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Finding the mass of precipitate by common ion effect

Submitted by ctr91 on Wed, 10/26/2011 - 19:50
A volume of 10.0mL of a 0.00550 M solution of Cl^- ions are reacted with 0.500 M solution of AgNO3. What is the maximum mass of AgCl that precipitates?

The problem could have asked how much AgCl will form (in mass units, not moles).

You know 1 mol Cl- will react with 1 mol Ag+ (or, in other words 1 mol of AgNO3) to yield 1 mol AgCl.

Usually, this kind of problem is a question of calculating number of moles of each reactant, and figuring out which one is the limiting reactant. As you posted it, there is no given amount of silver nitrate solution, so you may add an excess of it, and know it will not be the limiting reactant.

(10.0mL Cl- solution)(0.00550mol Cl-/L Cl- solution) = 0.0550 millimoles Cl-

(Your instructor may want it written as 5.50·10-5mol Cl-).

From 0.0550mmol Cl-, 0.0550mmol AgCl will form. You need the molar mass of AgCl to calculate the mass.

You could find the formula weight for AgCl in a table, but you are probably expecte to add atomic weights for Ag and Cl to calculate it as 107.87+35.45=143.3.

(5.50·10-5mol Cl-)(1mol AgCl/mol Cl-)(143.3g AgCl/mol AgCl) = 7.88·10-3g AgCl

The answer is 7.88mg AgCl.


It is obvious from the question that Cl- ions are limiting as no volume is given for AgNO3.

Ag+ (aq) (from AgNO3) + Cl-(aq) ----> AgCl(s)

(1 mole Ag+) + 1 mole Cl- ----> 1 mole AgCl. Find the mass of 1 mole AgCl (Molar mass). Then find the number of moles of Cl- from volume in L x molarity. Then find the mass of AgCl you can get from this number of moles.