A monoprotic acid in `1.00 M` solution is `0.01%` ionised. The dissociation constant of this acid is

Options

A. `1 xx 10^(-8)`
B. `1 xx 10^(-4)`
C. `1 xx 10^(-6)`
D. `1 xx 10^(-5)`

Answer

To find the dissociation constant (K) of a monoprotic acid that is 0.01% ionized in a 1.00 M solution, we can follow these steps:

### Step 1: Understand the given information
- The concentration of the acid (C) = 1.00 M
- The percentage ionization (α) = 0.01%

### Step 2: Convert percentage ionization to a fraction
To convert the percentage ionization to a decimal fraction:
\[ \alpha = \frac{0.01}{100} = 0.0001 \]

### Step 3: Use the formula for the dissociation constant
For a monoprotic acid, the dissociation constant (K) can be calculated using the formula:
\[ K = C \cdot \alpha^2 \]
Where:
- C is the concentration of the acid
- α is the fraction of the acid that is ionized

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ K = 1.00 \, \text{M} \cdot (0.0001)^2 \]

### Step 5: Calculate α²
Calculating \( (0.0001)^2 \):
\[ (0.0001)^2 = 0.00000001 = 1 \times 10^{-8} \]

### Step 6: Calculate K
Now substituting back into the equation for K:
\[ K = 1.00 \cdot 1 \times 10^{-8} = 1 \times 10^{-8} \]

### Final Answer
Thus, the dissociation constant \( K \) of the monoprotic acid is:
\[ K = 1 \times 10^{-8} \]

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