`K_(a)` for ascorbic acid `(HASc)` is `5 xx 10^(-5)`. Calculate the `[H^(o+)]` in an aqueous solution in which the concentration of `Asc^(Theta)` ions is `0.02M`.

Options

A. `2 xx 10^(-6)`
B. `2 xx 10^(-7)`
C. `5 xx 10^(-9)`
D. `5 xx 10^(-10)`

Explanation

To solve the problem, we need to find the concentration of hydrogen ions \([H^+]\) in an aqueous solution where the concentration of ascorbate ions \([Asc^-]\) is given as \(0.02 \, M\). The dissociation constant \(K_a\) for ascorbic acid \((HA)\) is \(5 \times 10^{-5}\).

### Step-by-Step Solution:

1. **Write the Hydrolysis Reaction**:
The ascorbate ion \((Asc^-)\) can undergo hydrolysis in water:
\[ Asc^- + H_2O \rightleftharpoons HA + OH^- \]

2. **Set Up the Expression for Hydrolysis**:
The equilibrium expression for the hydrolysis of the ascorbate ion can be written as:
\[ K_b = \frac{[HA][OH^-]}{[Asc^-]} \]
Where \(K_b\) is the base dissociation constant for the ascorbate ion. We can relate \(K_a\) and \(K_b\) using the equation:
\[ K_w = K_a \cdot K_b \]
Where \(K_w\) (the ion product of water) at room temperature is \(1 \times 10^{-14}\).

3. **Calculate \(K_b\)**:
Rearranging the equation gives us:
\[ K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} = 2 \times 10^{-10} \]

4. **Substitute Values into the Hydrolysis Expression**:
Let \(x\) be the concentration of \([OH^-]\) produced at equilibrium. The concentration of \([HA]\) will also be \(x\), and the concentration of \([Asc^-]\) will be:
\[ [Asc^-] = 0.02 - x \approx 0.02 \quad (\text{since } x \text{ is small}) \]
Thus, the expression becomes:
\[ K_b = \frac{x^2}{0.02} \]
Substituting the value of \(K_b\):
\[ 2 \times 10^{-10} = \frac{x^2}{0.02} \]

5. **Solve for \(x\)**:
Rearranging gives:
\[ x^2 = 2 \times 10^{-10} \times 0.02 = 4 \times 10^{-12} \]
Taking the square root:
\[ x = \sqrt{4 \times 10^{-12}} = 2 \times 10^{-6} \]
Therefore, \([OH^-] = 2 \times 10^{-6} \, M\).

6. **Calculate \([H^+]\)**:
Now, we can find \([H^+]\) using the relationship:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{2 \times 10^{-6}} = 5 \times 10^{-9} \, M \]

### Final Answer:
The concentration of hydrogen ions \([H^+]\) in the solution is:
\[ [H^+] = 5 \times 10^{-9} \, M \]