A solution of a weak monoprotic acid has dissociation constant `K_(a)`. The minimum initial concentration `C` such that the concentration of the undissociated acid can be equated to `C` within an error of `1%` would be

Options

A. `9900 K_(a)`
B. `10000K_(a)`
C. `99K_(a)`
D. `K_(a)`

Explanation

To find the minimum initial concentration \( C \) of a weak monoprotic acid such that the concentration of the undissociated acid can be equated to \( C \) within an error of 1%, we can follow these steps:

### Step 1: Understand the dissociation of the weak acid
For a weak monoprotic acid \( HA \), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Let \( C \) be the initial concentration of the acid. The degree of dissociation is represented by \( \alpha \).

### Step 2: Define the degree of dissociation
Since we want the concentration of undissociated acid to be equated to \( C \) within an error of 1%, we can set:
\[ \alpha = 0.01 \quad (\text{1% of } C) \]

### Step 3: Write the expression for the dissociation constant \( K_a \)
The dissociation constant \( K_a \) for the weak acid can be expressed as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
At equilibrium, the concentrations can be expressed as:
- Concentration of \( H^+ \) = \( C \alpha \)
- Concentration of \( A^- \) = \( C \alpha \)
- Concentration of undissociated acid \( [HA] \) = \( C(1 - \alpha) \)

Substituting these into the expression for \( K_a \):
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]

### Step 4: Substitute \( \alpha \) into the equation
Now substituting \( \alpha = 0.01 \):
\[ K_a = \frac{C (0.01)^2}{1 - 0.01} = \frac{C \cdot 0.0001}{0.99} \]

### Step 5: Rearranging to find \( C \)
Rearranging the equation gives:
\[ C = \frac{K_a \cdot 0.99}{0.0001} \]
\[ C = 9900 K_a \]

### Conclusion
Thus, the minimum initial concentration \( C \) such that the concentration of the undissociated acid can be equated to \( C \) within an error of 1% is:
\[ C = 9900 K_a \]