Question

Specific conductance of a saturated solution of AgBr is 8.486×10–7 ohm–1cm–1 at 250C. Specific conductance of pure water at 25°C is 0.75 ×10–6 ohm–1 cm–2 . m for KBr , AgNO3 and KNO3 are 137.4 , 133 , 131 ( S cm2 mol–1) respectively. Calculate the solubility of AgBr in gm/litre.

Answer 1

KBr+AgNO3→AgBr+KNO3

∴Am∞AgBr=Am∞(KBr)+(AgNO3)−KNO3

                               =137.4+133−131

Am∞=139.4Scm2mol−1

K of AgBR= saturated solution − pure water

                      =8.486×16−7−0.75×10−6

C=7.07×10

−7

mol/L

                      =9.86×10−8ohm−1cm−1

we know, Am=K/C×1000

    =9.86×10−8×1000/139.4

C=7.07×10 -⁷ mol/L