It has been found that the `pH` of a `0.01 M` solution of an organic acid is `4.15`. Calculate the concentration of the anion, the ionization constant of the acid and its `pK_(a)`.

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations.

### Step 1: Calculate the concentration of H⁺ ions
We are given the pH of the solution, which is 4.15. The concentration of hydrogen ions \([H^+]\) can be calculated using the formula:

\[ \text{pH} = -\log[H^+] \]

Rearranging this gives:

\[ [H^+] = 10^{-\text{pH}} \]

Substituting the given pH value:

\[ [H^+] = 10^{-4.15} \approx 7.08 \times 10^{-5} \, \text{M} \]

### Step 2: Determine the concentration of the anion (A⁻)
For a weak acid (HA) that dissociates into H⁺ and A⁻, we can write the dissociation as:

\[ HA \rightleftharpoons H^+ + A^- \]

From the dissociation, we can see that the concentration of the anion \([A^-]\) will be equal to the concentration of hydrogen ions \([H^+]\) since each molecule of acid that dissociates produces one ion of H⁺ and one ion of A⁻.

Thus:

\[ [A^-] = [H^+] = 7.08 \times 10^{-5} \, \text{M} \]

### Step 3: Calculate the ionization constant \(K_a\)
The ionization constant \(K_a\) can be calculated using the formula:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Where:
- \([H^+] = 7.08 \times 10^{-5} \, \text{M}\)
- \([A^-] = 7.08 \times 10^{-5} \, \text{M}\)
- \([HA]\) is the concentration of the undissociated acid. Since the initial concentration of the acid is 0.01 M and a small amount dissociates, we can approximate:

\[ [HA] \approx 0.01 - [H^+] \approx 0.01 - 7.08 \times 10^{-5} \approx 0.01 \, \text{M} \]

Now substituting the values into the \(K_a\) expression:

\[ K_a = \frac{(7.08 \times 10^{-5})(7.08 \times 10^{-5})}{0.01} \]

Calculating this gives:

\[ K_a = \frac{(7.08 \times 10^{-5})^2}{0.01} \approx 5.05 \times 10^{-7} \]

### Step 4: Calculate the \(pK_a\)
The \(pK_a\) can be calculated using the formula:

\[ pK_a = -\log K_a \]

Substituting the value of \(K_a\):

\[ pK_a = -\log(5.05 \times 10^{-7}) \approx 6.3 \]

### Final Answers:
- Concentration of the anion \([A^-] = 7.08 \times 10^{-5} \, \text{M}\)
- Ionization constant \(K_a = 5.05 \times 10^{-7}\)
- \(pK_a = 6.3\)