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 (Ka) of \(1 \times 10^{-5}\), we can follow these steps:

### Step 1: Understand the dissociation of the acid
The acid HA dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
Let the initial concentration of HA be \(C = 0.1 \, M\).

### Step 2: Set up the expression for the dissociation constant
The expression for the dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
At equilibrium, if \(x\) is the amount that dissociates, we have:
- \([H^+] = x\)
- \([A^-] = x\)
- \([HA] = C - x \approx C\) (since \(x\) will be small compared to \(C\))

Thus, we can approximate:
\[ K_a = \frac{x \cdot x}{C} = \frac{x^2}{C} \]

### Step 3: Substitute the known values into the equation
Given \(K_a = 1 \times 10^{-5}\) and \(C = 0.1\):
\[ 1 \times 10^{-5} = \frac{x^2}{0.1} \]

### Step 4: Solve for \(x\)
Rearranging the equation gives:
\[ x^2 = 1 \times 10^{-5} \times 0.1 \]
\[ x^2 = 1 \times 10^{-6} \]
Taking the square root of both sides:
\[ x = \sqrt{1 \times 10^{-6}} = 1 \times 10^{-3} \]

### Step 5: Calculate the concentration of hydrogen ions
Thus, the concentration of hydrogen ions \([H^+]\) is:
\[ [H^+] = x = 1 \times 10^{-3} \, M \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(1 \times 10^{-3}) \]
\[ \text{pH} = 3 \]

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

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