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The limiting equivalent conductivity of...

The limiting equivalent conductivity of `NaCl, KCl` and `KBr` are `126.5,150.0` and `151.5 S cm^(2) eq^(-1)` , respectively. The limiting equivalent ionic conductance for `Br^(-)` is `78 S cm^(2) eq^(-1)`. The limiting equivalent ionic conductance for `Na^(+)` ions would be `:`

A

128

B

125

C

49

D

50

Text Solution

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To find the limiting equivalent ionic conductance for Na⁺ ions, we can follow these steps: ### Step 1: Understand the Problem We are given the limiting equivalent conductivities of NaCl, KCl, and KBr, as well as the limiting equivalent ionic conductance for Br⁻. We need to find the limiting equivalent ionic conductance for Na⁺. ### Step 2: Use Kohlrausch's Law According to Kohlrausch's law, the limiting equivalent conductivity (Λ) of an electrolyte at infinite dilution can be expressed as the sum of the limiting equivalent ionic conductivities of its constituent ions. For NaBr, this can be represented as: \[ Λ_{NaBr} = Λ_{Na^+} + Λ_{Br^-} \] ### Step 3: Calculate the Limiting Equivalent Conductivity for NaBr To find \(Λ_{NaBr}\), we can use the values given for NaCl, KCl, and KBr: \[ Λ_{NaBr} = Λ_{NaCl} + Λ_{KBr} - Λ_{KCl} \] Substituting the values: - \(Λ_{NaCl} = 126.5 \, S \, cm^{-2} \, eq^{-1}\) - \(Λ_{KBr} = 151.5 \, S \, cm^{-2} \, eq^{-1}\) - \(Λ_{KCl} = 150.0 \, S \, cm^{-2} \, eq^{-1}\) Now, substituting these values into the equation: \[ Λ_{NaBr} = 126.5 + 151.5 - 150.0 \] Calculating this gives: \[ Λ_{NaBr} = 128.0 \, S \, cm^{-2} \, eq^{-1} \] ### Step 4: Substitute into the Equation for Na⁺ Now that we have \(Λ_{NaBr}\), we can substitute this value into the equation from Step 2: \[ 128.0 = Λ_{Na^+} + 78 \] Where \(78 \, S \, cm^{-2} \, eq^{-1}\) is the limiting equivalent ionic conductance for Br⁻. ### Step 5: Solve for \(Λ_{Na^+}\) Rearranging the equation to solve for \(Λ_{Na^+}\): \[ Λ_{Na^+} = 128.0 - 78 \] Calculating this gives: \[ Λ_{Na^+} = 50.0 \, S \, cm^{-2} \, eq^{-1} \] ### Final Answer The limiting equivalent ionic conductance for Na⁺ ions is: \[ \boxed{50.0 \, S \, cm^{-2} \, eq^{-1}} \] ---
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