Question

The specific conductance of a saturated solution of AgCl is 1.55*10-6

ohm-1 cm-1

and 

that of water is 4.3*10-7

ohm-1 cm-1

. In a field of 1 volt per cm. mobility of chloride and 

silver ions are 5.6*10-4 and 6.8*10-4 cm/sec. calculate solubility and solubility product 

of AgCl.

Answer 1

$K_{AgCl} = K_{AgCl}(solution)-K_{H_2O}$

$= 1.55 \times 10^{-6} - 4.3 \times 10^{-7}$

$= 1.12 \times 10^{-6}$

Therefore,

$S = \dfrac{K \times 1000 }{137.2}$

$S = \dfrac{1.12 \times 10^{-6} \times 1000}{137.2}$

$S = 0.00816 \times 10^{-3}$