What is the `[H^(+)]` in 0.40 M solution of , HOCl , `K_a = 3.5xx10^(-8)` ?

Options

A. `1.4xx10^(-8)M`
B. `1.2xx10^(-4)M`
C. `1.9xx10^(-4)M`
D. `3.7xx10^(-4)M`

Answer

To find the concentration of hydrogen ions \([H^+]\) in a 0.40 M solution of HOCl, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of hypochlorous acid (HOCl) in water can be represented as:
\[ \text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^- \]

### Step 2: Set up the initial concentrations
At the start (t = 0), we have:
- \([HOCl] = C = 0.40 \, M\)
- \([H^+] = 0\)
- \([OCl^-] = 0\)

### Step 3: Define the change in concentrations
Let \(\alpha\) be the degree of dissociation. At equilibrium, the concentrations will be:
- \([HOCl] = C - C\alpha = 0.40 - 0.40\alpha\)
- \([H^+] = C\alpha = 0.40\alpha\)
- \([OCl^-] = C\alpha = 0.40\alpha\)

### Step 4: Write the expression for \(K_a\)
The acid dissociation constant \(K_a\) for HOCl is given by:
\[ K_a = \frac{[H^+][OCl^-]}{[HOCl]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.40\alpha)(0.40\alpha)}{0.40 - 0.40\alpha} \]

### Step 5: Simplify the expression
This simplifies to:
\[ K_a = \frac{0.16\alpha^2}{0.40(1 - \alpha)} \]
Given that \(K_a = 3.5 \times 10^{-8}\), we can set up the equation:
\[ 3.5 \times 10^{-8} = \frac{0.16\alpha^2}{0.40(1 - \alpha)} \]

### Step 6: Assume \(\alpha\) is small
Since \(K_a\) is small, we can assume that \(\alpha\) is much less than 1, which means \(1 - \alpha \approx 1\). Thus, the equation simplifies to:
\[ 3.5 \times 10^{-8} = \frac{0.16\alpha^2}{0.40} \]

### Step 7: Solve for \(\alpha^2\)
Rearranging gives:
\[ \alpha^2 = \frac{3.5 \times 10^{-8} \times 0.40}{0.16} \]
Calculating the right side:
\[ \alpha^2 = \frac{1.4 \times 10^{-8}}{0.16} = 8.75 \times 10^{-8} \]

### Step 8: Calculate \(\alpha\)
Taking the square root:
\[ \alpha = \sqrt{8.75 \times 10^{-8}} \approx 9.35 \times 10^{-4} \]

### Step 9: Calculate \([H^+]\)
Now, we can find \([H^+]\):
\[ [H^+] = C\alpha = 0.40 \times 9.35 \times 10^{-4} \approx 3.74 \times 10^{-4} \, M \]

### Conclusion
Thus, the concentration of hydrogen ions \([H^+]\) in a 0.40 M solution of HOCl is approximately \(3.74 \times 10^{-4} \, M\).

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