Which of the following concentration of `NH_(4)^(+)` will be sufficient to present the precipitation of `Mg(OH)_(2)` form a solution which is 0.01 M `MgCl_(2)`and `0.1 M NH_(3)(aq)`. Given that `K_(sp)Mg(OH)_(2)=2.5xx10^(-11)` and `K_(b)` for `NH_(3) = 2xx10^(-5)`.

Answer

To determine the concentration of \( NH_4^+ \) required to precipitate \( Mg(OH)_2 \) from a solution containing \( 0.01 \, M \, MgCl_2 \) and \( 0.1 \, M \, NH_3 \), we can follow these steps:

### Step 1: Write the Ksp expression for \( Mg(OH)_2 \)

The dissociation of \( Mg(OH)_2 \) can be represented as:

\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2 OH^{-} (aq) \]

The solubility product constant \( K_{sp} \) for this equilibrium is given by:

\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]

Given \( K_{sp} = 2.5 \times 10^{-11} \).

### Step 2: Determine the concentration of \( Mg^{2+} \)

From the solution, we know that the concentration of \( MgCl_2 \) is \( 0.01 \, M \). Therefore, the concentration of \( Mg^{2+} \) is:

\[ [Mg^{2+}] = 0.01 \, M \]

### Step 3: Set up the Ksp expression with known values

Substituting the known concentration of \( Mg^{2+} \) into the \( K_{sp} \) expression:

\[ 2.5 \times 10^{-11} = (0.01)[OH^{-}]^2 \]

### Step 4: Solve for \( [OH^{-}] \)

Rearranging the equation to solve for \( [OH^{-}]^2 \):

\[ [OH^{-}]^2 = \frac{2.5 \times 10^{-11}}{0.01} = 2.5 \times 10^{-9} \]

Taking the square root to find \( [OH^{-}] \):

\[ [OH^{-}] = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5} \, M \]

### Step 5: Write the dissociation equation for \( NH_3 \)

The dissociation of \( NH_3 \) can be represented as:

\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^{-} \]

### Step 6: Set up the expression for \( K_b \)

The base dissociation constant \( K_b \) for \( NH_3 \) is given by:

\[ K_b = \frac{[NH_4^+][OH^{-}]}{[NH_3]} \]

Given \( K_b = 2 \times 10^{-5} \) and \( [NH_3] = 0.1 \, M \).

### Step 7: Substitute known values into the \( K_b \) expression

Substituting the known values into the \( K_b \) expression:

\[ 2 \times 10^{-5} = \frac{[NH_4^+](5 \times 10^{-5})}{0.1} \]

### Step 8: Solve for \( [NH_4^+] \)

Rearranging the equation to solve for \( [NH_4^+] \):

\[ [NH_4^+] = \frac{2 \times 10^{-5} \times 0.1}{5 \times 10^{-5}} = 0.04 \, M \]

### Conclusion

The concentration of \( NH_4^+ \) required to precipitate \( Mg(OH)_2 \) is:

\[ \boxed{0.04 \, M} \]