In aqueous solution the ionization constants for carbonic acid are:
`K_(1)=4.2xx10^(-7)and K_(2)=4.8xx10^(-11)`
Select the correct statement for a saturated `0.034M` solution of the carbonic acid.

Options

A. The concentration of `CO_(3)^(2-)` is 0.034 M.
B. The cocentration of `CO_(3)^(-)` is greater than that of `HCO_(3)^(-).`
C. The concentration of `H^(+)` and `HCO_(3)^(-)` are approximately equal.
D. The concentration of `H^(+)` is double that of `CO_(3)^(2-)`.

Answer

To solve the problem regarding the ionization of carbonic acid in a saturated solution, we can follow these steps:

### Step 1: Understand the Ionization of Carbonic Acid
Carbonic acid (H2CO3) ionizes in two steps:
1. **First Ionization**:
\[ \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \]
with ionization constant \( K_1 = 4.2 \times 10^{-7} \).

2. **Second Ionization**:
\[ \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \]
with ionization constant \( K_2 = 4.8 \times 10^{-11} \).

### Step 2: Analyze the Ionization Constants
Since \( K_1 \) is significantly larger than \( K_2 \), we can conclude that the first ionization step is the predominant reaction. This means that most of the hydrogen ions (\( \text{H}^+ \)) and bicarbonate ions (\( \text{HCO}_3^- \)) in the solution come from the first ionization.

### Step 3: Concentration of Ions
Given that the concentration of the saturated solution of carbonic acid is \( 0.034 \, M \):
- Let \( x \) be the concentration of \( \text{H}^+ \) and \( \text{HCO}_3^- \) produced from the first ionization.
- From the first ionization, we can assume:
\[ [\text{H}^+] \approx [\text{HCO}_3^-] \approx x \]
- The total concentration of carbonic acid that remains un-ionized is approximately \( 0.034 - x \).

### Step 4: Apply the First Ionization Constant
Using the expression for \( K_1 \):
\[ K_1 = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \approx \frac{x^2}{0.034 - x} \]
Since \( K_1 \) is small, we can assume \( x \) is much smaller than \( 0.034 \), simplifying the equation to:
\[ K_1 \approx \frac{x^2}{0.034} \]
Substituting \( K_1 \):
\[ 4.2 \times 10^{-7} = \frac{x^2}{0.034} \]
Solving for \( x \):
\[ x^2 = 4.2 \times 10^{-7} \times 0.034 \]
\[ x^2 = 1.428 \times 10^{-8} \]
\[ x \approx \sqrt{1.428 \times 10^{-8}} \approx 3.78 \times 10^{-5} \, M \]

### Step 5: Concentration of Carbonate Ions
Now, for the second ionization:
Using \( K_2 \):
\[ K_2 = \frac{[\text{H}^+][\text{CO}_3^{2-}]}{[\text{HCO}_3^-]} \approx \frac{x \cdot y}{x} = y \]
where \( y \) is the concentration of \( \text{CO}_3^{2-} \). Since \( K_2 \) is very small, we can assume:
\[ y = K_2 \approx 4.8 \times 10^{-11} \]

### Conclusion
From the analysis, we can summarize:
- The concentration of \( \text{H}^+ \) and \( \text{HCO}_3^- \) are approximately equal.
- The concentration of \( \text{CO}_3^{2-} \) is significantly smaller than both \( \text{H}^+ \) and \( \text{HCO}_3^- \).

### Correct Statement
The correct statement for the saturated \( 0.034 \, M \) solution of carbonic acid is that the concentration of \( \text{H}^+ \) and \( \text{HCO}_3^- \) are approximately equal.