If the `pH` of `0.26 M HNO_(2)` is `2.5`, what will be its dissociation constant.

Explanation

To find the dissociation constant (Ka) of the weak acid HNO2 given its concentration and pH, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions from pH
The pH is given as 2.5. We can find the concentration of hydrogen ions [H⁺] using the formula:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the value of pH:
\[ [H^+] = 10^{-2.5} \]

### Step 2: Calculate [H⁺]
Calculating the above expression:
\[ [H^+] = 10^{-2.5} \approx 0.00316 \, \text{M} \]

### Step 3: Set up the equilibrium expression
For the dissociation of HNO2:
\[ HNO_2 \rightleftharpoons H^+ + NO_2^- \]
Let the initial concentration of HNO2 be \( C = 0.26 \, \text{M} \). At equilibrium, the concentration of HNO2 will be:
\[ [HNO_2] = C - [H^+] \]
Thus,
\[ [HNO_2] = 0.26 - 0.00316 \approx 0.25684 \, \text{M} \]

### Step 4: Write the expression for Ka
The dissociation constant (Ka) can be expressed as:
\[ K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \]
Since the concentration of H⁺ and NO2⁻ produced will be equal:
\[ K_a = \frac{[H^+]^2}{[HNO_2]} \]

### Step 5: Substitute the values into the Ka expression
Substituting the values we have:
\[ K_a = \frac{(0.00316)^2}{0.25684} \]

### Step 6: Calculate Ka
Calculating the above expression:
\[ K_a = \frac{0.0000100}{0.25684} \approx 3.90 \times 10^{-5} \]

### Final Answer
Thus, the dissociation constant \( K_a \) for HNO2 is approximately:
\[ K_a \approx 3.90 \times 10^{-5} \]

---