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Question: Conductivity of \(2.5 \times {10^{ - 4}}{\text{ M}}\) methanoic acid is \(5.25 \times {10^{ - 5}}{\t...

Conductivity of 2.5×10−4 M2.5 \times {10^{ - 4}}{\text{ M}} methanoic acid is 5.25×10−5 S cm−15.25 \times {10^{ - 5}}{\text{ S c}}{{\text{m}}^{ - 1}}. Calculate its molar conductivity and degree of dissociation.
(Given: λ0(H+)=349.5 S cm2 mol−1{\lambda ^0}\left( {{{\text{H}}^ + }} \right) = 349.5{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} and λ0(HCOO−)=50.5 S cm2 mol−1{\lambda ^0}\left( {{\text{HCO}}{{\text{O}}^ - }} \right) = 50.5{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}})

Explanation

Solution

The conductance of a unit cube material is known as conductivity. The units of conductivity are S cm−1{\text{S c}}{{\text{m}}^{ - 1}} or S m−1{\text{S }}{{\text{m}}^{ - 1}}. The ratio of conductivity to the molar concentration of the dissolved electrolyte is known as its molar conductivity. The units of molar conductivity are S cm2 mol−1{\text{S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} or S m2 mol−1{\text{S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}.

Formulae used: Λ=kC\Lambda = \dfrac{k}{C}
Λ∘=λ0(cation)+λ0(anion){\Lambda _ \circ } = {\lambda ^0}\left( {{\text{cation}}} \right) + {\lambda ^0}\left( {{\text{anion}}} \right)
α=ΛΛ∘\alpha = \dfrac{\Lambda }{{{\Lambda _ \circ }}}

Complete step by step answer:
Calculate the molar conductivity of methanoic acid using the equation as follows:
Λ=kC\Lambda = \dfrac{k}{C}
Where Λ\Lambda is the molar conductivity of the solution,
k is the conductivity of the solution,
C is the molar concentration of the solution.
The number of moles of a solute per litre of solution is known as the molarity of the solution. Thus, the molarity of the solution is 2.5×10−4 M=2.5×10−4 mol L−12.5 \times {10^{ - 4}}{\text{ M}} = 2.5 \times {10^{ - 4}}{\text{ mol }}{{\text{L}}^{ - 1}}.
Substitute 5.25×10−5 S cm−15.25 \times {10^{ - 5}}{\text{ S c}}{{\text{m}}^{ - 1}} for the conductivity of the solution, 2.5×10−4 mol L−12.5 \times {10^{ - 4}}{\text{ mol }}{{\text{L}}^{ - 1}} for the molarity of the solution. Thus,
λ=5.25×10−5 S cm−12.5×10−4 mol L−1\lambda = \dfrac{{5.25 \times {{10}^{ - 5}}{\text{ S c}}{{\text{m}}^{ - 1}}}}{{2.5 \times {{10}^{ - 4}}{\text{ mol }}{{\text{L}}^{ - 1}}}}
But 1 L=103 cm3{\text{1 L}} = {10^3}{\text{ c}}{{\text{m}}^3}. Thus,
λ=5.25×10−5 S cm−12.5×10−4 mol L−1×103 cm3L\lambda = \dfrac{{5.25 \times {{10}^{ - 5}}{\text{ S c}}{{\text{m}}^{ - 1}}}}{{2.5 \times {{10}^{ - 4}}{\text{ mol }}{{\text{L}}^{ - 1}}}} \times \dfrac{{{{10}^3}{\text{ c}}{{\text{m}}^3}}}{{\text{L}}}
λ=210 S cm2 mol−1\lambda = 210{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}
Thus, the molar conductivity of methanoic acid is 210 S cm2 mol−1210{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}.
Calculate the molar conductivity of methanoic acid at infinite dilution using the Kohlrausch law as follows:
Kohlrausch law states that at infinite dilution, each ion migrates independently of its co-ion and makes its contribution to the total molar conductivity. Thus,
Λ∘(HCOOH)=λ0(H+)+λ0(HCOO−){\Lambda _ \circ }\left( {{\text{HCOOH}}} \right) = {\lambda ^0}\left( {{{\text{H}}^ + }} \right) + {\lambda ^0}\left( {{\text{HCO}}{{\text{O}}^ - }} \right)
Where, Λ∘(HCOOH){\Lambda _ \circ }\left( {{\text{HCOOH}}} \right) is the molar conductivity of methanoic acid at infinite dilution,
λ0(H+){\lambda ^0}\left( {{{\text{H}}^ + }} \right) is the molar conductivity of H+{{\text{H}}^ + } ion at infinite dilution,
λ0(HCOO−){\lambda ^0}\left( {{\text{HCO}}{{\text{O}}^ - }} \right) is the molar conductivity of HCOO−{\text{HCO}}{{\text{O}}^ - } ion at infinite dilution.
Substitute 349.5 S cm2 mol−1349.5{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} for the molar conductivity of H+{{\text{H}}^ + } ion at infinite dilution, 50.5 S cm2 mol−150.5{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} for the molar conductivity of HCOO−{\text{HCO}}{{\text{O}}^ - } ion at infinite dilution. Thus,
Λ∘(HCOOH)=(349.5+50.5) S cm2 mol−1{\Lambda _ \circ }\left( {{\text{HCOOH}}} \right) = \left( {349.5 + 50.5} \right){\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}
Λ∘(HCOOH)=400 S cm2 mol−1{\Lambda _ \circ }\left( {{\text{HCOOH}}} \right) = 400{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}
Thus, molar conductivity of methanoic acid at infinite dilution is 400 S cm2 mol−1400{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}.
Calculate the degree of dissociation of methanoic acid using the equation as follows:
The degree of dissociation of a weak electrolyte is the ratio of its molar conductivity to its molar conductivity at infinite dilution. Thus,
α=ΛΛ∘\alpha = \dfrac{\Lambda }{{{\Lambda _ \circ }}}
Where α\alpha is the degree of dissociation,
Λ\Lambda is the molar conductivity of the solution,
Λ∘{\Lambda _ \circ } is the molar conductivity at infinite dilution,
Substitute 210 S cm2 mol−1210{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} for the molar conductivity, 400 S cm2 mol−1400{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} for the molar conductivity at infinite dilution. Thus,
α=210 S cm2 mol−1400 S cm2 mol−1\alpha = \dfrac{{210{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}{{400{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}
α\alpha = 0.525

Thus, the degree of dissociation of methanoic acid is 0.525.

Note: The conducting power of all the ions that are produced by dissolving one mole of any electrolyte in the solution is known as the molar conductivity of the solution.
As the temperature of the solution increases, the molar conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases.
The molar conductivity of both strong and weak electrolytes increases as the dilution increases. This is because dilution increases the degree of dissociation and the total number of ions that carry current increases.