What will be the degree of dissociation of 0.005M `NH_(4)OH`solution. If `pK_(b)` for `NH_(4)OH` is 4.7

Options

A. `2%`
B. `0.03%`
C. `1.4%`
D. `6. 32%`

Explanation

To find the degree of dissociation of a 0.005 M NH₄OH solution given that the pKₑ for NH₄OH is 4.7, we can follow these steps:

### Step 1: Calculate Kb from pKb
The relationship between pKb and Kb is given by:
\[ \text{pK}_b = -\log K_b \]
To find Kb, we can rearrange this equation:
\[ K_b = 10^{-\text{pK}_b} \]
Substituting the given value:
\[ K_b = 10^{-4.7} \]

### Step 2: Calculate the value of Kb
Using a calculator, we find:
\[ K_b \approx 2.0 \times 10^{-5} \]

### Step 3: Set up the expression for Kb
For the dissociation of NH₄OH:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
Let α be the degree of dissociation. The initial concentration (C) is 0.005 M. At equilibrium, the concentrations will be:
- [NH₄OH] = C(1 - α)
- [NH₄⁺] = Cα
- [OH⁻] = Cα

Thus, we can express Kb as:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}]} = \frac{Cα \cdot Cα}{C(1 - α)} \]
This simplifies to:
\[ K_b = \frac{C^2α^2}{C(1 - α)} = \frac{Cα^2}{1 - α} \]

### Step 4: Substitute values into the Kb expression
Substituting the known values:
\[ 2.0 \times 10^{-5} = \frac{0.005α^2}{1 - α} \]

### Step 5: Assume α is small
Since NH₄OH is a weak base, we can assume that α is small compared to 1, thus:
\[ 1 - α \approx 1 \]
So the equation simplifies to:
\[ 2.0 \times 10^{-5} = 0.005α^2 \]

### Step 6: Solve for α
Rearranging gives:
\[ α^2 = \frac{2.0 \times 10^{-5}}{0.005} \]
\[ α^2 = 4.0 \times 10^{-3} \]
Taking the square root:
\[ α = \sqrt{4.0 \times 10^{-3}} \]
\[ α \approx 0.0632455532 \]

### Step 7: Convert to percentage
To express α as a percentage:
\[ \alpha \times 100 \approx 6.32\% \]

### Final Answer
The degree of dissociation of the 0.005 M NH₄OH solution is approximately **6.32%**.

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