Calculate the pH of a 2.0 M solution of `NH_4Cl. [K_b(NH_3)= 1.8 xx 10^(-5)]`

Answer

To calculate the pH of a 2.0 M solution of \( NH_4Cl \), we will follow these steps:

### Step 1: Identify the nature of the salt
\( NH_4Cl \) is a salt formed from a strong acid (HCl) and a weak base (NH3). When dissolved in water, it dissociates into \( NH_4^+ \) and \( Cl^- \). The \( NH_4^+ \) ion can hydrolyze to produce \( H^+ \) ions, thus affecting the pH of the solution.

### Step 2: Write the hydrolysis reaction
The hydrolysis of \( NH_4^+ \) can be represented as:
\[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \]

### Step 3: Use the formula for \( [H^+] \) concentration
For a weak base and strong acid salt, the concentration of \( H^+ \) can be calculated using the formula:
\[ [H^+] = \sqrt{\frac{K_w \cdot C}{K_b}} \]
where:
- \( K_w \) is the ion product of water (\( 1.0 \times 10^{-14} \) at 25°C),
- \( C \) is the concentration of the salt (2.0 M),
- \( K_b \) is the base dissociation constant for ammonia (\( 1.8 \times 10^{-5} \)).

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[ [H^+] = \sqrt{\frac{(1.0 \times 10^{-14}) \cdot (2.0)}{1.8 \times 10^{-5}}} \]

### Step 5: Calculate the value
Calculating the numerator:
\[ (1.0 \times 10^{-14}) \cdot (2.0) = 2.0 \times 10^{-14} \]
Now, divide by \( K_b \):
\[ \frac{2.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 1.11 \times 10^{-9} \]
Now take the square root:
\[ [H^+] = \sqrt{1.11 \times 10^{-9}} \approx 3.33 \times 10^{-5} \, M \]

### Step 6: Calculate the pH
Now, we can find the pH using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ pH = -\log(3.33 \times 10^{-5}) \]
Calculating this gives:
\[ pH \approx 4.478 \]

### Final Answer
The pH of a 2.0 M solution of \( NH_4Cl \) is approximately **4.48**.

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