The dissociations constant of HCN is ` 7.24 xx 10 ^(-10) ` Calculate its degree of dissociation and ` [H_3O^(+) ]` in 0.01 M solutions.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- The dissociation constant \( K_a \) of HCN is \( 7.24 \times 10^{-10} \).
- The concentration of the HCN solution \( C \) is \( 0.01 \, \text{M} \).

### Step 2: Write the dissociation equation
The dissociation of HCN in water can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CN}^- \]

### Step 3: Write the expression for the dissociation constant
The dissociation constant \( K_a \) for this reaction can be expressed as:
\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} \]

### Step 4: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of HCN. At equilibrium:
- The concentration of HCN will be \( C(1 - \alpha) \).
- The concentration of \( \text{H}_3\text{O}^+ \) and \( \text{CN}^- \) will be \( C\alpha \).

### Step 5: Substitute into the \( K_a \) expression
Substituting the equilibrium concentrations into the \( K_a \) expression gives:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]

### Step 6: Approximate for small \( \alpha \)
Since HCN is a weak acid, we can assume \( \alpha \) is small, allowing us to neglect \( \alpha \) in \( 1 - \alpha \):
\[ K_a \approx C \alpha^2 \]

### Step 7: Solve for \( \alpha \)
Rearranging gives:
\[ \alpha^2 = \frac{K_a}{C} \]
Substituting the known values:
\[ \alpha^2 = \frac{7.24 \times 10^{-10}}{0.01} = 7.24 \times 10^{-8} \]
Taking the square root:
\[ \alpha = \sqrt{7.24 \times 10^{-8}} \approx 2.69 \times 10^{-4} \]

### Step 8: Calculate the concentration of \( \text{H}_3\text{O}^+ \)
The concentration of \( \text{H}_3\text{O}^+ \) at equilibrium is given by:
\[ [\text{H}_3\text{O}^+] = C \alpha \]
Substituting the values:
\[ [\text{H}_3\text{O}^+] = 0.01 \times 2.69 \times 10^{-4} \approx 2.69 \times 10^{-6} \, \text{M} \]

### Final Results
- Degree of dissociation \( \alpha \approx 2.69 \times 10^{-4} \)
- Concentration of \( \text{H}_3\text{O}^+ \approx 2.69 \times 10^{-6} \, \text{M} \)