What concentration of `HCOO^(-)` is present in a solution of weak of `0.01` M HCOOH `(K_(a)=1.8xx10^(-4)` and `0.01` M HCl?

Options

A. `1.8xx10^(-3)`
B. `10^(-2)`
C. `1.8xx10^(-4)`
D. `10^(-4)`

Answer

To find the concentration of `HCOO^(-)` in a solution of `0.01 M HCOOH` (formic acid) and `0.01 M HCl`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- We have `0.01 M HCOOH` (a weak acid) and `0.01 M HCl` (a strong acid).
- The dissociation of HCl in water will produce `H^+` and `Cl^-` ions completely.

2. **Dissociation of HCl**:
- Since HCl is a strong acid, it will dissociate completely:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
- At equilibrium, the concentration of `H^+` from HCl will be `0.01 M`.

3. **Dissociation of HCOOH**:
- The dissociation of formic acid can be represented as:
\[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]
- Let the degree of dissociation of HCOOH be represented by `α`. The initial concentration of HCOOH is `0.01 M`.

4. **Setting Up the Equilibrium Expression**:
- At equilibrium, the concentration of `H^+` from HCOOH will be `α`, and the total concentration of `H^+` will be `α + 0.01 M`.
- The concentration of `HCOO^-` will also be `α`.

5. **Using the Acid Dissociation Constant (Ka)**:
- The expression for `Ka` is given by:
\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]
- Substituting the concentrations at equilibrium:
\[ K_a = \frac{(α + 0.01)(α)}{(0.01 - α)} \]
- Given that `K_a = 1.8 \times 10^{-4}`.

6. **Assuming α is Small**:
- Since HCOOH is a weak acid, we can assume `α` is much smaller than `0.01 M`. Therefore, we can simplify the equation:
\[ K_a \approx \frac{(0.01)(α + 0.01)}{0.01} \]
- This simplifies to:
\[ K_a \approx (α + 0.01) \]

7. **Solving for α**:
- Rearranging gives:
\[ α + 0.01 = 1.8 \times 10^{-4} \]
- Thus:
\[ α = 1.8 \times 10^{-4} - 0.01 \approx 1.8 \times 10^{-4} \]
- Since `α` is very small compared to `0.01`, we can neglect it in the subtraction.

8. **Conclusion**:
- The concentration of `HCOO^-` is equal to `α`, which is:
\[ [\text{HCOO}^-] = 1.8 \times 10^{-4} \text{ M} \]

### Final Answer:
The concentration of `HCOO^-` in the solution is `1.8 x 10^{-4} M`.