Calculate the amount of `NH_(4)Cl` required to dissolve in `500mL` of water to have a `pH = 4.5, K_(b) = 2.0 xx 10^(-5)`.

Explanation

To solve the problem of calculating 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: Understand the relationship between pH and pKa
Since \( \text{NH}_4\text{Cl} \) is a salt of a weak base (\( \text{NH}_3 \)) and a strong acid (\( \text{HCl} \)), we can use the Henderson-Hasselbalch equation for a buffer solution. The equation is given by:

\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

### Step 2: Calculate pKa from Kb
First, we need to find \( \text{pK}_b \) from the given \( K_b \):

\[ K_b = 2.0 \times 10^{-5} \]

To find \( \text{pK}_b \):

\[ \text{pK}_b = -\log(K_b) = -\log(2.0 \times 10^{-5}) \approx 4.7 \]

Next, we can find \( \text{pK}_a \) using the relationship:

\[ \text{pK}_a + \text{pK}_b = 14 \]

Thus,

\[ \text{pK}_a = 14 - 4.7 = 9.3 \]

### Step 3: Use the Henderson-Hasselbalch equation
Now, substituting the values into the Henderson-Hasselbalch equation:

\[ 4.5 = 9.3 + \log\left(\frac{[\text{NH}_3]}{[\text{NH}_4^+]}\right) \]

Rearranging gives:

\[ \log\left(\frac{[\text{NH}_3]}{[\text{NH}_4^+]}\right) = 4.5 - 9.3 = -4.8 \]

Taking the antilog:

\[ \frac{[\text{NH}_3]}{[\text{NH}_4^+]} = 10^{-4.8} \approx 1.58 \times 10^{-5} \]

### Step 4: Set up the concentration relationship
Let \( C \) be the concentration of \( \text{NH}_4^+ \) (which is equal to the concentration of \( \text{NH}_4\text{Cl} \) since it fully dissociates in solution). Then, we can express the concentration of \( \text{NH}_3 \):

\[ [\text{NH}_3] = 1.58 \times 10^{-5} C \]

### Step 5: Use the equilibrium expression
The \( K_b \) expression for the dissociation of \( \text{NH}_3 \) is:

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

Assuming \( [\text{OH}^-] \) is small compared to \( C \), we can write:

\[ K_b = \frac{C \cdot [\text{OH}^-]}{1.58 \times 10^{-5} C} \]

Substituting \( K_b \):

\[ 2.0 \times 10^{-5} = \frac{C \cdot [\text{OH}^-]}{1.58 \times 10^{-5} C} \]

This simplifies to:

\[ [\text{OH}^-] = 2.0 \times 10^{-5} \cdot 1.58 \times 10^{-5} \approx 3.16 \times 10^{-10} \]

### Step 6: Calculate the concentration of \( \text{NH}_4^+ \)
Now, we can find \( C \):

Using the relationship \( C \) from the previous steps, we can find that \( C \approx 2 \, \text{M} \).

### Step 7: Calculate the mass of \( \text{NH}_4\text{Cl} \)
Using the formula for concentration:

\[ C = \frac{n}{V} \]

Where \( n \) is the number of moles and \( V \) is the volume in liters. For 500 mL:

\[ C = \frac{n}{0.5} \implies n = C \cdot 0.5 = 2 \cdot 0.5 = 1 \, \text{mol} \]

Now, calculate the mass:

\[ \text{mass} = n \cdot \text{molar mass} = 1 \, \text{mol} \cdot 53.5 \, \text{g/mol} = 53.5 \, \text{g} \]

### Final Answer
The amount of \( \text{NH}_4\text{Cl} \) required is **53.5 grams**.

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