The dissociation constant of a weak monoprotic acid, which is 0.01 % ioniosed in 1 .00M solution , is

Options

A. `1xx10^(-8)`
B. `1xx10^(-4)`
C. `1xx10^(-6)`
D. `10^(-5)`

Answer

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

### Step 1: Understand the dissociation of the weak acid
Let’s denote the weak monoprotic acid as HA. The dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Define the initial concentration
Given that the concentration of the acid (C) is 1.00 M, we can set up the initial concentrations:
- \([HA] = C = 1.00 \, \text{M}\)
- \([H^+] = 0\)
- \([A^-] = 0\)

### Step 3: Define the change in concentration at equilibrium
Let α be the degree of ionization. Since the acid is 0.01% ionized:
\[ \alpha = \frac{0.01}{100} = 0.0001 \]

At equilibrium, the concentrations will be:
- \([HA] = C - C\alpha = 1.00 - 1.00 \cdot 0.0001 = 1.00 - 0.0001 = 0.9999 \, \text{M}\)
- \([H^+] = C\alpha = 1.00 \cdot 0.0001 = 0.0001 \, \text{M}\)
- \([A^-] = C\alpha = 1.00 \cdot 0.0001 = 0.0001 \, \text{M}\)

### Step 4: Write the expression for the dissociation constant (Ka)
The dissociation constant for the weak acid is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.0001)(0.0001)}{0.9999} \]

### Step 5: Simplify the expression
Since \(0.9999\) is very close to \(1\), we can approximate:
\[ K_a \approx \frac{(0.0001)(0.0001)}{1} = 0.00000001 = 1 \times 10^{-8} \]

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