A solution of `0.01M Cd^(2+)` contains `0.01IM NH_(4)OH`. What conc. of `NH _(4)^(+)` form `NH_(4)Cl` is necessry to prevent precipitation of `Cd(OH)_(2)` ? `K_(sp)`, of `Cd(OH)_(2), = 2.0 xx 10^(-14) , K_(b) ` of `NH_(4)OH 1.8 xx 10^(-5)` if answer is `1.272xx 10^(-x)` mol/ltr then x =__________ ?

Answer

To solve the problem step by step, we need to determine the concentration of \( NH_4^+ \) ions required to prevent the precipitation of \( Cd(OH)_2 \) in a solution containing \( 0.01M \) \( Cd^{2+} \) and \( 0.01IM \) \( NH_4OH \).

### Step 1: Write the dissociation equation for \( Cd(OH)_2 \)
The dissociation of cadmium hydroxide can be represented as:
\[ Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^- \]

### Step 2: Write the expression for the solubility product \( K_{sp} \)
The solubility product \( K_{sp} \) for \( Cd(OH)_2 \) is given by:
\[ K_{sp} = [Cd^{2+}][OH^-]^2 \]
Given \( K_{sp} = 2.0 \times 10^{-14} \) and \( [Cd^{2+}] = 0.01M = 1.0 \times 10^{-2}M \).

### Step 3: Substitute the known values into the \( K_{sp} \) expression
Substituting the known values into the \( K_{sp} \) expression:
\[ 2.0 \times 10^{-14} = (1.0 \times 10^{-2})[OH^-]^2 \]

### Step 4: Solve for \( [OH^-] \)
Rearranging the equation gives:
\[ [OH^-]^2 = \frac{2.0 \times 10^{-14}}{1.0 \times 10^{-2}} = 2.0 \times 10^{-12} \]
Taking the square root:
\[ [OH^-] = \sqrt{2.0 \times 10^{-12}} = 1.414 \times 10^{-6} M \]

### Step 5: Write the dissociation equation for \( NH_4OH \)
The dissociation of ammonium hydroxide can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]

### Step 6: Write the expression for the base dissociation constant \( K_b \)
The base dissociation constant \( K_b \) for \( NH_4OH \) is given by:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Given \( K_b = 1.8 \times 10^{-5} \) and \( [NH_4OH] = 0.01M = 1.0 \times 10^{-2}M \).

### Step 7: Substitute the known values into the \( K_b \) expression
Substituting the known values into the \( K_b \) expression gives:
\[ 1.8 \times 10^{-5} = \frac{[NH_4^+][1.414 \times 10^{-6}]}{1.0 \times 10^{-2}} \]

### Step 8: Solve for \( [NH_4^+] \)
Rearranging the equation gives:
\[ [NH_4^+] = \frac{1.8 \times 10^{-5} \times 1.0 \times 10^{-2}}{1.414 \times 10^{-6}} \]
Calculating this gives:
\[ [NH_4^+] = \frac{1.8 \times 10^{-7}}{1.414 \times 10^{-6}} = 1.272 \times 10^{-1} M \]

### Step 9: Compare with the given answer format
The answer is given in the form \( 1.272 \times 10^{-x} \). Here, we have:
\[ 1.272 \times 10^{-1} = 1.272 \times 10^{-x} \]
This implies \( x = 1 \).

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{1} \]