0.01 M monoprotonic acid is 10% ionised, what is `K_(a)` of the solution ?

Options

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

Answer

To find the dissociation constant \( K_a \) of the monoprotonic acid, we can follow these steps:

### Step 1: Understand the given information
We have a 0.01 M solution of a monoprotonic acid that is 10% ionized. This means that 10% of the acid has dissociated into its ions.

### Step 2: Calculate the degree of ionization (\( \alpha \))
The degree of ionization \( \alpha \) can be calculated from the percentage ionization:
\[ \alpha = \frac{\text{Percentage Ionization}}{100} = \frac{10}{100} = 0.1 \]

### Step 3: Set up the equilibrium expression
For the dissociation of the acid \( HA \):
\[ HA \rightleftharpoons H^+ + A^- \]
Let \( C \) be the initial concentration of the acid, which is 0.01 M. At equilibrium, the concentrations will be:
- \( [HA] = C(1 - \alpha) = 0.01(1 - 0.1) = 0.01 \times 0.9 = 0.009 \, \text{M} \)
- \( [H^+] = C \alpha = 0.01 \times 0.1 = 0.001 \, \text{M} \)
- \( [A^-] = C \alpha = 0.01 \times 0.1 = 0.001 \, \text{M} \)

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

### Step 5: Substitute the equilibrium concentrations into the \( K_a \) expression
Substituting the values we calculated:
\[ K_a = \frac{(0.001)(0.001)}{0.009} = \frac{0.000001}{0.009} \]

### Step 6: Calculate \( K_a \)
Now, performing the calculation:
\[ K_a = \frac{1 \times 10^{-6}}{9 \times 10^{-3}} = \frac{1}{9} \times 10^{-3} \approx 1.11 \times 10^{-4} \]

### Final Answer
Thus, the dissociation constant \( K_a \) of the solution is approximately:
\[ K_a \approx 1.11 \times 10^{-4} \]

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