pH of `0.1` M monobasic acid is found to be 2 . Hence its osmotic pressure at a given temp. T K is :

Options

A. `0.1 RT`
B. `0.11 RT`
C. `1.1 RT`
D. `0.01 RT`

Explanation

To find the osmotic pressure of a 0.1 M monobasic acid given that its pH is 2, we can follow these steps:

### Step 1: Calculate the concentration of hydrogen ions \([H^+]\)
The pH is given as 2. The concentration of hydrogen ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

### Step 2: Determine the degree of dissociation (\(\alpha\))
For a monobasic acid, the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
If the initial concentration of the acid \(HA\) is 0.1 M, and it dissociates to produce \(0.01 \, \text{M}\) of \(H^+\), we can express \(\alpha\) as:
\[ \alpha = \frac{[H^+]}{C} = \frac{0.01}{0.1} = 0.1 \]

### Step 3: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \(i\) for a weak acid can be calculated using the formula:
\[ i = 1 + \alpha \]
Substituting the value of \(\alpha\):
\[ i = 1 + 0.1 = 1.1 \]

### Step 4: Use the formula for osmotic pressure (\(\Pi\))
The osmotic pressure can be calculated using the formula:
\[ \Pi = iCRT \]
Where:
- \(i\) = van 't Hoff factor = 1.1
- \(C\) = concentration of the solution = 0.1 M
- \(R\) = universal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin (T is given as a variable)

Substituting the values into the equation:
\[ \Pi = 1.1 \times 0.1 \times R \times T \]
\[ \Pi = 0.11RT \]

### Final Answer
Thus, the osmotic pressure of the solution at temperature \(T\) is:
\[ \Pi = 0.11RT \, \text{atm} \]

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