1.

How would you calculate the solubility of sparingly soluble salt using Kohlrausch's law?

Answer»

Solution :(i) SUBSTANCES like `AgCl, PbSO_4` are sparingly soluble in water. The solubility PRODUCT can be determined using conductivity experiments.
(ii) Let us CONSIDER AgCl as an example
`AgCl_((s)) `K_(SP) = [Ag^+][Cl^-]`
(iii) Let the concetration of `[Ag^+]` be `.C.` mol `L^(-1)`
If `[Ag^+] = C`, then `[Cl^-]`is also equal to C mol `L^(-1)`
`:. K_(sp) = C.C`
`K_(sp) = C^2`
(iv) The relationship between molar conductance and equivalent conductance is
`Lambda_@ = (K xx 10^(-3))/(C (mol L^(-1))) (or ) C = (k xx 10^(-3))/(Lambda^@)`
Substitute the concentration value in the relation `K_(sp) = C^2`
`K_(sp) = [(k xx 10^(-3))/(Lambda^@)]^(2)`.


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