Question

A 1.96% solution of H₂SO₄ in water freezes at – 0.794°C. If Kf of water is 1.86 K kg/mol and density of solution is 1.0 g/mL, the value of ionization constant for the following reaction is

HSO₄ ⇌ H+ + SO₄2-

• A. 0.146
• B. 3.3×10-3
• C. 2×10-4
• D. 5×10-6

Answer 1

Answer: (B)

3.3×10-3

Answer 2

The freezing point depression is $\Delta T_f = 0^\circ C - (-0.794^\circ C) = 0.794^\circ C$.

The percentage concentration of H₂SO₄ is 1.96% by weight.

Mass of H₂SO₄ = 1.96 g. Mass of solution = 100 g. Mass of water = 100 - 1.96 = 98.04 g = 0.09804 kg.

Molar mass of H₂SO₄ = 98 g/mol. Moles of H₂SO₄ = $\frac{1.96}{98} = 0.02$ mol.

Molality $m_0 = \frac{0.02 \text{ mol}}{0.09804 \text{ kg}} \approx 0.203998 \text{ mol/kg}$.

The freezing point depression is related to molality by $\Delta T_f = i \times K_f \times m_0$.

$0.794 = i \times 1.86 \times 0.203998$.

$i = \frac{0.794}{1.86 \times 0.203998} \approx 2.0926$.

H₂SO₄ dissociates as H₂SO₄ → H⁺ + HSO₄⁻ (complete) and HSO₄⁻ ⇌ H⁺ + SO₄²⁻ (partial, degree of dissociation $\alpha$).

The van't Hoff factor is $i = 2 + \alpha$.

$2.0926 = 2 + \alpha \implies \alpha = 0.0926$.

The ionization constant for HSO₄⁻ ⇌ H⁺ + SO₄²⁻ is $K_a = \frac{[H⁺][SO₄²⁻]}{[HSO₄⁻]}$.

Using molalities at equilibrium: $[H⁺] = m_0(1+\alpha)$, $[SO₄²⁻] = m_0\alpha$, $[HSO₄⁻] = m_0(1-\alpha)$.

$K_a = \frac{m_0(1+\alpha)m_0\alpha}{m_0(1-\alpha)} = \frac{m_0\alpha(1+\alpha)}{1-\alpha}$.

$K_a = \frac{0.203998 \times 0.0926 \times (1+0.0926)}{1-0.0926} = \frac{0.203998 \times 0.0926 \times 1.0926}{0.9074} \approx 0.02277$.

This calculated value does not match any of the given options. Let's check which option for $K_a$ results in a freezing point depression closest to the given value (0.794).

Option (b): $K_a = 3.3 \times 10^{-3} = 0.0033$.

Let's find $\alpha$ using $K_a = \frac{m_0\alpha(1+\alpha)}{1-\alpha}$ with $m_0 = 0.203998$.

$0.0033 = \frac{0.203998 \alpha(1+\alpha)}{1-\alpha}$.

$0.0033(1-\alpha) = 0.203998\alpha + 0.203998\alpha^2$.

$0.203998\alpha^2 + 0.207298\alpha - 0.0033 = 0$.

Solving for $\alpha$: $\alpha \approx 0.01583$.

$i = 2 + \alpha = 2.01583$.

$\Delta T_f = i \times K_f \times m_0 = 2.01583 \times 1.86 \times 0.203998 \approx 0.7636$.

The difference from the given $\Delta T_f = 0.794$ is $|0.794 - 0.7636| = 0.0304$.

Let's check option (a): $K_a = 0.146$.

Solving $0.146 = \frac{0.203998 \alpha(1+\alpha)}{1-\alpha}$ gives $\alpha \approx 0.3468$.

$i = 2 + \alpha = 2.3468$.

$\Delta T_f = i \times K_f \times m_0 = 2.3468 \times 1.86 \times 0.203998 \approx 0.888$.

The difference from the given $\Delta T_f = 0.794$ is $|0.794 - 0.888| = 0.094$.

Comparing the differences, option (b) provides a $\Delta T_f$ value that is closer to the given value than option (a). While the calculated $K_a$ (0.02277) from the given data does not match any option, option (b) is the closest fit in terms of resulting colligative property. This suggests that option (b) is the intended answer, assuming a slight error in the problem data.

The final answer is $\boxed{3.3×10^{-3}}$.