The value of α is _______ . | TestHub

Passage / Comprehension

At $298 K$ , the limiting molar conductivity of a weak monobasic acid is $4 \times 1 0^{2} S cm^{2} mol^{- 1}$ . At $298 K$ , for an aqueous solution of the acid the degree of dissociation is $α$ and the molar conductivity is $y \times 1 0^{2} S cm^{2} mol^{- 1}$ . At $298 K$ , upon 20 times dilution with water, the molar conductivity of the solution becomes $3 y \times 1 0^{2} S cm^{2} mol^{- 1}$

The value of $α$ is _______ .

Answer:

0.22

Solution:

$α = \frac{λ_{m}}{λ_{m}^{\circ}}$

$α_{1} = \frac{y \times 10^{+ 2}}{4 \times 10^{2}} = \frac{y}{4} . . . (1)$

$k_{a} = \frac{C_{1} α_{1}^{2}}{1 - α_{1}} . . . (2)$

After dilution

$α_{2} = \frac{3 y \times 10^{2}}{4 \times 10^{2}} = \frac{3 y}{4} . . . (3)$

$k_{a} = \frac{C_{2} α_{2}^{2}}{1 - α_{2}} . . . (4)$

eq. $(1) / (3)$

$\Rightarrow \frac{α_{1}}{α_{2}} = \frac{1}{3}$

eq $(2) / (4)$

$\Rightarrow 1 = \frac{C_{1} α_{1}^{2} (1 - α_{2})}{(1 - α_{1}) C_{2} α_{2}^{2}}$

$\frac{α_{1}^{2} (1 - α_{2})}{α_{2}^{2} (1 - α_{1})} = \frac{C_{2}}{C_{1}}$

$C_{1} = C_{2} 20 y$

$\frac{1}{2} = \frac{C_{2}}{C_{1}}$

$\frac{α_{1}^{2} (1 - α_{2})}{α_{2}^{2} (1 - α_{1})} = \frac{1}{20}$

$\frac{20 α_{1}^{2}}{(1 - α_{1})} = \frac{α_{2}^{2}}{(1 - α_{2})}$

$\frac{20}{(1 - α_{1})} = \frac{9}{(1 - α_{2})}$

$20 - 60 α_{1} = 9 - 9 α_{1}$

$11 = 51 α_{1}$

$α_{1} = \frac{11}{51} = 0 .22$

Chemistry - Electrochemistry Question with Solution | TestHub

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