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: Calculate the degree of dissociation (α)
The degree of dissociation (α) is defined as the fraction of the acid that dissociates. Given that the acid is 0.01% ionized, we can express α as:

\[ \alpha = \frac{\text{Percent Ionization}}{100} = \frac{0.01}{100} = 0.0001 \]

### Step 2: Set up the equilibrium expression
For a weak monoprotic acid (HA) dissociating in water, the dissociation can be represented as:

\[ HA \rightleftharpoons H^+ + A^- \]

If we start with a concentration \( C \) of the acid (1.00 M in this case), at equilibrium, the concentrations will be:

- \([H^+] = C \cdot \alpha = 1.00 \cdot 0.0001 = 0.0001 \, \text{M}\)
- \([A^-] = C \cdot \alpha = 1.00 \cdot 0.0001 = 0.0001 \, \text{M}\)
- \([HA] = C - C \cdot \alpha = 1.00 - 1.00 \cdot 0.0001 = 0.9999 \, \text{M}\)

### Step 3: Write the expression for the dissociation constant (Ka)
The dissociation constant \( K_a \) is given by the expression:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations into this expression:

\[ K_a = \frac{(0.0001)(0.0001)}{0.9999} \]

### Step 4: Calculate Ka
Calculating the numerator:

\[ (0.0001)(0.0001) = 1 \times 10^{-8} \]

Now substituting back into the Ka expression:

\[ K_a = \frac{1 \times 10^{-8}}{0.9999} \approx 1 \times 10^{-8} \quad (\text{since } 0.9999 \approx 1) \]

Thus, the dissociation constant \( K_a \) is approximately:

\[ K_a \approx 1 \times 10^{-8} \]

### Final Answer
The dissociation constant \( K_a \) of the weak monoprotic acid is \( 1 \times 10^{-8} \).

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