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 \(λ^θ\) AgNO_{3  =} \(200 Ω^{-1}cm^{2}mole^{-1}\)

\(λ^θ\)NaNO₃ = \(310 Ω^{-1}cm^{2}mole^{-1}\) 

calculate K_sp of AgCl

Answer 1

 Here, λ⁰_Agcl = 140

Total conductance = 10^-5 

S = 140x4x10^-5x1000/140

= 1.4x10^-4/14

= 5.4x10^-4

Now, S² = 1x10^-8