An acid solution of 0.005 M has a pH of 5. The degree of ionisation of acid is

Options

A. `0.1 xx 10^(-2)`
B. `0.2 xx 10^(-2)`
C. `0.5 xx 10^(-2)`
D. `0.6 xx 10^(-2)`

Answer

To find the degree of ionization of the acid, 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 \([H^+]\) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 5, we can find the concentration of hydrogen ions:
\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \text{ M} \]

### Step 2: Set up the dissociation equation for the acid.
Assume the acid is represented as HA. The dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Initially, the concentration of HA is \(C = 0.005 \text{ M}\). Upon dissociation, if \(\alpha\) is the degree of ionization, the concentrations at equilibrium will be:
- \([HA] = C(1 - \alpha)\)
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)

### Step 3: Relate the concentration of hydrogen ions to the degree of ionization.
From the dissociation, we know that:
\[ [H^+] = C\alpha \]
Substituting the values we have:
\[ 10^{-5} = 0.005 \cdot \alpha \]

### Step 4: Solve for \(\alpha\).
Rearranging the equation gives:
\[ \alpha = \frac{10^{-5}}{0.005} \]
Calculating this:
\[ \alpha = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{1}{5} \times 10^{-2} = 0.2 \times 10^{-2} \]
Thus, \(\alpha = 0.002\) or 0.2%.

### Step 5: Conclusion
The degree of ionization of the acid is \(0.002\) or \(0.2\%\).

---