The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below: ∧°CH3COONa=91.0 S cm2 / equiv ∧°HCl=426.2cm2 / equiv What additional information/quantity one needs to calculate ∧° of an aqueous solution of acetic acid?

Question

The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below: ∧°CH3COONa=91.0 S cm2 / equiv ∧°HCl=426.2cm2 / equiv What additional information/quantity one needs to calculate ∧° of an aqueous solution of acetic acid? (A)  ∧° of NaCl (B)  ∧° of CH3COOK (C) The limiting equivalent conductance of H+( ∧° H+)  (D)  ∧° of chloroacetic acid (ClCH2COOH)

Tardigrade Admin · Accepted Answer

From Kohlrausch’s law Lambda° CH3COOH= Lambda°CH3COONa+ Lambda° HCl- LambdaNaCl° Hence, (1) is the correct answer.

$\land^{\circ}$ of $N a Cl$

$\land^{\circ}$ of $C H_{3} COO K$

The limiting equivalent conductance of $H^{+} (\land_{H^{+}}^{\circ})$

$\land^{\circ}$ of chloroacetic acid $(ClC H_{2} COO H)$

From Kohlrausch’s law
$Λ_{C H_{3} COO H}^{\circ} = Λ_{C H_{3} COON a}^{\circ} + Λ_{H Cl}^{\circ} - Λ_{N a Cl}^{\circ}$
Hence, (1) is the correct answer.

∧∘ of NaCl

∧∘ of CH3​COOK

The limiting equivalent conductance of H+(∧H+∘​)

∧∘ of chloroacetic acid (ClCH2​COOH)

From Kohlrausch’s law ΛCH3​COOH∘​=ΛCH3​COONa∘​+ΛHCl∘​−ΛNaCl∘​ Hence, (1) is the correct answer.

$\land^{\circ}$ of $N a Cl$

$\land^{\circ}$ of $C H_{3} COO K$

The limiting equivalent conductance of $H^{+} (\land_{H^{+}}^{\circ})$

$\land^{\circ}$ of chloroacetic acid $(ClC H_{2} COO H)$

From Kohlrausch’s law
$Λ_{C H_{3} COO H}^{\circ} = Λ_{C H_{3} COON a}^{\circ} + Λ_{H Cl}^{\circ} - Λ_{N a Cl}^{\circ}$
Hence, (1) is the correct answer.