If the concentration of`[NH_(4)^(+)]` in a solution having 0.02 M `NH_(3)`and 0.005 M Ca `( OH)_(2)` is `a xx 10^(-6) ` M,determine a.
`[k_(b) ( NH_(3)) = 1.8 xx 10^(-5) ]`

Answer

To determine the concentration of the ammonium ion \([NH_4^+]\) in a solution containing \(0.02 \, M \, NH_3\) and \(0.005 \, M \, Ca(OH)_2\), we can follow these steps:

### Step 1: Understand the dissociation of \(Ca(OH)_2\)
Calcium hydroxide dissociates in water as follows:
\[ Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \]
Given that the concentration of \(Ca(OH)_2\) is \(0.005 \, M\), the concentration of \(OH^-\) ions produced will be:
\[ [OH^-] = 2 \times 0.005 \, M = 0.01 \, M \]

### Step 2: Write the equilibrium expression for \(NH_4OH\)
Ammonium hydroxide (\(NH_4OH\)) dissociates in water as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
The equilibrium expression for this reaction can be written as:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Where \(K_b\) for ammonia (\(NH_3\)) is given as \(1.8 \times 10^{-5}\).

### Step 3: Determine the concentration of \(NH_4OH\)
Since \(NH_3\) is present in the solution, we can assume that the concentration of \(NH_4OH\) is approximately equal to the initial concentration of \(NH_3\), which is \(0.02 \, M\).

### Step 4: Substitute known values into the equilibrium expression
We can substitute the known values into the equilibrium expression:
\[ 1.8 \times 10^{-5} = \frac{[NH_4^+][0.01]}{0.02} \]

### Step 5: Solve for \([NH_4^+]\)
Rearranging the equation gives:
\[ [NH_4^+] = \frac{(1.8 \times 10^{-5}) \times 0.02}{0.01} \]
\[ [NH_4^+] = 2 \times 1.8 \times 10^{-5} = 3.6 \times 10^{-5} \, M \]

### Step 6: Express \([NH_4^+]\) in the required format
We need to express the concentration of \([NH_4^+]\) in the form \(a \times 10^{-6} \, M\):
\[ 3.6 \times 10^{-5} = 36 \times 10^{-6} \, M \]
Thus, \(a = 36\).

### Final Answer:
The value of \(a\) is \(36\).

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