Question

What is the solubility (M) of silver chloride in a 0.10 M solution of HCl?.

Answer 1

The formula weight of silver chloride is 107.8 + 35.4 = 143.3. The solubility of AgCl is

Silver chloride

AgCl

1.8×10–10

Solubilty   product of AgCl =    1.8*10^-10   = (Ag + conc) *( Cl - ion conc )    = s * s

so s = (1.8*10^-10)1/2 s = 1.3416 *10^-5

So Ag + = Cl -   = 1.3416 *10^-5

Here as 0.10 M sol of HCl   will contain Cl- conc ().10 M   common with AgCl

So ionisation of Cl -ion from AgCl is less by this common ion    = 1.3416*10^-5 -.10 - .10   =

so solubilty now   = Ag+ ion conc * Cl- ion conc 1.3416*10^-5 * (1.3416*10^-5) -0.10)