A solution contains 1.77 g of dissolved silver. How many moles of potassium chloride must be added to the solution to completely precipitate all of the silver? What mass of potassium chloride must be added?

1 Answer
May 22, 2016

Answer:

#Ag^+ + Cl^(-) rarr AgCl(s)darr#

We need to add over #1# #g# #"KCl"#.

Explanation:

#"Moles of silver in solution"# #=# #(1.77*g)/(107.87*g*mol^-1`)# #=# #0.0164*mol#

Clearly we need an equivalent quantity of chloride, thus we add a mass of potassium chloride:

#=0.0164*molxx74.55*g*mol^-1# #=# #??*g#

Of course addition of the 1 equiv will precipitate most of the silver ion. Silver chloride will possess some (low) aqueous solubility, according to its solubility product. The result is a curdy white precipitate of silver chloride, that will be very hard to handle.