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 \( K_a \) of the solution, we can follow these steps:

### Step 1: Understand the Ionization of the Acid
Given a monoprotonic acid (HA), it ionizes in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set Up Initial Concentrations
The initial concentration of the acid (HA) is given as 0.01 M. Since no products are formed initially, we can represent the initial concentrations as:
- \([HA] = 0.01 \, M\)
- \([H^+] = 0 \, M\)
- \([A^-] = 0 \, M\)

### Step 3: Determine the Degree of Ionization
It is given that the acid is 10% ionized. This means that 10% of the initial concentration of the acid dissociates. Therefore:
\[ \alpha = \frac{10}{100} = 0.1 \]

### Step 4: Calculate the Change in Concentration
Since 10% of the acid ionizes, the change in concentration for the products at equilibrium will be:
- Change in \([H^+]\) and \([A^-]\) = \(0.01 \times 0.1 = 0.001 \, M\)
- Remaining concentration of \([HA]\) at equilibrium = \(0.01 - 0.001 = 0.009 \, M\)

### Step 5: Write the Equilibrium Concentrations
At equilibrium, the concentrations will be:
- \([HA] = 0.009 \, M\)
- \([H^+] = 0.001 \, M\)
- \([A^-] = 0.001 \, M\)

### Step 6: 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 7: Substitute the Equilibrium Concentrations into the \( K_a \) Expression
Substituting the equilibrium concentrations into the \( K_a \) expression:
\[ K_a = \frac{(0.001)(0.001)}{0.009} \]

### Step 8: Calculate \( K_a \)
Calculating the above expression:
\[ K_a = \frac{0.000001}{0.009} \]
\[ K_a = 0.0001111 \, M \]
\[ K_a \approx 1.11 \times 10^{-4} \]

### Conclusion
Thus, the \( K_a \) of the solution is approximately \( 1.11 \times 10^{-4} \).

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