Calculate the pH of 0.15 M solution of hypochlorous acid HClO `(K_(a)=9.6xx10^(-6))`.

Answer

To calculate the pH of a 0.15 M solution of hypochlorous acid (HClO) with a given \( K_a = 9.6 \times 10^{-6} \), we can follow these steps:

### Step 1: Write the dissociation equation
Hypochlorous acid (HClO) dissociates in water as follows:
\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \]

### Step 2: Set up the expression for \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \]

### Step 3: Define the concentrations at equilibrium
Let \( x \) be the concentration of \( \text{H}^+ \) ions produced at equilibrium. Initially, the concentration of HClO is 0.15 M, and at equilibrium:
- \([\text{HClO}] = 0.15 - x\)
- \([\text{H}^+] = x\)
- \([\text{ClO}^-] = x\)

### Step 4: Substitute into the \( K_a \) expression
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{0.15 - x} = \frac{x^2}{0.15 - x} \]

### Step 5: Assume \( x \) is small
Since \( K_a \) is small, we can assume that \( x \) is much smaller than 0.15, so:
\[ K_a \approx \frac{x^2}{0.15} \]

### Step 6: Solve for \( x \)
Now, substituting the value of \( K_a \):
\[ 9.6 \times 10^{-6} = \frac{x^2}{0.15} \]
\[ x^2 = 9.6 \times 10^{-6} \times 0.15 \]
\[ x^2 = 1.44 \times 10^{-6} \]
\[ x = \sqrt{1.44 \times 10^{-6}} \]
\[ x \approx 0.0012 \, \text{M} \]

### Step 7: Calculate the pH
The concentration of \( \text{H}^+ \) ions is \( 0.0012 \, \text{M} \). Now, we can calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] \]
\[ \text{pH} = -\log(0.0012) \]
\[ \text{pH} \approx 2.92 \]

### Final Answer
The pH of the 0.15 M solution of hypochlorous acid is approximately **2.92**.

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