Find moles of `NH_(4)Cl` required to prevent `Mg(OH)_(2)` from precipitating in a litre of solution which contains 0.02 mole `NH_(3)` and 0.001 mole `Mg^(2+)` ions.
Given : `K_(b)(NH_(3))=10^(-5),` `K_(sp)[Mg(OH)_(2)]=10^(-11)`.

Options

A. `10^(-4)`
B. `2xx10^(-3)`
C. `0.02`
D. `0.1`

Answer

To solve the problem of finding the moles of `NH₄Cl` required to prevent `Mg(OH)₂` from precipitating, we will follow these steps:

### Step 1: Determine the concentration of `Mg²⁺` ions
Given that there are 0.001 moles of `Mg²⁺` ions in 1 liter of solution, the concentration of `Mg²⁺` ions is:
\[ \text{Concentration of } Mg^{2+} = \frac{0.001 \text{ moles}}{1 \text{ L}} = 0.001 \text{ M} \]

### Step 2: Calculate the concentration of `OH⁻` ions using the solubility product constant (`Ksp`)
The solubility product constant for `Mg(OH)₂` is given as \( K_{sp} = 10^{-11} \).
The relationship for the solubility product is:
\[ K_{sp} = [Mg^{2+}][OH^-]^2 \]
Substituting the known concentration of `Mg²⁺`:
\[ 10^{-11} = (0.001)[OH^-]^2 \]
Rearranging to find the concentration of `OH⁻`:
\[ [OH^-]^2 = \frac{10^{-11}}{0.001} = 10^{-8} \]
Taking the square root:
\[ [OH^-] = \sqrt{10^{-8}} = 10^{-4} \text{ M} \]

### Step 3: Calculate the pOH
The pOH can be calculated from the concentration of `OH⁻`:
\[ pOH = -\log[OH^-] = -\log(10^{-4}) = 4 \]

### Step 4: Calculate the pKb of ammonia
The given \( K_b \) for ammonia is \( 10^{-5} \). Thus:
\[ pK_b = -\log(K_b) = -\log(10^{-5}) = 5 \]

### Step 5: Use the Henderson-Hasselbalch equation to find the ratio of concentrations
Using the relationship:
\[ pOH = pK_b + \log\left(\frac{[NH₄Cl]}{[NH₃]}\right) \]
Substituting the known values:
\[ 4 = 5 + \log\left(\frac{[NH₄Cl]}{[NH₃]}\right) \]
Rearranging gives:
\[ \log\left(\frac{[NH₄Cl]}{[NH₃]}\right) = 4 - 5 = -1 \]
This implies:
\[ \frac{[NH₄Cl]}{[NH₃]} = 10^{-1} = 0.1 \]

### Step 6: Calculate the concentration of `NH₄Cl`
Let the concentration of ammonia, `[NH₃]`, be 0.02 M (given). Thus:
\[ [NH₄Cl] = 0.1 \times [NH₃] = 0.1 \times 0.02 = 0.002 \text{ M} \]

### Step 7: Calculate the moles of `NH₄Cl` required
Since we are considering a 1-liter solution:
\[ \text{Moles of } NH₄Cl = [NH₄Cl] \times \text{Volume} = 0.002 \text{ moles/L} \times 1 \text{ L} = 0.002 \text{ moles} \]

### Final Answer
The moles of `NH₄Cl` required to prevent `Mg(OH)₂` from precipitating is:
\[ \boxed{2 \times 10^{-3} \text{ moles}} \]