The pH of 0.1 M solution of an organic acid is 3.0. Calculate the dissociation constant of the acid.

Explanation

To solve the problem of calculating the dissociation constant (Ka) of an organic acid given its pH and concentration, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The concentration of the organic acid (HA) = 0.1 M
- The pH of the solution = 3.0

### Step 2: Calculate the concentration of H⁺ ions
The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the formula:
\[ \text{pH} = -\log[\text{H}^+] \]

Given that the pH is 3.0, we can find the concentration of H⁺ ions:
\[ 3.0 = -\log[\text{H}^+] \]

To find [H⁺]:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 3: Set up the dissociation equation
The dissociation of the organic acid (HA) can be represented as:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

From the dissociation, we can see that:
- The concentration of H⁺ ions produced is equal to the concentration of A⁻ ions produced, which is also [H⁺] = \(10^{-3} \, \text{M}\).

### Step 4: Determine the equilibrium concentrations
Initially, we have:
- [HA] = 0.1 M (initial concentration)
- [H⁺] = 0 M (initial concentration)
- [A⁻] = 0 M (initial concentration)

At equilibrium, we have:
- [H⁺] = \(10^{-3} \, \text{M}\)
- [A⁻] = \(10^{-3} \, \text{M}\)
- [HA] = Initial concentration - [H⁺] = \(0.1 - 10^{-3} \approx 0.1 \, \text{M}\) (since \(10^{-3}\) is negligible compared to 0.1)

### Step 5: Write the expression for the dissociation constant (Ka)
The dissociation constant (Ka) is given by the expression:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

Substituting the equilibrium concentrations into the expression:
\[ K_a = \frac{(10^{-3})(10^{-3})}{0.1} \]

### Step 6: Calculate Ka
Now we can calculate Ka:
\[ K_a = \frac{10^{-6}}{0.1} = 10^{-5} \]

### Final Answer
The dissociation constant (Ka) of the organic acid is:
\[ K_a = 10^{-5} \]

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