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 hypochlorous acid (HOCl) with a dissociation constant \(K_a = 3.5 \times 10^{-8}\), we can follow these steps:

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

### Step 2: Set up the initial concentrations
Let the initial concentration of HOCl be \(C = 0.40 \, \text{M}\). At equilibrium, if \(x\) is the amount that dissociates, we have:
- \([H^+] = x\)
- \([OCl^-] = x\)
- \([HOCl] = C - x = 0.40 - x\)

### Step 3: Write the expression for \(K_a\)
The expression for the acid dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][OCl^-]}{[HOCl]} \]
Substituting the equilibrium concentrations into the expression, we get:
\[ K_a = \frac{x \cdot x}{0.40 - x} = \frac{x^2}{0.40 - x} \]

### Step 4: Substitute the value of \(K_a\)
Substituting \(K_a = 3.5 \times 10^{-8}\):
\[ 3.5 \times 10^{-8} = \frac{x^2}{0.40 - x} \]

### Step 5: Assume \(x\) is small
Since \(K_a\) is very small, we can assume that \(x\) is much smaller than 0.40 M, allowing us to simplify \(0.40 - x \approx 0.40\):
\[ 3.5 \times 10^{-8} \approx \frac{x^2}{0.40} \]

### Step 6: Solve for \(x^2\)
Rearranging gives:
\[ x^2 = 3.5 \times 10^{-8} \times 0.40 \]
\[ x^2 = 1.4 \times 10^{-8} \]

### Step 7: Calculate \(x\)
Taking the square root of both sides:
\[ x = \sqrt{1.4 \times 10^{-8}} \approx 1.2 \times 10^{-4} \, \text{M} \]

### Step 8: Conclusion
Thus, the concentration of hydrogen ions \([H^+]\) in the solution is:
\[ [H^+] = 1.2 \times 10^{-4} \, \text{M} \]