Carbonic acid `(H_(2)CO_(3)),` a diprotic acid has `K_(a1)=4.0xx10^(-7)` and `K_(a2)=7.0xx10^(-11).` What is the `[HCO_(3)^(-)]` of a 0.025 M solution of carbonic acid?

Options

A. `7.8xx10^(-3)`
B. `6.6xx10^(-4)`
C. `10^(-10)`
D. `1.0xx10^(-4)`

Answer

To find the concentration of bicarbonate ion \([HCO_3^-]\) in a 0.025 M solution of carbonic acid \((H_2CO_3)\), we can follow these steps:

### Step 1: Write the dissociation reactions
Carbonic acid is a diprotic acid, which means it can donate two protons (H⁺). The dissociation reactions are:

1. \( H_2CO_3 \rightleftharpoons HCO_3^- + H^+ \) (with \( K_{a1} = 4.0 \times 10^{-7} \))
2. \( HCO_3^- \rightleftharpoons CO_3^{2-} + H^+ \) (with \( K_{a2} = 7.0 \times 10^{-11} \))

### Step 2: Set up the equilibrium expressions
For the first dissociation reaction, we can express the equilibrium constant \( K_{a1} \) as follows:

\[ K_{a1} = \frac{[HCO_3^-][H^+]}{[H_2CO_3]} \]

Let \( C \) be the initial concentration of \( H_2CO_3 \) (0.025 M). At equilibrium, if \( x \) is the amount that dissociates, we have:

- \([H_2CO_3] = C - x\)
- \([HCO_3^-] = x\)
- \([H^+] = x\)

Thus, the expression becomes:

\[ K_{a1} = \frac{x \cdot x}{C - x} = \frac{x^2}{0.025 - x} \]

### Step 3: Make approximations
Since \( K_{a1} \) is much larger than \( K_{a2} \), we can assume that \( x \) is small compared to \( C \), allowing us to approximate \( C - x \approx C \):

\[ K_{a1} \approx \frac{x^2}{0.025} \]

### Step 4: Substitute the known values
Substituting the value of \( K_{a1} \):

\[ 4.0 \times 10^{-7} = \frac{x^2}{0.025} \]

### Step 5: Solve for \( x \)
Rearranging gives:

\[ x^2 = 4.0 \times 10^{-7} \times 0.025 \]

Calculating the right side:

\[ x^2 = 1.0 \times 10^{-8} \]

Taking the square root:

\[ x = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \]

### Step 6: Determine the concentration of bicarbonate ion
Since \( x \) represents the concentration of \( HCO_3^- \):

\[ [HCO_3^-] = 1.0 \times 10^{-4} \, \text{M} \]

### Final Answer
The concentration of bicarbonate ion \([HCO_3^-]\) in a 0.025 M solution of carbonic acid is \( \mathbf{1.0 \times 10^{-4} \, M} \).

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