Answer the fast the tell the please thank you the - Chemistry - Equilibrium

Question

Answer the fast the tell the please thank you the

Goutam Pradhan · Accepted Answer

Solution:Solubility product of AgCl is 4×10-10 at 298KWe have to find solubility of AgCl in 0
04M CaCl2[Cl(aq)-]=0
08M will have common ion effect with Cl(aq)- of AgClAgCl(s) ⇌Ag(aq)++Cl(aq)-                    s            sCaCl2 → Ca(aq)2++2Cl(aq)-0
04M                          0
08MKsp=[Ag(aq)+][Cl(aq)-] 4×10-10=s×(0
08)s=5×10-9 MOption(c) is correct