Question

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?

Answer 1

Answer:

Explanation:

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.