Question

Depression in freezing point of 0.01 molal aqueous HCOOH solution is 0.02046.1 molal aqueous urea solution freezes at - 186°C. Assuming molality equal to molarity, pH of HCOOH solution is:

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Answer 2

The problem requires us to determine the pH of a weak acid solution by first calculating its degree of dissociation using colligative properties.

Step 1: Calculate the cryoscopic constant ($K_f$) for water.

The freezing point depression for a 1 molal aqueous urea solution is given. Urea is a non-electrolyte, so its van't Hoff factor ($i$) is 1.

Given:

Molality of urea solution ($m_{\text{urea}}$) = 1 molal

Freezing point of urea solution = $-1.86^\circ C$

Depression in freezing point ($\Delta T_f^{\text{urea}}$) = $0^\circ C - (-1.86^\circ C) = 1.86^\circ C$

Using the formula for depression in freezing point:

$\Delta T_f = i \cdot K_f \cdot m$

For urea solution:

$1.86^\circ C = 1 \cdot K_f \cdot 1 \text{ molal}$

$K_f = 1.86 \text{ K kg mol}^{-1}$

Step 2: Calculate the van't Hoff factor ($i$) for HCOOH solution.

Given:

Molality of HCOOH solution ($m_{\text{HCOOH}}$) = 0.01 molal

Depression in freezing point ($\Delta T_f^{\text{HCOOH}}$) = 0.02046°C

Using the $K_f$ value calculated above:

$\Delta T_f^{\text{HCOOH}} = i \cdot K_f \cdot m_{\text{HCOOH}}$

$0.02046 = i \cdot 1.86 \cdot 0.01$

$i = \frac{0.02046}{1.86 \cdot 0.01}$

$i = \frac{0.02046}{0.0186}$

$i = 1.1$

Step 3: Determine the degree of dissociation ($\alpha$) for HCOOH.

Formic acid (HCOOH) is a weak acid that dissociates as follows:

$\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-$

For a weak electrolyte dissociating into 'n' ions, the van't Hoff factor ($i$) is related to the degree of dissociation ($\alpha$) by the formula:

$i = 1 + (n-1)\alpha$

In this case, HCOOH dissociates into 2 ions (H$^+$ and HCOO$^-$), so $n=2$.

$i = 1 + (2-1)\alpha$

$i = 1 + \alpha$

We found $i = 1.1$:

$1.1 = 1 + \alpha$

$\alpha = 1.1 - 1 = 0.1$

Step 4: Calculate the concentration of H$^+$ ions.

The initial concentration of HCOOH is given as 0.01 molal. Assuming molality is equal to molarity for dilute aqueous solutions, the initial concentration ($C$) is 0.01 M.

The concentration of H$^+$ ions at equilibrium is given by:

$[\text{H}^+] = C \alpha$

$[\text{H}^+] = 0.01 \text{ M} \times 0.1$

$[\text{H}^+] = 0.001 \text{ M}$

$[\text{H}^+] = 10^{-3} \text{ M}$

Step 5: Calculate the pH of the HCOOH solution.

pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:

$\text{pH} = -\log[\text{H}^+]$

$\text{pH} = -\log(10^{-3})$

$\text{pH} = 3$

The pH of the HCOOH solution is 3.