`[H^(+)]` concentration in 0.01 M `H_2O_2` solution `(K_(a_(1))=3xx10^(-12) and K_(a_2)~~0)` is `xxM`.Fill first two digits of `10^(8)x` as answer.

Explanation

To solve the problem of finding the concentration of \([H^+]\) in a 0.01 M \(H_2O_2\) solution, we will follow these steps:

### Step 1: Understand the dissociation of \(H_2O_2\)
The dissociation of hydrogen peroxide (\(H_2O_2\)) can be represented as:
\[ H_2O_2 \rightleftharpoons H^+ + HO_2^- \]
This reaction has an acid dissociation constant \(K_{a1}\).

### Step 2: Identify the given values
From the problem, we know:
- \(C_0 = 0.01 \, M\) (initial concentration of \(H_2O_2\))
- \(K_{a1} = 3 \times 10^{-12}\)
- \(K_{a2} \approx 0\) (indicating that the second dissociation is negligible)

### Step 3: Consider the dissociation of water
Water also dissociates, contributing to the \([H^+]\) concentration:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
The dissociation constant for water is:
\[ K_w = 10^{-14} \]

### Step 4: Set up the equation for \([H^+]\)
The total concentration of \([H^+]\) can be expressed as:
\[ [H^+] = \sqrt{K_{a1} \cdot C_0} + \sqrt{K_w} \]

### Step 5: Calculate \([H^+]\)
Substituting the known values into the equation:
1. Calculate \(\sqrt{K_{a1} \cdot C_0}\):
\[ \sqrt{K_{a1} \cdot C_0} = \sqrt{(3 \times 10^{-12}) \cdot (0.01)} = \sqrt{3 \times 10^{-14}} = \sqrt{3} \times 10^{-7} \]
Approximating \(\sqrt{3} \approx 1.732\), we get:
\[ \sqrt{K_{a1} \cdot C_0} \approx 1.732 \times 10^{-7} \, M \]

2. Calculate \(\sqrt{K_w}\):
\[ \sqrt{K_w} = \sqrt{10^{-14}} = 10^{-7} \, M \]

3. Combine the two contributions:
\[ [H^+] \approx (1.732 \times 10^{-7}) + (1.0 \times 10^{-7}) \approx 2.732 \times 10^{-7} \, M \]

### Step 6: Express \([H^+]\) in the required format
To find \(x\) such that \(10^8 \times x\):
\[ x = 2.732 \times 10^{-7} \, M \]
Calculating \(10^8 \times x\):
\[ 10^8 \times (2.732 \times 10^{-7}) = 2.732 \times 10^1 = 27.32 \]

### Step 7: Identify the first two digits
The first two digits of \(27.32\) are \(27\).

### Final Answer
The answer is **27**.