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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
MCQ
Plotting $1 / \Lambda_{\mathrm{m}}$ against $\mathrm{c} \Lambda_{\mathrm{m}}$ for aqueous solutions of a monobasic weak acid ($HX$) resulted in a straight line with $y$-axis intercept of $\mathrm{P}$ and slope of $S$. The ratio $\mathrm{P} / \mathrm{S}$ is

$\left[\Lambda_{\mathrm{m}}=\right.$ molar conductivity

$\Lambda_{\mathrm{m}}^{\circ}=$ limiting molar conductivity

$\mathrm{c}=$ molar concentration

$\mathrm{K}_{\mathrm{a}}=$ dissociation constant of $\mathrm{HX}$ ]

  • ✓
    $\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$
  • B
    $\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ} / 2$
  • C
    $2 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$
  • D
    $1 /\left(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}\right)$
✓

Answer

Correct option: A.
$\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$
a
For weak acid, $\alpha=\frac{\Lambda_m}{\Lambda_0}$

$\mathrm{K}_\alpha=\frac{C \alpha^2}{1-\alpha} \Rightarrow \mathrm{K}_{\mathrm{s}}(1-\alpha)=C \alpha^2$

$\Rightarrow \mathrm{K}_\alpha\left(1-\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)=C\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)^2$

$\Rightarrow \mathrm{K}_2-\frac{\Lambda_{\mathrm{m}} \mathrm{K}_{\mathrm{a}}}{\Lambda_0}=\frac{C \Lambda_{\mathrm{m}}^2}{\left(\Lambda_0\right)^2}$

Diride by ' $\Lambda_{\mathrm{m}}$ '

$\Rightarrow \frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\left(\Lambda_0\right)^2}+\frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_0}$

$\Rightarrow \frac{1}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\Lambda_0\right)^2}+\frac{1}{\Lambda_0}$

Plot $\frac{1}{\Lambda_m}$ vi $C \Lambda_m$ has

Slope $=\frac{1}{\mathrm{~K}_{\mathrm{u}}\left(\Lambda_0\right)^2}=\mathrm{S}$

$\mathrm{y} \text {-intercept }=\frac{1}{\Lambda_0}=\mathrm{P}$

Then, $\frac{P}{S}=\frac{\frac{1}{\Lambda_0}}{\frac{1}{K_2\left(\Lambda_0\right)^2}}=\mathrm{K}_2 \Lambda_0$

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