What concentration of `FCH_(2)COOH` `(K_(a)=2.6xx10^(-3))` is needed so that `[H^(+)]=2xx10^(-3)`?

Options

A. `2xx10^(-3)` M
B. `2.6xx10^(-3)` M
C. `5.2xx10^(-3)` M
D. `3.53xx10^(-3)` M

Answer

To find the concentration of `FCH₂COOH` needed to achieve a specific hydrogen ion concentration `[H⁺] = 2 × 10^(-3)`, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the weak acid `FCH₂COOH` can be represented as:
\[ FCH₂COOH \rightleftharpoons FCH₂COO^- + H^+ \]

### Step 2: Define initial concentration and change
Let the initial concentration of `FCH₂COOH` be \( C \). At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- The concentration of `FCH₂COOH` at equilibrium will be \( C(1 - \alpha) \)
- The concentration of `FCH₂COO^-` and \( [H^+] \) will both be \( C\alpha \)

Given that \( [H^+] = 2 \times 10^{-3} \), we can set:
\[ C\alpha = 2 \times 10^{-3} \]

### Step 3: Write the expression for \( K_a \)
The acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[FCH₂COO^-][H^+]}{[FCH₂COOH]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 4: Substitute known values
We know \( K_a = 2.6 \times 10^{-3} \) and \( [H^+] = 2 \times 10^{-3} \). From the earlier step, we have:
\[ \alpha = \frac{[H^+]}{C} = \frac{2 \times 10^{-3}}{C} \]

Now substitute \( \alpha \) into the \( K_a \) expression:
\[ 2.6 \times 10^{-3} = \frac{C\left(\frac{2 \times 10^{-3}}{C}\right)^2}{1 - \frac{2 \times 10^{-3}}{C}} \]

### Step 5: Simplify the equation
This simplifies to:
\[ 2.6 \times 10^{-3} = \frac{C \cdot 4 \times 10^{-6}}{C(1 - \frac{2 \times 10^{-3}}{C})} \]
\[ 2.6 \times 10^{-3} = \frac{4 \times 10^{-6}}{1 - \frac{2 \times 10^{-3}}{C}} \]

### Step 6: Solve for C
Cross-multiplying gives:
\[ 2.6 \times 10^{-3} \left(1 - \frac{2 \times 10^{-3}}{C}\right) = 4 \times 10^{-6} \]
Expanding this:
\[ 2.6 \times 10^{-3} - \frac{5.2 \times 10^{-6}}{C} = 4 \times 10^{-6} \]
Rearranging gives:
\[ 2.6 \times 10^{-3} = 4 \times 10^{-6} + \frac{5.2 \times 10^{-6}}{C} \]
\[ 2.6 \times 10^{-3} - 4 \times 10^{-6} = \frac{5.2 \times 10^{-6}}{C} \]

### Step 7: Calculate C
Calculating the left side:
\[ 2.6 \times 10^{-3} - 4 \times 10^{-6} = 2.596 \times 10^{-3} \]
Now, rearranging gives:
\[ C = \frac{5.2 \times 10^{-6}}{2.596 \times 10^{-3}} \]
Calculating this gives:
\[ C \approx 2.00 \times 10^{-3} \]

### Step 8: Total concentration
Finally, we need to add the concentration of \( [H^+] \) to the concentration of undissociated acid:
\[ \text{Total concentration} = C + [H^+] = 2.00 \times 10^{-3} + 2.00 \times 10^{-3} = 4.00 \times 10^{-3} \]

### Conclusion
The concentration of `FCH₂COOH` needed is approximately \( 4.00 \times 10^{-3} \).

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