HClO is a weak acid. The concentration of `H^(+)` ions in 0.1 M solution of HClO `(K_(a) = 5 xx 10^(-8))` will be equal to

Options

A. `7.07 xx 10^(-5) m`
B. `5 xx 10^(-9) m`
C. `5 xx 10^(-7) m`
D. `7 xx 10^(-4) m`

Explanation

To find the concentration of \( H^+ \) ions in a 0.1 M solution of HClO, we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the weak acid HClO can be represented as:
\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \]

### Step 2: Set up the equilibrium expression
The equilibrium constant \( K_a \) for the dissociation of HClO is given by:
\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \]
Let \( x \) be the concentration of \( H^+ \) ions at equilibrium. Since HClO is a weak acid, we can assume that it only partially dissociates.

### Step 3: Define the initial concentrations
Initially, the concentration of HClO is 0.1 M, and the concentrations of \( H^+ \) and \( ClO^- \) are both 0 M:
- \([\text{HClO}] = 0.1\)
- \([\text{H}^+] = 0\)
- \([\text{ClO}^-] = 0\)

At equilibrium, the concentrations will be:
- \([\text{HClO}] = 0.1 - x\)
- \([\text{H}^+] = x\)
- \([\text{ClO}^-] = x\)

### Step 4: Substitute into the \( K_a \) expression
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]
Given that \( K_a = 5 \times 10^{-8} \), we can set up the equation:
\[ 5 \times 10^{-8} = \frac{x^2}{0.1 - x} \]

### Step 5: Assume \( x \) is small compared to 0.1 M
Since \( HClO \) is a weak acid, we can assume that \( x \) is much smaller than 0.1 M, allowing us to simplify the equation:
\[ 5 \times 10^{-8} \approx \frac{x^2}{0.1} \]
This simplifies to:
\[ x^2 = 5 \times 10^{-8} \times 0.1 = 5 \times 10^{-9} \]

### Step 6: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \sqrt{5 \times 10^{-9}} \approx 7.07 \times 10^{-5} \text{ M} \]

### Conclusion
The concentration of \( H^+ \) ions in a 0.1 M solution of HClO is approximately \( 7.07 \times 10^{-5} \) M.

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