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 \( NH_4^+ \) that will be sufficient 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 solubility product expression for \( Mg(OH)_2 \)
The solubility product constant (\( K_{sp} \)) for \( Mg(OH)_2 \) can be expressed as:
\[ K_{sp} = [Mg^{2+}][OH^-]^2 \]
Given that \( K_{sp} = 2.5 \times 10^{-11} \).

### Step 2: Substitute the concentration of \( Mg^{2+} \)
From the problem, the concentration of \( Mg^{2+} \) from \( MgCl_2 \) is \( 0.01 \, M \). Therefore, we can substitute this value into the \( K_{sp} \) expression:
\[ 2.5 \times 10^{-11} = (0.01)[OH^-]^2 \]

### Step 3: 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 4: Use the \( K_b \) expression for \( NH_3 \)
The base dissociation constant (\( K_b \)) for \( NH_3 \) is given as \( 2 \times 10^{-5} \). The dissociation of \( NH_3 \) can be represented as:
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
The expression for \( K_b \) is:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \]
Substituting the known values:
\[ 2 \times 10^{-5} = \frac{[NH_4^+](5 \times 10^{-5})}{0.1} \]

### Step 5: Solve for \( [NH_4^+] \)
Rearranging the equation to solve for \( [NH_4^+] \):
\[ [NH_4^+] = \frac{2 \times 10^{-5} \times 0.1}{5 \times 10^{-5}} = \frac{2 \times 10^{-6}}{5 \times 10^{-5}} = 0.04 \, M \]

### Final Answer
The concentration of \( NH_4^+ \) required to precipitate \( Mg(OH)_2 \) is \( 0.04 \, M \).

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