The pH of a solution obtained by dissolving 0.025 mole of an acid (HA) in 250 ml of the aqueous solution was found to be 4.0. Calculate the dissociation constant of HA.

Answer

To calculate the dissociation constant (Ka) of the weak acid (HA) given the pH of the solution, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from pH
The pH of the solution is given as 4.0. We can use the formula:
\[ \text{pH} = -\log[H^+] \]
To find [H⁺]:
\[ [H^+] = 10^{-\text{pH}} = 10^{-4} \, \text{M} \]

### Step 2: Calculate the initial concentration of the acid (HA)
We know that 0.025 moles of HA is dissolved in 250 mL of solution. First, convert the volume from mL to L:
\[ 250 \, \text{mL} = 0.250 \, \text{L} \]
Now, calculate the concentration (C) of HA:
\[ C = \frac{\text{moles}}{\text{volume}} = \frac{0.025 \, \text{moles}}{0.250 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Set up the dissociation equation
The dissociation of the acid HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Let α be the degree of dissociation. At equilibrium:
- The concentration of H⁺ ions = Cα = \( 10^{-4} \, \text{M} \)
- The concentration of A⁻ ions = Cα = \( 10^{-4} \, \text{M} \)
- The concentration of undissociated HA = C(1 - α)

### Step 4: Relate α to the concentration of H⁺
From the previous step, we know:
\[ Cα = 10^{-4} \]
Substituting C = 0.1 M:
\[ 0.1α = 10^{-4} \]
\[ α = \frac{10^{-4}}{0.1} = 10^{-3} \]

### Step 5: Calculate the dissociation constant (Ka)
The expression for the dissociation constant (Ka) is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the values:
\[ K_a = \frac{(Cα)(Cα)}{C(1 - α)} = \frac{(0.1)(10^{-3})(0.1)(10^{-3})}{0.1(1 - 10^{-3})} \]
Since α is very small, we can approximate:
\[ 1 - α \approx 1 \]
Thus:
\[ K_a = \frac{(0.1)(10^{-3})^2}{0.1} = (10^{-3})^2 = 10^{-6} \]

### Final Answer
The dissociation constant (Ka) of the acid HA is:
\[ K_a = 10^{-6} \]

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