The pH of `1 xx 10^(-3) M H_(2)O_(2)` solution `(K_(a)=2.2 xx 10^(-12))` is

Options

A. `approx 3`
B. slightly less than 7
C. slightly greater than 7
D. `=7`

Answer

To find the pH of a `1 x 10^(-3) M H2O2` solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Concentration of H2O2: `C = 1 x 10^(-3) M`
- Acid dissociation constant (Ka) for H2O2: `Ka = 2.2 x 10^(-12)`

2. **Set Up the Dissociation Equation**:
- The dissociation of hydrogen peroxide (H2O2) can be represented as:
\[ H2O2 \rightleftharpoons H^+ + HO2^- \]
- Initially, we have:
- [H2O2] = `C = 1 x 10^(-3) M`
- [H+] = 0
- [HO2^-] = 0

3. **Change in Concentration at Equilibrium**:
- Let `x` be the amount of H2O2 that dissociates at equilibrium.
- At equilibrium, the concentrations will be:
- [H2O2] = `C - x = 1 x 10^(-3) - x`
- [H+] = `x`
- [HO2^-] = `x`

4. **Write the Expression for Ka**:
- The expression for the acid dissociation constant is:
\[ Ka = \frac{[H^+][HO2^-]}{[H2O2]} = \frac{x^2}{1 x 10^{-3} - x} \]
- Substituting the value of Ka:
\[ 2.2 x 10^{-12} = \frac{x^2}{1 x 10^{-3} - x} \]

5. **Assume x is Small**:
- Since Ka is very small, we can assume that `x` is much smaller than `1 x 10^(-3)`, thus:
\[ 1 x 10^{-3} - x \approx 1 x 10^{-3} \]
- This simplifies our equation to:
\[ 2.2 x 10^{-12} = \frac{x^2}{1 x 10^{-3}} \]

6. **Solve for x**:
- Rearranging gives:
\[ x^2 = 2.2 x 10^{-12} \times 1 x 10^{-3} \]
\[ x^2 = 2.2 x 10^{-15} \]
\[ x = \sqrt{2.2 x 10^{-15}} = 4.69 x 10^{-8} M \]

7. **Calculate pH**:
- The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
- Substituting the value of `x`:
\[ pH = -\log(4.69 x 10^{-8}) \approx 7.33 \]

### Final Answer:
The pH of the `1 x 10^(-3) M H2O2` solution is approximately **7.33**.