When salt `NH_(4)Cl` is hydrolysed at `25^(@)C`, the `pH` is

Options

A. 7
B. `lt7`
C. `gt7`
D. 0

Explanation

To find the pH of the hydrolysis of the salt \( NH_4Cl \) at \( 25^\circ C \), we can follow these steps:

### Step 1: Understand the Hydrolysis of \( NH_4Cl \)
When \( NH_4Cl \) is dissolved in water, it dissociates into \( NH_4^+ \) and \( Cl^- \) ions. The \( Cl^- \) ion is the conjugate base of a strong acid (HCl) and does not affect the pH significantly. However, the \( NH_4^+ \) ion is the conjugate acid of a weak base (NH3) and will undergo hydrolysis.

### Step 2: Write the Hydrolysis Reaction
The hydrolysis of the ammonium ion can be represented as:
\[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \]

### Step 3: Determine the \( K_a \) of \( NH_4^+ \)
The equilibrium constant for this reaction is the acid dissociation constant \( K_a \) for \( NH_4^+ \). We can find \( K_a \) using the relationship:
\[ K_a \cdot K_b = K_w \]
Where:
- \( K_w \) is the ion product of water (\( 1.0 \times 10^{-14} \) at \( 25^\circ C \)).
- \( K_b \) is the base dissociation constant for \( NH_3 \) (approximately \( 1.8 \times 10^{-5} \)).

Calculating \( K_a \):
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

### Step 4: Set Up the Equilibrium Expression
Let \( x \) be the concentration of \( H_3O^+ \) produced at equilibrium. The equilibrium expression for the hydrolysis reaction is:
\[ K_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} = \frac{x \cdot x}{C - x} \approx \frac{x^2}{C} \]
where \( C \) is the initial concentration of \( NH_4Cl \).

### Step 5: Solve for \( x \)
Assuming \( C \) is not very small compared to \( x \), we can simplify:
\[ K_a \approx \frac{x^2}{C} \]
Rearranging gives:
\[ x^2 = K_a \cdot C \]
\[ x = \sqrt{K_a \cdot C} \]

### Step 6: Calculate the pH
The concentration of \( H_3O^+ \) is equal to \( x \), so:
\[ pH = -\log[H_3O^+] = -\log(x) \]

### Step 7: Substitute Values
If we assume a concentration \( C \) of \( 0.1 \, M \) for \( NH_4Cl \):
\[ x = \sqrt{5.56 \times 10^{-10} \cdot 0.1} = \sqrt{5.56 \times 10^{-11}} \approx 7.45 \times 10^{-6} \]
Then,
\[ pH = -\log(7.45 \times 10^{-6}) \approx 5.13 \]

### Final Answer
The pH of the hydrolysis of \( NH_4Cl \) at \( 25^\circ C \) is approximately **5.13**.

---