The dissociation constant of an acid, HA is `1 x 10^-5` The pH of 0.1 M solution of the acid will be

Options

A. 3
B. 5
C. 4
D. 2

Answer

To find the pH of a 0.1 M solution of the acid HA with a dissociation constant (K_a) of \(1 \times 10^{-5}\), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the acid HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the expression for the dissociation constant
The expression for the dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Let the degree of dissociation be represented by \(\alpha\).

### Step 3: Define initial concentrations
For a 0.1 M solution of HA:
- Initial concentration of HA = \(C = 0.1\) M
- Change in concentration due to dissociation = \(-C\alpha\)
- At equilibrium:
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)
- \([HA] = C(1 - \alpha)\)

### Step 4: Substitute into the \(K_a\) expression
Substituting the equilibrium concentrations into the \(K_a\) expression:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]
Given \(K_a = 1 \times 10^{-5}\) and \(C = 0.1\):
\[ 1 \times 10^{-5} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 5: Assume \(\alpha\) is small
Since \(K_a\) is small, we can assume that \(\alpha\) is small compared to 1, so \(1 - \alpha \approx 1\):
\[ 1 \times 10^{-5} \approx 0.1\alpha^2 \]
\[ \alpha^2 = \frac{1 \times 10^{-5}}{0.1} = 1 \times 10^{-4} \]
\[ \alpha = \sqrt{1 \times 10^{-4}} = 1 \times 10^{-2} \]

### Step 6: Calculate \([H^+]\)
Now, we can find the concentration of hydrogen ions:
\[ [H^+] = C\alpha = 0.1 \times 1 \times 10^{-2} = 1 \times 10^{-3} \text{ M} \]

### Step 7: Calculate pH
Finally, we can calculate the pH:
\[ pH = -\log[H^+] = -\log(1 \times 10^{-3}) = 3 \]

### Final Answer
The pH of the 0.1 M solution of the acid HA is **3**.

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