Consider the dissociation of the weak acid HX as given below HX(aq) leftharpoons H^+(aq)+X^-(aq), Ka=1.2 \times 10^-5 [ K_a : dissociation constant] The osmotic pressure of 0.03 M aqueous solution of HX at 300 ~K is \underline\hspace2cm \times 10^-2 bar (nearest integer). [Given : R=0.083 ~L ~bar ~mol^-1 ~K^-1 ]

Ionic Equilibrium PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q23 with Solution

View written solutionFree

Correct answer: 76

Given data

$\mathrm{HX}(aq) \rightleftharpoons \mathrm{H^+}(aq) + \mathrm{X^-}(aq), \qquad K_a = 1.2\times 10^{-5}$

Concentration of acid: $C = 0.03\,\text{M}$

Temperature: $T = 300\,\text{K}$

Gas constant: $R = 0.083\,\text{L bar mol}^{-1}\text{K}^{-1}$
Find degree of dissociation

Let degree of dissociation be $\alpha$ .

Initial and equilibrium concentrations:
$& \mathrm{HX} & \mathrm{H^+} & \mathrm{X^-} \\ \hline \text{Initial} & C & 0 & 0 \\ \text{Change} & -C\alpha & +C\alpha & +C\alpha \\ \text{Equilibrium} & C(1-\alpha) & C\alpha & C\alpha \end{array}$$ Therefore, $$K_a = \frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$ Since HX is a weak acid, $\alpha$ is small, so $1-\alpha \approx 1$. Hence, $$K_a \approx C\alpha^2$$ $$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2\times 10^{-5}}{0.03}}$$ $$\alpha = \sqrt{4\times 10^{-4}} = 2\times 10^{-2} = 0.02$$$
Van't Hoff factor

For $\mathrm{HX} \to \mathrm{H^+} + \mathrm{X^-}$ one particle gives 2 particles on dissociation.

Therefore, $i = 1 + \alpha = 1 + 0.02 = 1.02$
Osmotic pressure formula

$\pi = iCRT$

Substitute the values: $\pi = (1.02)(0.03)(0.083)(300)$

First, $0.083\times 300 = 24.9$

Then, $0.03\times 24.9 = 0.747$

Finally, $\pi = 1.02\times 0.747 = 0.76194\,\text{bar}$
Express in the asked form

We need $\pi = \underline{\hspace{1cm}} \times 10^{-2}\,\text{bar}$

$0.76194\,\text{bar} = 76.194\times 10^{-2}\,\text{bar}$

Nearest integer $= 76$ .
Final answer

$\boxed{76}$
Comparison with stored answer

Stored correct answer = $76$ .

My derived answer matches the stored answer.

More from Ionic Equilibrium

View written solutionFree

Correct answer: 76

Given data

$\mathrm{HX}(aq) \rightleftharpoons \mathrm{H^+}(aq) + \mathrm{X^-}(aq), \qquad K_a = 1.2\times 10^{-5}$

Concentration of acid: $C = 0.03\,\text{M}$

Temperature: $T = 300\,\text{K}$

Gas constant: $R = 0.083\,\text{L bar mol}^{-1}\text{K}^{-1}$
Find degree of dissociation

Let degree of dissociation be $\alpha$ .

Initial and equilibrium concentrations:
$& \mathrm{HX} & \mathrm{H^+} & \mathrm{X^-} \\ \hline \text{Initial} & C & 0 & 0 \\ \text{Change} & -C\alpha & +C\alpha & +C\alpha \\ \text{Equilibrium} & C(1-\alpha) & C\alpha & C\alpha \end{array}$$ Therefore, $$K_a = \frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$ Since HX is a weak acid, $\alpha$ is small, so $1-\alpha \approx 1$. Hence, $$K_a \approx C\alpha^2$$ $$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2\times 10^{-5}}{0.03}}$$ $$\alpha = \sqrt{4\times 10^{-4}} = 2\times 10^{-2} = 0.02$$$
Van't Hoff factor

For $\mathrm{HX} \to \mathrm{H^+} + \mathrm{X^-}$ one particle gives 2 particles on dissociation.

Therefore, $i = 1 + \alpha = 1 + 0.02 = 1.02$
Osmotic pressure formula

$\pi = iCRT$

Substitute the values: $\pi = (1.02)(0.03)(0.083)(300)$

First, $0.083\times 300 = 24.9$

Then, $0.03\times 24.9 = 0.747$

Finally, $\pi = 1.02\times 0.747 = 0.76194\,\text{bar}$
Express in the asked form

We need $\pi = \underline{\hspace{1cm}} \times 10^{-2}\,\text{bar}$

$0.76194\,\text{bar} = 76.194\times 10^{-2}\,\text{bar}$

Nearest integer $= 76$ .
Final answer

$\boxed{76}$
Comparison with stored answer

Stored correct answer = $76$ .

My derived answer matches the stored answer.

Ionic Equilibrium question

Ionic Equilibrium question

More from Ionic Equilibrium

Ionic Equilibrium question

Ionic Equilibrium question

More from Ionic Equilibrium