The dissociation constant of weak acid `HA` is `1 xx 10^(-5)`. Its concentration is `0.1 M`. pH of that solution is

Options

A. `1.0`
B. `2.0`
C. `3.0`
D. `5.0`

Explanation

To find the pH of the weak acid solution, we can follow these steps:

### Step 1: Write the dissociation equation for the weak acid HA.
The weak acid HA dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Write the expression for the dissociation constant (Ka).
The dissociation constant \( K_a \) for the weak acid is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

### Step 3: Set up the equilibrium expression.
Let \( x \) be the concentration of \( H^+ \) ions that dissociate from the weak acid. At equilibrium, we have:
- \([H^+] = x\)
- \([A^-] = x\)
- \([HA] = 0.1 - x\)

Substituting these into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{0.1 - x} \]
\[ K_a = \frac{x^2}{0.1 - x} \]

### Step 4: Substitute the known values into the equation.
Given that \( K_a = 1 \times 10^{-5} \) and the initial concentration of the acid \( C = 0.1 \, M \):
\[ 1 \times 10^{-5} = \frac{x^2}{0.1 - x} \]

### Step 5: Make an assumption to simplify the equation.
Since \( K_a \) is small, we can assume \( x \) is much smaller than 0.1, so \( 0.1 - x \approx 0.1 \):
\[ 1 \times 10^{-5} = \frac{x^2}{0.1} \]

### Step 6: Solve for \( x \).
Rearranging gives:
\[ x^2 = 1 \times 10^{-5} \times 0.1 \]
\[ x^2 = 1 \times 10^{-6} \]
\[ x = \sqrt{1 \times 10^{-6}} \]
\[ x = 1 \times 10^{-3} \]

### Step 7: Calculate the pH.
The concentration of \( H^+ \) ions is \( 1 \times 10^{-3} \, M \). The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(1 \times 10^{-3}) \]
\[ \text{pH} = 3 \]

### Final Answer:
The pH of the solution is **3**.

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