A hand book states that the solubility of `RNH_(2)` (g) in water at 1 atm and `0^(@)`C is 22.41 litres volumes of `RNH_(2)`(g) per volume of water. `(pK_(b) of RNH_(2)=4)` Find the max. pOH that can be attained by dissolving `RNH_(2)` in water:

Options

A. 1
B. 2
C. 4
D. 6

Answer

To solve the problem of finding the maximum pOH that can be attained by dissolving `RNH2` in water, we will follow these steps:

### Step 1: Understand the Given Information
- The solubility of `RNH2` in water is 22.41 liters of gas per volume of water at 1 atm and 0°C.
- The pKb of `RNH2` is given as 4.

### Step 2: Convert Temperature to Kelvin
- The temperature in Celsius is 0°C, which is equivalent to 273 K.

### Step 3: Use the Ideal Gas Law to Find Moles of `RNH2`
Using the ideal gas equation \( PV = nRT \):
- \( P = 1 \, \text{atm} \)
- \( V = 22.41 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \)

Substituting these values into the equation:
\[ n = \frac{PV}{RT} = \frac{1 \times 22.41}{0.0821 \times 273} \approx 1 \, \text{mol} \]

### Step 4: Calculate the Concentration of `RNH2`
Since the volume of water is 1 liter (as per the problem statement), the concentration \( C \) of `RNH2` will be:
\[ C = \frac{n}{\text{Volume}} = \frac{1 \, \text{mol}}{1 \, \text{L}} = 1 \, \text{M} \]

### Step 5: Relate pOH to Concentration of OH⁻ Ions
The relationship between pOH and the concentration of hydroxide ions \([OH^-]\) is given by:
\[ \text{pOH} = \frac{1}{2} \text{pK}_b - \log C \]

### Step 6: Substitute Values into the pOH Equation
Substituting \( \text{pK}_b = 4 \) and \( C = 1 \, \text{M} \):
\[ \text{pOH} = \frac{1}{2} \times 4 - \log(1) \]
Since \( \log(1) = 0 \):
\[ \text{pOH} = 2 - 0 = 2 \]

### Conclusion
The maximum pOH that can be attained by dissolving `RNH2` in water is:
\[ \text{pOH} = 2 \]

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