A sample of solution has a mixture of 0.01 M Aluminum Chloride and 0.05 M Magnesium Chloride. What is the molarity of Silver that is required to precipitate all the chloride ions present in the solution?

AlCl₃ ⇒Al⁺³ + 3Cl⁻¹

.01M 3 x. 01 M

So. 01 M AlCl₃ will give 3 x. 01 M Cl⁻¹

MgCl₂ ⇒ Mg⁺² + 2Cl⁻¹

.05 M 2 x. 05M

.05M MgCl₂ will give 2 x. 05 M Cl⁻¹

Total Cl⁻¹ formed = 3 x. 01 + 2 x. 05 M

=.03 +.1 M

=.13 M

AgCl ⇒ Ag⁺ + Cl⁻¹

So AgCl required =.13 M. to precipitate all chloride ions.