First slide

Applications of Kohlraush law

Question

The limiting equivalent conductivity of NaCl, KCI and KBr are 126.s, 150.0 and151.5 scm²eq^-1, respectively. The limiting equivalent ionic conductance for Br^- is 78 S cm² eq^-1. The limiting equivalent ionic conductance for Na⁺ ions would be :

Easy

A

128
B

125
C

49
D

50

Solution

$\begin{array}{l} Λ_{m}^{\infty} (NaBr) = Λ_{m}^{\infty} (NaCl) + Λ_{m}^{\infty} KBr - Λ_{m}^{\infty} (KCl) \\ λ_{m}^{\infty} ({Na}^{+}) + λ_{m}^{\infty} ({Br}^{-}) = 126 .5 + 151 .5 - 150 \\ λ_{m}^{\infty} ({Na}^{+}) = 50 - 78 = 50 \end{array}$

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