when 0.2 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. The solution is found to have a hydroxide ion concentration of `1.34 xx 10^(-3)` . The dissociation constant of ammonia is

Options

A. `1.8 xx 10^(-5)`
B. `1.6 xx 10^(-6)`
C. `1.34 xx 10^(-3)`
D. `1.8 xx 10^(-4)`

Answer

To find the dissociation constant (Kb) of ammonia (NH3) when 0.2 moles of it are dissolved in 1 liter of water and the hydroxide ion concentration is given as \(1.34 \times 10^{-3}\) M, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of ammonia in water can be represented as:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

### Step 2: Identify the concentrations
From the problem, we know:
- Initial concentration of NH3 = 0.2 M (since it's dissolved in 1 L)
- Concentration of OH⁻ = \(1.34 \times 10^{-3}\) M

### Step 3: Determine the concentration of NH4⁺
Since each molecule of NH3 that dissociates produces one molecule of NH4⁺ and one molecule of OH⁻, the concentration of NH4⁺ will also be \(1.34 \times 10^{-3}\) M.

### Step 4: Calculate the concentration of undissociated NH3
Let \(x\) be the amount of NH3 that dissociates. From the dissociation, we have:
- Concentration of NH3 at equilibrium = Initial concentration - Amount dissociated
\[ [\text{NH}_3] = 0.2 - x \]
Since \(x = 1.34 \times 10^{-3}\):
\[ [\text{NH}_3] = 0.2 - 1.34 \times 10^{-3} = 0.19866 \text{ M} \]

### Step 5: Write the expression for Kb
The dissociation constant (Kb) for ammonia can be expressed as:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Substituting the values we have:
\[ K_b = \frac{(1.34 \times 10^{-3})(1.34 \times 10^{-3})}{0.19866} \]

### Step 6: Calculate Kb
Now, calculate Kb:
\[ K_b = \frac{(1.34 \times 10^{-3})^2}{0.19866} \]
Calculating the numerator:
\[ (1.34 \times 10^{-3})^2 = 1.7956 \times 10^{-6} \]
Now, divide by 0.19866:
\[ K_b = \frac{1.7956 \times 10^{-6}}{0.19866} \approx 9.04 \times 10^{-6} \]

### Final Answer
The dissociation constant of ammonia (Kb) is approximately:
\[ K_b \approx 9.04 \times 10^{-6} \]