Which is the constant equation for hydrolysis constant of `NH_(4)CN`?

Options

A. `K_(a) xx (K_(b))/((K_(w))`
B. `sqrt((K_(w))/(K_(a)))`
C. `(K_(b))/(K_(a) xx K_(w))`
D. `(K_(w))/(K_(a) xx K_(b))`

Answer

To find the hydrolysis constant of ammonium cyanide (NH₄CN), we need to understand the nature of the salt and the reactions involved in its hydrolysis. Here’s a step-by-step solution:

### Step 1: Identify the Components
Ammonium cyanide (NH₄CN) is a salt formed from a weak acid (hydrogen cyanide, HCN) and a weak base (ammonium hydroxide, NH₄OH).

### Step 2: Write the Hydrolysis Reaction
When NH₄CN undergoes hydrolysis, it can be represented by the following reaction:
\[ \text{NH}_4^+ + \text{CN}^- + \text{H}_2O \rightleftharpoons \text{NH}_4OH + \text{HCN} \]

### Step 3: Write the Dissociation Constants
For the weak base (ammonium hydroxide):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4OH]} \]

For the weak acid (hydrogen cyanide):
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

### Step 4: Combine the Reactions
The hydrolysis of NH₄CN can be thought of as the combination of the dissociation of NH₄OH and HCN. Thus, we can express the hydrolysis constant (K_h) as:
\[ K_h = \frac{[\text{NH}_4OH][\text{HCN}]}{[\text{NH}_4^+][\text{CN}^-]} \]

### Step 5: Relate Hydrolysis Constant to Ionization Constants
The hydrolysis constant can be related to the ionization constants (Kw, Ka, and Kb) as follows:
\[ K_h = \frac{K_w}{K_a \cdot K_b} \]
Where:
- \( K_w \) is the ion product of water (1.0 x 10⁻¹⁴ at 25°C),
- \( K_a \) is the dissociation constant of the weak acid (HCN),
- \( K_b \) is the dissociation constant of the weak base (NH₄OH).

### Conclusion
Thus, the constant equation for the hydrolysis constant of NH₄CN is:
\[ K_h = \frac{K_w}{K_a \cdot K_b} \]