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 containing 0.05 M ammonium hydroxide (\( NH_4OH \)) and 0.001 M ammonium chloride (\( NH_4Cl \)), we can use the concept of a buffer solution. Here are the steps to solve the problem:

### Step 1: Identify the components of the buffer solution
We have:
- A weak base: Ammonium hydroxide (\( NH_4OH \))
- A salt of the weak base: Ammonium chloride (\( NH_4Cl \))

### Step 2: Write the expression for \( pOH \)
For a buffer solution, the \( pOH \) can be calculated using the following formula:
\[ pOH = pK_b + \log\left(\frac{[NH_4^+]}{[NH_4OH]}\right) \]
Where:
- \( pK_b = -\log(K_b) \)
- \( [NH_4^+] \) is the concentration of the conjugate acid (from \( NH_4Cl \))
- \( [NH_4OH] \) is the concentration of the weak base.

### Step 3: Calculate \( pK_b \)
Given \( K_b \) for \( NH_4OH \) is \( 1.8 \times 10^{-5} \):
\[ pK_b = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 4: Substitute the concentrations into the equation
Substituting the known values:
- \( [NH_4^+] = 0.001 \, M \)
- \( [NH_4OH] = 0.05 \, M \)

The equation becomes:
\[ pOH = 4.74 + \log\left(\frac{0.001}{0.05}\right) \]

### Step 5: Calculate the logarithm
Calculating the logarithm:
\[ \log\left(\frac{0.001}{0.05}\right) = \log(0.001) - \log(0.05) = -3 - (-1.301) = -3 + 1.301 = -1.699 \]

### Step 6: Calculate \( pOH \)
Now substituting back into the \( pOH \) equation:
\[ pOH = 4.74 - 1.699 = 3.041 \]

### Step 7: Calculate \( OH^- \) concentration
Now, we can find the concentration of \( OH^- \):
\[ [OH^-] = 10^{-pOH} = 10^{-3.041} \approx 9.1 \times 10^{-4} \, M \]

### Final Answer
Thus, the concentration of \( OH^- \) ions in the solution is approximately \( 9.1 \times 10^{-4} \, M \).

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