Calculate $\Lambda _m^o$ for $CaCl_2$ and $MgSO_4$ from the data given in the table below:
Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ $H^{+} : 349.6$ $OH^{-} : 199.1$ $Na^{+} : 50.1$ $Cl^{-} : 76.3$ $K^{+} : 73.5$ $Br^{-} : 78.1$ $Ca^{2+} : 119.0$ $CH_3COO^{-} : 40.9$ $Mg^{2+} : 106.0$ $SO_4^{2-} : 160.0$
| Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ | Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ |
|---|---|
| $H^{+} : 349.6$ | $OH^{-} : 199.1$ |
| $Na^{+} : 50.1$ | $Cl^{-} : 76.3$ |
| $K^{+} : 73.5$ | $Br^{-} : 78.1$ |
| $Ca^{2+} : 119.0$ | $CH_3COO^{-} : 40.9$ |
| $Mg^{2+} : 106.0$ | $SO_4^{2-} : 160.0$ |
According to Kohlrausch law of independent migration of ions:
$\Lambda _{m(CaCl_2)}^o = \lambda _{Ca^{2+}}^o + 2\lambda _{Cl^{-}}^o$
$= 119.0 \ S \ cm^2 \ mol^{-1} + 2(76.3 \ S \ cm^2 \ mol^{-1})$
$= 119.0 + 152.6 = 271.6 \ S \ cm^2 \ mol^{-1}$
$\Lambda _{m(MgSO_4)}^o = \lambda _{Mg^{2+}}^o + \lambda _{SO_4^{2-}}^o$
$= 106.0 \ S \ cm^2 \ mol^{-1} + 160.0 \ S \ cm^2 \ mol^{-1}$
$= 266.0 \ S \ cm^2 \ mol^{-1}$
$\Lambda _{m(CaCl_2)}^o = \lambda _{Ca^{2+}}^o + 2\lambda _{Cl^{-}}^o$
$= 119.0 \ S \ cm^2 \ mol^{-1} + 2(76.3 \ S \ cm^2 \ mol^{-1})$
$= 119.0 + 152.6 = 271.6 \ S \ cm^2 \ mol^{-1}$
$\Lambda _{m(MgSO_4)}^o = \lambda _{Mg^{2+}}^o + \lambda _{SO_4^{2-}}^o$
$= 106.0 \ S \ cm^2 \ mol^{-1} + 160.0 \ S \ cm^2 \ mol^{-1}$
$= 266.0 \ S \ cm^2 \ mol^{-1}$
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