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प्रश्न
The conductivity of a 0.01 M solution of a 1 : 1 weak electrolyte at 298 K is 1.5 × 10−4 S cm−1.
i) molar conductivity of the solution
ii) degree of dissociation and the dissociation constant of the weak electrolyte
Given that
\[\ce{λ^∘_{cation}}\] = 248.2 S cm2 mol−1
\[\ce{λ^∘_{anion}}\] = 51.8 S cm2 mol−1
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उत्तर
Given C = 0.01 M
K = 1.5 × 10−4 S cm−1
\[\ce{λ^∘_{cation}}\] = 248.2 S cm2 mol−1
\[\ce{λ^∘_{anion}}\] = 51.8 S cm2 mol−1
i) Molar conductivity
K = 1.5 × 10−4 S cm−1
1 cm−1 = 102 m−1
= 1.5 × 102
`Λ_"m"^∘ = ("K"("sm"^-1) xx 10^-3)/("C"("in M"))` mol−1 m3
= `(1.5 xx 10^2 xx 10^-3)/0.01` S mol−1 m2
= 1.5 × 10−3 S m2 mol−1
ii) Degree of dissociation α = `(Λ^∘)/(Λ_∞^∘)`
`(Λ_∞^∘) = λ_"cation"^∘ + λ_"anion"^∘`
= (248.2 + 51.8) S cm2 mol−1
= 300 S cm2 mol−1
= 300 × 10−4 S m2 mol−1
α = `(1.5 xx 10^-3 "S m"^2 "mol"^-1)/(300 xx 10^-4 "S m"^2 "mol"^-1)`
α = 0.05
Ka = `(α^2"C")/(1 - α)`
= `((0.05)^2 (0.01))/(1 - 0.05)`
= `(25 xx 10^-4 xx 10^-2)/(95 xx 10^-2)`
= 0.26 × 10−4
= 2.6 × 10−5
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