0.05 M ammonium hydroxide solution is dissolved in 0.001 M ammonium chloride solution. What will be the `OH^(-)` ion concentration of this solution ?
`(K_(b) (NH_(4) OH) = 1.8 xx 10^(-5)`

Options

A. `3.0 xx 10^(-3)`
B. `9.0 xx 10^(-4)`
C. `9.0 xx 10^(-3)`
D. `3.0 xx 10^(-4)`

Answer

To find the concentration of `OH^(-)` ions in a solution of 0.05 M ammonium hydroxide (NH₄OH) mixed with 0.001 M ammonium chloride (NH₄Cl), we can follow these steps:

### Step 1: Identify the components of the solution
We have a weak base (NH₄OH) and a salt (NH₄Cl) in the solution. The NH₄Cl will dissociate in water to give NH₄⁺ ions, which is the conjugate acid of the weak base NH₄OH.

### Step 2: Use the Henderson-Hasselbalch equation
For a basic buffer solution, we can use the Henderson-Hasselbalch equation:
\[ pOH = pK_b + \log \left( \frac{[Base]}{[Acid]} \right) \]
Where:
- \( pK_b = -\log(K_b) \)
- \([Base] = [NH₄OH]\)
- \([Acid] = [NH₄^+]\)

### Step 3: Calculate \( pK_b \)
Given \( K_b = 1.8 \times 10^{-5} \):
\[ pK_b = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 4: Determine the concentrations of the base and acid
- The concentration of the base (NH₄OH) is 0.05 M.
- The concentration of the acid (NH₄⁺ from NH₄Cl) is 0.001 M.

### Step 5: Substitute values into the Henderson-Hasselbalch equation
\[ pOH = 4.74 + \log \left( \frac{0.05}{0.001} \right) \]
Calculating the log term:
\[ \log \left( \frac{0.05}{0.001} \right) = \log(50) \approx 1.699 \]
Thus,
\[ pOH = 4.74 + 1.699 \approx 6.439 \]

### Step 6: Calculate the concentration of \( OH^- \)
Using the relationship between pOH and \( OH^- \) concentration:
\[ pOH = -\log[OH^-] \]
We can find \( [OH^-] \):
\[ [OH^-] = 10^{-pOH} = 10^{-6.439} \approx 3.63 \times 10^{-7} \text{ M} \]

### Step 7: Final concentration adjustment
Since we have a buffer solution, we can also consider the effect of dilution when mixing two solutions. However, in this case, we can assume that the concentrations remain approximately the same due to the small amount of NH₄Cl.

### Conclusion
The concentration of \( OH^- \) ions in the solution is approximately:
\[ \text{[OH^-]} \approx 3.63 \times 10^{-7} \text{ M} \]