Calculate the amount of `(NH_(4))_(2)SO_(4)` in grams which must be added to `500 ml` of `0.2 M NH_(3)` to yield a solution of `pH=9`, `K_(b)` for `NH_(3)=2xx10^(-5)`

Options

A. `3.248 g`
B. `4.248 g`
C. `1.320 g`
D. `6.248 g`

Explanation

To solve the problem of calculating the amount of `(NH₄)₂SO₄` that must be added to `500 ml` of `0.2 M NH₃` to yield a solution of `pH = 9`, we will follow these steps:

### Step 1: Calculate the number of moles of NH₃
Given:
- Molarity of NH₃ = 0.2 M
- Volume of NH₃ = 500 ml = 0.5 L

Using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Number of moles of NH₃} = 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]

### Step 2: Calculate the pOH from the given pH
Given:
- pH = 9

Using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pOH} = 14 - 9 = 5 \]

### Step 3: Calculate pKₐ from Kb
Given:
- \( K_b \) for NH₃ = \( 2 \times 10^{-5} \)

Using the formula:
\[ pK_b = -\log(K_b) \]
\[ pK_b = -\log(2 \times 10^{-5}) = -(\log(2) + \log(10^{-5})) = -0.301 - (-5) = 4.699 \]

### Step 4: Use the Henderson-Hasselbalch equation
For a buffer solution, the equation is:
\[ \text{pOH} = pK_b - \log\left(\frac{[\text{Base}]}{[\text{Salt}]}\right) \]
Substituting the known values:
\[ 5 = 4.699 - \log\left(\frac{0.2}{[\text{Salt}]}\right) \]

Rearranging gives:
\[ \log\left(\frac{0.2}{[\text{Salt}]}\right) = 4.699 - 5 = -0.301 \]
Taking antilog:
\[ \frac{0.2}{[\text{Salt}]} = 10^{-0.301} \approx 0.5 \]
Thus:
\[ [\text{Salt}] = \frac{0.2}{0.5} = 0.1 \, \text{M} \]

### Step 5: Calculate the number of moles of (NH₄)₂SO₄ needed
Since the concentration of the salt (NH₄)₂SO₄ is 0.1 M in 0.5 L:
\[ \text{Number of moles of (NH₄)₂SO₄} = 0.1 \, \text{mol/L} \times 0.5 \, \text{L} = 0.05 \, \text{mol} \]

### Step 6: Calculate the mass of (NH₄)₂SO₄
The molar mass of (NH₄)₂SO₄ is calculated as follows:
- Nitrogen (N): 14 g/mol × 2 = 28 g/mol
- Hydrogen (H): 1 g/mol × 8 = 8 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Total molar mass:
\[ 28 + 8 + 32 + 64 = 132 \, \text{g/mol} \]

Now, calculate the mass:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 132 \, \text{g/mol} = 6.6 \, \text{g} \]

### Final Answer
The amount of `(NH₄)₂SO₄` that must be added is approximately **6.6 grams**.

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