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Kohlrausch Law

The ionic equ...

Question

The ionic equivalent conductivities of $K^{+}$ , Al^{3+}, and $S O_{4}^{2 -}$ ions are x,y and z $S c m^{2} e q^{- 1}$ , respectively. Calculate $Λ_{m}^{0}$ and $Λ_{e q}^{0}$ for $(K_{2} S O_{4} . A l_{2} (S O_{4})_{3} . 24 H_{2} O)$ (Potash alum).

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Solution

$Λ_{m}^{0} (A l u m) = 2 Λ_{m}^{0} (K^{+}) + 2 Λ_{m}^{0} ({A l}_{3 +}) + 4 Λ_{m}^{0} (S O_{4}^{2 -}) = 2 x + 2 y + 4 z = (2 x + 2 y + 4 z) S C m^{2} {m o l}^{- 1}$
and $∵ Λ_{e q}^{0} = \frac{Λ_{m}^{0}}{Λ_{f}} = (\frac{2 x = 2 y + 4 z}{8}) S c m^{2} e q^{- 1}$
Since $^{n} f a c t o r o f A l u m i s 8 (2 {A l}^{3 +}, 2 K^{+} = 8 + v e c h a r g e)$

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Similar questions

K^{+} A l^{3 +}, and S O_{4}^{2 -}

S c m^{2} E q^{- 1}

\land_{e q}^{0}

(K_{2} S O_{4} . A l_{2} {(S O_{4})}_{3} . 24 H_{2} O)

Given the ionic equivalent conductivities for the following ions :

$λ_{e q}^{\circ} K^{\oplus} = 73.5 c m^{2} o h m^{- 1} e q^{- 1}$

$λ_{e q}^{\circ} A l^{3 +} = 189 c m^{2} o h m^{- 1} e q^{- 1}$

$λ_{e q}^{\circ} S O_{4}^{2 -} = 85.8 c m^{2} o h m^{- 1} e q^{- 1}$

The $Λ_{e q}^{\circ}$ for potash alum $(K_{2} S O_{4} . A l_{2} (S O_{4})_{3} . 24 H_{2} O)$ is :

C_{2} O_{4}^{2 -}, K^{+},

N a^{+}

c m^{2} E q^{- 1}

A_{e q}^{0}

(N a O O C - C O O K)

1

N_{A}

= K_{2} S O_{4} . A l_{2} (S O_{4})_{3} . 24 H_{2} O

A_{M}^{\infty}

A_{e q .}^{\infty}

λ

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