The `P^(H)` of a weak mono basic acid is 5. The degree of ionisation of acid in 0.1 M solution is

Options

A. `10^(-4)`
B. `10^(-3)`
C. `10^(-2)`
D. `10^(-1)`

Answer

To solve the problem of finding the degree of ionization of a weak monobasic acid given its pH, we can follow these steps:

### Step 1: Understand the relationship between pH and hydrogen ion concentration
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) in the solution by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 5, we can calculate the concentration of hydrogen ions.

### Step 2: Calculate the concentration of hydrogen ions
Using the pH value:
\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

### Step 3: Set up the dissociation equation for the weak acid
Let the weak monobasic acid be represented as \(HA\). The dissociation of the acid can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Initially, the concentration of \(HA\) is 0.1 M, and the concentrations of \(H^+\) and \(A^-\) are both 0 M.

### Step 4: Define the change in concentration
Let \(\alpha\) be the degree of ionization. At equilibrium, the concentrations will be:
- \([HA] = 0.1(1 - \alpha)\)
- \([H^+] = [A^-] = 0 + \alpha \cdot 0.1 = 0.1\alpha\)

### Step 5: Relate the concentration of hydrogen ions to the degree of ionization
Since we already calculated \([H^+] = 10^{-5} \, \text{M}\), we can set up the equation:
\[ 0.1\alpha = 10^{-5} \]

### Step 6: Solve for \(\alpha\)
Rearranging the equation gives:
\[ \alpha = \frac{10^{-5}}{0.1} = 10^{-4} \]

### Step 7: Conclusion
The degree of ionization of the weak monobasic acid in a 0.1 M solution is:
\[ \alpha = 10^{-4} \]