([F^-]\) in the mixture is \(1:6.6\). Therefore, the correct answer is option (a) \(1:6.6\).