To solve the problem of determining the amount of ammonium sulfate \((NH_4)_2SO_4\) needed to achieve a pH of 9.35 in a 500 mL solution of 0.2 M ammonium hydroxide \((NH_4OH)\), we can follow these steps:
### Step 1: Convert Volume to Liters
The volume of the solution is given as 500 mL. We need to convert this to liters:
\[
\text{Volume (L)} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L}
\]
**Hint:** Remember that 1 L = 1000 mL for conversion.
### Step 2: Calculate Moles of Ammonium Hydroxide
Using the molarity and volume, we can calculate the number of moles of ammonium hydroxide:
\[
\text{Moles of } NH_4OH = \text{Molarity} \times \text{Volume} = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{moles}
\]
**Hint:** Moles can be calculated using the formula: Moles = Molarity × Volume.
### Step 3: Determine pOH from pH
Since we are given the pH, we can find the pOH:
\[
\text{pOH} = 14 - \text{pH} = 14 - 9.35 = 4.65
\]
**Hint:** The relationship between pH and pOH is given by the equation: pH + pOH = 14.
### Step 4: Calculate pKb
Given that the pKa of the ammonium ion \((NH_4^+)\) is 9.26, we can find pKb:
\[
\text{pKb} = 14 - \text{pKa} = 14 - 9.26 = 4.74
\]
**Hint:** pKb can be found using the relationship between pKa and pKb.
### Step 5: Use the Henderson-Hasselbalch Equation
Since we are dealing with a buffer solution, we can use the Henderson-Hasselbalch equation:
\[
\text{pOH} = \text{pKb} + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)
\]
Substituting the known values:
\[
4.65 = 4.74 + \log\left(\frac{[\text{Salt}]}{0.1}\right)
\]
**Hint:** The base in this case is the concentration of \(NH_4OH\).
### Step 6: Rearranging the Equation
Rearranging the equation to solve for the ratio of salt to base:
\[
\log\left(\frac{[\text{Salt}]}{0.1}\right) = 4.65 - 4.74 = -0.09
\]
Taking antilog:
\[
\frac{[\text{Salt}]}{0.1} = 10^{-0.09} \approx 0.8187
\]
Thus,
\[
[\text{Salt}] \approx 0.1 \times 0.8187 \approx 0.08187 \, \text{M}
\]
**Hint:** Remember that the antilogarithm can be calculated as \(10^{\text{value}}\).
### Step 7: Calculate Moles of Ammonium Ion
Since the salt is ammonium sulfate \((NH_4)_2SO_4\), each mole of salt provides 2 moles of ammonium ions:
\[
\text{Moles of } NH_4^+ = 0.08187 \, \text{M} \times 0.5 \, \text{L} = 0.040935 \, \text{moles}
\]
**Hint:** Moles of ions can be calculated using the molarity of the salt and the volume of the solution.
### Step 8: Calculate Moles of Ammonium Sulfate
Since each mole of ammonium sulfate produces 2 moles of ammonium ions:
\[
\text{Moles of } (NH_4)_2SO_4 = \frac{0.040935}{2} \approx 0.0204675 \, \text{moles}
\]
**Hint:** Divide the moles of ammonium ions by 2 to find the moles of ammonium sulfate.
### Step 9: Calculate the Mass of Ammonium Sulfate
The molar mass of ammonium sulfate \((NH_4)_2SO_4\) is calculated as follows:
- Nitrogen: \(14 \times 2 = 28\)
- Hydrogen: \(1 \times 4 \times 2 = 8\)
- Sulfur: \(32\)
- Oxygen: \(16 \times 4 = 64\)
Total molar mass:
\[
28 + 8 + 32 + 64 = 132 \, \text{g/mol}
\]
Now, we can calculate the mass:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.0204675 \, \text{moles} \times 132 \, \text{g/mol} \approx 2.70 \, \text{g}
\]
**Hint:** Use the formula: Mass = Moles × Molar Mass to find the mass of the compound.
### Final Answer
The amount of ammonium sulfate \((NH_4)_2SO_4\) required is approximately **2.70 g**.
