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

Options

A. `5.5xx10^(-9)`
B. `5.5xx10^(-8)`
C. `7.0xx10^(-9)`
D. `7.0xx10^(-11)`

Answer

To find the concentration of carbonate ions \([CO_3^{2-}]\) in a 0.025 M solution of carbonic acid \((H_2CO_3)\), we need to consider the dissociation of the diprotic acid in two steps:

1. **Dissociation of Carbonic Acid**:
\[ H_2CO_3 \rightleftharpoons HCO_3^- + H^+ \quad (K_{a1} = 4.0 \times 10^{-7}) \]
\[ HCO_3^- \rightleftharpoons CO_3^{2-} + H^+ \quad (K_{a2} = 7.0 \times 10^{-11}) \]

2. **Initial Concentration**:
The initial concentration of \(H_2CO_3\) is given as 0.025 M. At equilibrium, we will denote the change in concentration due to dissociation as \(x\) for the first dissociation and \(y\) for the second dissociation.

3. **Equilibrium Concentrations**:
- For the first dissociation:
- \([H_2CO_3] = 0.025 - x\)
- \([HCO_3^-] = x\)
- \([H^+] = x\)

- For the second dissociation:
- \([HCO_3^-] = x - y\)
- \([CO_3^{2-}] = y\)
- \([H^+] = x + y\)

4. **Assumption**:
Since \(K_{a1}\) is much greater than \(K_{a2}\), we can assume that \(x\) (the concentration of \(H^+\) from the first dissociation) is much greater than \(y\) (the concentration of \(CO_3^{2-}\) from the second dissociation). Therefore, we can neglect \(y\) in the expressions for \(HCO_3^-\) and \(H^+\).

5. **Using \(K_{a2}\)**:
The expression for \(K_{a2}\) is given by:
\[ K_{a2} = \frac{[CO_3^{2-}][H^+]}{[HCO_3^-]} \]
Substituting the equilibrium concentrations:
\[ K_{a2} = \frac{y(x)}{(x)} \quad \text{(since we neglect } y \text{)} \]
This simplifies to:
\[ K_{a2} = y \]

6. **Substituting the Value of \(K_{a2}\)**:
Given that \(K_{a2} = 7.0 \times 10^{-11}\), we can conclude:
\[ [CO_3^{2-}] = y = 7.0 \times 10^{-11} \text{ M} \]

Thus, the concentration of carbonate ions \([CO_3^{2-}]\) in a 0.025 M solution of carbonic acid is \(7.0 \times 10^{-11} \text{ M}\).