Calculate the amount of `NH_(4)Cl` required to dissolve in 500 mL of water to have a pH of `4.5 . K_(b) = 1.8 xx 10^(-5)`

Answer

To calculate the amount of NH₄Cl required to dissolve in 500 mL of water to achieve a pH of 4.5, we can follow these steps:

### Step 1: Understand the relationship between pH and the concentration of the salt
Since NH₄Cl is a salt of a weak base (NH₄OH) and a strong acid (HCl), we can use the formula for pH in terms of the concentration of the salt:

\[ \text{pH} = \frac{pK_w - pK_b}{2} - \log C \]

Where:
- \( pK_w = 14 \)
- \( pK_b = -\log K_b \)
- \( C \) is the concentration of NH₄Cl in mol/L.

### Step 2: Calculate \( pK_b \)
Given \( K_b = 1.8 \times 10^{-5} \):

\[ pK_b = -\log(1.8 \times 10^{-5}) = -\log(1.8) - \log(10^{-5}) \]

Calculating \( -\log(1.8) \) gives approximately 0.255, and \( -\log(10^{-5}) = 5 \). Thus,

\[ pK_b \approx 0.255 + 5 = 5.255 \]

### Step 3: Substitute values into the pH equation
Now substituting the values into the pH equation:

\[ 4.5 = \frac{14 - 5.255}{2} - \log C \]

Calculating \( \frac{14 - 5.255}{2} \):

\[ \frac{8.745}{2} = 4.3725 \]

So the equation becomes:

\[ 4.5 = 4.3725 - \log C \]

### Step 4: Solve for \( \log C \)
Rearranging gives:

\[ \log C = 4.3725 - 4.5 = -0.1275 \]

### Step 5: Calculate \( C \)
Now, finding \( C \):

\[ C = 10^{-0.1275} \approx 0.743 \text{ mol/L} \]

### Step 6: Calculate the number of moles in 500 mL
To find the number of moles in 500 mL (0.5 L):

\[ \text{Number of moles} = C \times \text{Volume in L} = 0.743 \, \text{mol/L} \times 0.5 \, \text{L} = 0.3715 \, \text{mol} \]

### Step 7: Calculate the mass of NH₄Cl
The molar mass of NH₄Cl is calculated as follows:

- Nitrogen (N): 14 g/mol
- Hydrogen (H): 4 g/mol (4 × 1)
- Chlorine (Cl): 35.5 g/mol

Total molar mass = 14 + 4 + 35.5 = 53.5 g/mol.

Now, calculate the mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 0.3715 \, \text{mol} \times 53.5 \, \text{g/mol} \approx 19.9 \, \text{g} \]

### Final Answer
The amount of NH₄Cl required to dissolve in 500 mL of water to achieve a pH of 4.5 is approximately **19.9 grams**.

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