Question

A sample of agcl was treated with 5 ml 1.5 m na2co3 to give ag2co3 remaining sol has .0026 cl- ksp

Answer 1

a product has formed , just answered for points

Answer 2

"We know that sodium carbonate is the strong electrolyte

The concentration of carbonate in sodium carbonate is given as,

$[CO_3]^{2-} = [Na_2CO_3] = 1.5M$

$Ksp(AgCO) = [Ag+] [CO_3]^{2-}$

The solubility product of $AgNO_3, Ksp(AgCO_3) is 8.2 x 10^{-12}$

$8.2 x 10^{-12} = [Ag^+] (1.5)$

$[Ag^+] =(8.5 x 10^{-12})/1.5 = 5.66 x 10^{-12}M$

$Ksp(AgCl)= [Ag^+] [Cl^-] =(5.66 x 10^{-12})(0.0026/35.5) = 4.144 x 10^{-16}$

(The solution has given Ksp of $Cl^-$ as 0.0026 and its atomic weight is 35.5)"