The dissociation constant of an acid is `1xx10^(-5)`. The `pH` of its `0.1 M` solution will be approximately

Options

A. Five
B. Four
C. Three
D. One

Answer

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

### Step 1: Understand the relationship between Ka, [H⁺], and concentration (C)
The dissociation constant (Ka) of an acid is related to the concentration of hydrogen ions ([H⁺]) produced when the acid dissociates. The relationship can be expressed as:
\[ K_a = \frac{[H^+]^2}{C - [H^+]} \]
For weak acids, we can assume that \([H^+]\) is much smaller than the initial concentration (C), so we can approximate:
\[ K_a \approx \frac{[H^+]^2}{C} \]

### Step 2: Substitute the known values into the equation
Given:
- \(K_a = 1 \times 10^{-5}\)
- \(C = 0.1\) M

Substituting these values into the equation gives:
\[ 1 \times 10^{-5} = \frac{[H^+]^2}{0.1} \]

### Step 3: Rearrange the equation to solve for [H⁺]
Rearranging the equation to find \([H^+]\):
\[ [H^+]^2 = 1 \times 10^{-5} \times 0.1 \]
\[ [H^+]^2 = 1 \times 10^{-6} \]

### Step 4: Take the square root to find [H⁺]
Taking the square root of both sides:
\[ [H^+] = \sqrt{1 \times 10^{-6}} = 1 \times 10^{-3} \text{ M} \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(1 \times 10^{-3}) = -(-3) = 3 \]

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

---