Question

If conductivity of water used to make saturated of AgCl is found to be $3.1\times10^{-5}Ω^{-1}cm^{-1}$ and conductance of the solution of AgCl = $4.5\times10^{-5}Ω^{-1}cm^{-1}$

If $λ^θ$ AgNO3 = $200 Ω^{-1}cm^{2}mole^{-1}$

$λ^θ$NaNO3 = $310 Ω^{-1}cm^{2}mole^{-1}$

calculate Ksp of AgCl

Answer 1

** ** Here, λ0Agcl = 140

Total conductance = 10-5

S = 140x4x10-5x1000/140

= 1.4x10-4/14

= 5.4x10-4

Now, S2 = 1x10-8