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 \( \text{NH}_4\text{Cl} \) required to dissolve in 500 mL of water to achieve a pH of 4.5, we can follow these steps:

### Step 1: Identify the nature of the salt
\( \text{NH}_4\text{Cl} \) is a salt formed from a weak base (ammonium hydroxide, \( \text{NH}_4\text{OH} \)) and a strong acid (hydrochloric acid, \( \text{HCl} \)). This means that in solution, it will produce \( \text{NH}_4^+ \) ions which can affect the pH.

### Step 2: Use the pH to find the concentration of \( \text{NH}_4^+ \)
Given that the pH is 4.5, we can find the concentration of hydrogen ions \( [\text{H}^+] \):
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-4.5} \approx 3.16 \times 10^{-5} \, \text{M} \]

### Step 3: Calculate \( K_a \) from \( K_b \)
We know that:
\[ K_w = K_a \times K_b \]
Where \( K_w = 1.0 \times 10^{-14} \) and \( K_b = 1.8 \times 10^{-5} \). Thus,
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

### Step 4: Set up the equilibrium expression
For the dissociation of \( \text{NH}_4^+ \):
\[ \text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+ \]
The \( K_a \) expression is:
\[ K_a = \frac{[\text{NH}_3][\text{H}^+]}{[\text{NH}_4^+]} \]
Let \( C \) be the concentration of \( \text{NH}_4^+ \). At equilibrium, we have:
\[ K_a = \frac{(x)(3.16 \times 10^{-5})}{C - x} \approx \frac{(3.16 \times 10^{-5})^2}{C} \]
Assuming \( x \) is small compared to \( C \).

### Step 5: Solve for \( C \)
Substituting \( K_a \):
\[ 5.56 \times 10^{-10} = \frac{(3.16 \times 10^{-5})^2}{C} \]
\[ C = \frac{(3.16 \times 10^{-5})^2}{5.56 \times 10^{-10}} \approx 0.018 \, \text{M} \]

### Step 6: Calculate the number of moles in 500 mL
To find the number of moles of \( \text{NH}_4\text{Cl} \):
\[ \text{Number of moles} = C \times \text{Volume in L} = 0.018 \, \text{M} \times 0.5 \, \text{L} = 0.009 \, \text{moles} \]

### Step 7: Calculate the mass of \( \text{NH}_4\text{Cl} \)
The molar mass of \( \text{NH}_4\text{Cl} \) is:
\[ \text{Molar mass} = 14 + 4 + 35.5 = 53.5 \, \text{g/mol} \]
Thus, the mass required is:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 0.009 \, \text{moles} \times 53.5 \, \text{g/mol} \approx 0.48 \, \text{g} \]

### Final Answer
The amount of \( \text{NH}_4\text{Cl} \) required to dissolve in 500 mL of water to achieve a pH of 4.5 is approximately **0.48 grams**.