pH of 0.1 M aqueous solution of weak monoprotic acid HA is 2. Calculate osmotic pressure of this solution at `27^(@)C`. Take solution constant `R="0.082 L atm/K-mol"`

Options

A. 2.46 atm
B. 3.5 atm
C. 3.05 atm
D. 2.7 atm

Answer

To calculate the osmotic pressure of a 0.1 M aqueous solution of a weak monoprotic acid (HA) with a given pH of 2, we will follow these steps:

### Step 1: Determine the concentration of H⁺ ions
Given that the pH is 2, we can find the concentration of hydrogen ions [H⁺] using the formula:
\[ \text{pH} = -\log[H^+] \]
From the pH:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \text{ M} = 0.01 \text{ M} \]

### Step 2: Calculate the degree of dissociation (α)
For a weak monoprotic acid HA that dissociates as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
Let the initial concentration of HA be \(C = 0.1 \text{ M}\). At equilibrium, if \(C\) is the initial concentration and \(C \cdot \alpha\) is the amount dissociated, we have:
- [H⁺] = \(C \cdot \alpha\)
- [A⁻] = \(C \cdot \alpha\)
- [HA] = \(C(1 - \alpha)\)

From the concentration of H⁺:
\[ C \cdot \alpha = 0.01 \text{ M} \]
Substituting \(C = 0.1 \text{ M}\):
\[ 0.1 \cdot \alpha = 0.01 \implies \alpha = \frac{0.01}{0.1} = 0.1 \]

### Step 3: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \(i\) for a weak acid is given by:
\[ i = 1 + \alpha \]
Substituting the value of \(\alpha\):
\[ i = 1 + 0.1 = 1.1 \]

### Step 4: Calculate the osmotic pressure (π)
The osmotic pressure can be calculated using the formula:
\[ \pi = iCRT \]
Where:
- \(i = 1.1\)
- \(C = 0.1 \text{ M}\)
- \(R = 0.082 \text{ L atm/K mol}\)
- \(T = 27^{\circ}C = 300 \text{ K}\)

Substituting the values:
\[ \pi = 1.1 \times 0.1 \times 0.082 \times 300 \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ \pi = 1.1 \times 0.1 \times 0.082 \times 300 = 2.718 \text{ atm} \]

### Final Answer
The osmotic pressure of the solution is approximately \(2.718 \text{ atm}\).

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