The ionization constant of propionic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05`M solution and also its pH. What will be its degree of ionization in the solution of `0.01N HCI` ?

Explanation

To solve the problem step-by-step, we will follow these procedures:

### Step 1: Understand the Ionization of Propionic Acid
Propionic acid (HA) ionizes in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Write the Expression for the Ionization Constant (Ka)
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

### Step 3: Set Up the Initial Concentrations
Let:
- The initial concentration of propionic acid, \( C = 0.05 \, M \)
- The degree of ionization, \( \alpha \)

At equilibrium:
- Concentration of \( HA = C(1 - \alpha) \)
- Concentration of \( H^+ = C\alpha \)
- Concentration of \( A^- = C\alpha \)

### Step 4: Substitute into the Ka Expression
Substituting the equilibrium concentrations into the \( K_a \) expression:
\[ K_a = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} \]
\[ K_a = \frac{C^2\alpha^2}{C(1 - \alpha)} \]
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Step 5: Assume \( 1 - \alpha \approx 1 \)
Since \( K_a \) is small, we can assume \( 1 - \alpha \approx 1 \):
\[ K_a \approx C\alpha^2 \]

### Step 6: Substitute Known Values
Substituting the known values into the equation:
\[ 1.32 \times 10^{-5} = 0.05 \cdot \alpha^2 \]

### Step 7: Solve for \( \alpha^2 \)
Rearranging gives:
\[ \alpha^2 = \frac{1.32 \times 10^{-5}}{0.05} \]
\[ \alpha^2 = 2.64 \times 10^{-4} \]

### Step 8: Calculate \( \alpha \)
Taking the square root:
\[ \alpha = \sqrt{2.64 \times 10^{-4}} \]
\[ \alpha \approx 0.0162 \text{ or } 1.62\% \]

### Step 9: Calculate the pH
The concentration of \( H^+ \) is:
\[ [H^+] = C\alpha = 0.05 \cdot 0.0162 = 8.1 \times 10^{-4} \, M \]
Now, calculate the pH:
\[ pH = -\log[H^+] = -\log(8.1 \times 10^{-4}) \]
\[ pH \approx 3.09 \]

### Step 10: Degree of Ionization in 0.01N HCl
In a solution of 0.01N HCl, the concentration of \( H^+ \) is 0.01M. The presence of strong acid will suppress the ionization of the weak acid.

Assuming \( x \) is the amount of dissociation of propionic acid:
\[ K_a = \frac{(0.01 + x)x}{0.05 - x} \]

Since \( x \) is negligible compared to 0.01:
\[ K_a \approx \frac{(0.01)(x)}{0.05} \]
\[ 1.32 \times 10^{-5} = \frac{0.01x}{0.05} \]
\[ x = \frac{1.32 \times 10^{-5} \cdot 0.05}{0.01} \]
\[ x = 6.6 \times 10^{-5} \]

### Step 11: Calculate the Degree of Ionization
The degree of ionization in this case is:
\[ \text{Degree of ionization} = \frac{x}{0.05} = \frac{6.6 \times 10^{-5}}{0.05} \]
\[ \text{Degree of ionization} \approx 0.00132 \text{ or } 0.132\% \]

### Summary of Results:
- Degree of ionization in 0.05M solution: \( 1.62\% \)
- pH of 0.05M solution: \( 3.09 \)
- Degree of ionization in 0.01N HCl: \( 0.132\% \)