Calculate `[H^(+)] " and "%` dissociation of `0.1` M solution of ammonium hydroxide solution . The ionisation constant for `NH_(4)OH " is " K_(b) = 2.0 xx xx 10^(-5)` .

Options

A. `7.09xx10^(-12)M, 3%`
B. `7.09xx10^(-12)M, 1.4%`
C. `9.02xx10^(-12)M, 2.4%`
D. `9.02xx10^(-12)M,3%`

Answer

To solve the problem of calculating the concentration of \([H^+]\) and the percentage dissociation of a \(0.1 \, M\) solution of ammonium hydroxide (\(NH_4OH\)), given that the ionization constant \(K_b = 2.0 \times 10^{-5}\), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of ammonium hydroxide can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \([NH_4OH] = 0.1 \, M\)
- \([NH_4^+] = 0 \, M\)
- \([OH^-] = 0 \, M\)

### Step 3: Define the change in concentration
Let \(\alpha\) be the degree of dissociation. At equilibrium, the concentrations will be:
- \([NH_4OH] = 0.1 - \alpha\)
- \([NH_4^+] = \alpha\)
- \([OH^-] = \alpha\)

### Step 4: Write the expression for \(K_b\)
The expression for the base dissociation constant \(K_b\) is given by:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Substituting the equilibrium concentrations into the equation gives:
\[ K_b = \frac{\alpha \cdot \alpha}{0.1 - \alpha} = \frac{\alpha^2}{0.1 - \alpha} \]

### Step 5: Substitute the known value of \(K_b\)
We know \(K_b = 2.0 \times 10^{-5}\), so we can set up the equation:
\[ 2.0 \times 10^{-5} = \frac{\alpha^2}{0.1 - \alpha} \]

### Step 6: Assume \(\alpha\) is small
Since \(K_b\) is small, we can assume that \(\alpha\) is much smaller than \(0.1\), allowing us to simplify \(0.1 - \alpha \approx 0.1\):
\[ 2.0 \times 10^{-5} = \frac{\alpha^2}{0.1} \]

### Step 7: Solve for \(\alpha^2\)
Rearranging gives:
\[ \alpha^2 = 2.0 \times 10^{-5} \times 0.1 = 2.0 \times 10^{-6} \]

### Step 8: Calculate \(\alpha\)
Taking the square root:
\[ \alpha = \sqrt{2.0 \times 10^{-6}} \approx 1.41 \times 10^{-3} \]

### Step 9: Calculate the percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \left(\frac{\alpha}{0.1}\right) \times 100 = \left(\frac{1.41 \times 10^{-3}}{0.1}\right) \times 100 \approx 1.41\% \]

### Step 10: Calculate \([OH^-]\) and \([H^+]\)
Since \([OH^-] = \alpha = 1.41 \times 10^{-3} \, M\), we can find \([H^+]\) using the ion product of water \(K_w = 1.0 \times 10^{-14}\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{1.41 \times 10^{-3}} \approx 7.09 \times 10^{-12} \, M \]

### Final Results
- \([H^+] \approx 7.09 \times 10^{-12} \, M\)
- Percentage dissociation \(\approx 1.41\%\)