What is the molar solubility of `Mn(OH)_(2)``(K_(sp)=4.5xx10^(-14))` in a buffer solution containing equal amounts of `NH_(4)^(+)` and `NH_(3)` `(K_(b)=1.8xx10^(-5))`?

Options

A. `3.0xx10^(-4)`
B. `1.38xx10^(-4)`
C. `1.38xx10^(-3)`
D. `7.3xx10^(-4)`

Answer

To find the molar solubility of \( \text{Mn(OH)}_2 \) in a buffer solution containing equal amounts of \( \text{NH}_4^+ \) and \( \text{NH}_3 \), we will follow these steps:

### Step 1: Understand the Buffer System
The buffer solution consists of \( \text{NH}_4^+ \) (the conjugate acid) and \( \text{NH}_3 \) (the base). Since they are in equal amounts, the pH of the solution will be determined by the \( K_b \) of ammonia.

### Step 2: Calculate \( pK_b \)
Given \( K_b = 1.8 \times 10^{-5} \):
\[ pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 3: Calculate \( pOH \)
Since the concentrations of \( \text{NH}_4^+ \) and \( \text{NH}_3 \) are equal, we can use the Henderson-Hasselbalch equation:
\[ pOH = pK_b + \log\left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right) \]
Since \( \frac{[\text{NH}_4^+]}{[\text{NH}_3]} = 1 \):
\[ pOH = pK_b \approx 4.74 \]

### Step 4: Calculate \( [OH^-] \)
Using \( pOH \):
\[ pOH = -\log[OH^-] \implies [OH^-] = 10^{-pOH} = 10^{-4.74} \approx 1.82 \times 10^{-5} \, \text{M} \]

### Step 5: Write the Dissociation Equation for \( \text{Mn(OH)}_2 \)
The dissociation of \( \text{Mn(OH)}_2 \) can be represented as:
\[ \text{Mn(OH)}_2 (s) \rightleftharpoons \text{Mn}^{2+} (aq) + 2 \text{OH}^- (aq) \]
Let the molar solubility of \( \text{Mn(OH)}_2 \) be \( S \). Thus, at equilibrium:
- \( [\text{Mn}^{2+}] = S \)
- \( [\text{OH}^-] = 2S \)

### Step 6: Write the Solubility Product Expression
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]
Substituting the expressions for the concentrations:
\[ K_{sp} = S(2S)^2 = S \cdot 4S^2 = 4S^3 \]
Given \( K_{sp} = 4.5 \times 10^{-14} \):
\[ 4S^3 = 4.5 \times 10^{-14} \]
\[ S^3 = \frac{4.5 \times 10^{-14}}{4} = 1.125 \times 10^{-14} \]

### Step 7: Solve for \( S \)
Taking the cube root:
\[ S = \sqrt[3]{1.125 \times 10^{-14}} \approx 2.24 \times 10^{-5} \, \text{M} \]

### Final Answer
The molar solubility of \( \text{Mn(OH)}_2 \) in the buffer solution is approximately:
\[ \boxed{2.24 \times 10^{-5} \, \text{M}} \]