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?

Question

Cameron Andrews · Accepted Answer

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

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

#"Moles of silver in solution"# #=# #(1.77*g)/(107.87*g*mol^-1`)# #=# #0.0164*mol# We add a mass of potassium chloride since it is obvious that we require an equivalent amount of chloride: #=0.0164*molxx74.55*g*mol^-1# #=# #??*g# The majority of the silver ion will naturally precipitate upon addition of the 1 equiv. Silver chloride will have some (low) aqueous solubility based on its solubility product. This will yield a precipitate of silver chloride that is curdy and difficult to handle.

Avery Adams · Answer

undefined To completely precipitate all of the silver, you need to add an equal number of moles of chloride ions. Since silver chloride has a 1:1 ratio of silver to chloride, you would need the same number of moles of potassium chloride as moles of silver in the solution.

First, calculate the moles of silver:
$$
\text{Moles of silver} = \frac{\text{Mass of silver}}{\text{Molar mass of silver}}
$$

Then, use the mole ratio between silver and chloride to find the moles of potassium chloride needed. Since potassium chloride has a 1:1 ratio of potassium to chloride, you would use the same number of moles for potassium chloride as for chloride.

Finally, calculate the mass of potassium chloride needed:
$$
\text{Mass of potassium chloride} = \text{Moles of potassium chloride} \times \text{Molar mass of potassium chloride}
$$