Question

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?

Answer 1

`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.