Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride:
Ag+ (aq) + Cl- (aq) →
AgCl (s)
Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.276 M AgNO3 solution to completely precipitate the silver?
First, we need to determine the moles of AgNO3 in the solution:
Moles of AgNO3 = Molarity x Volume
Moles of AgNO3 = 0.276 mol/L x 0.025 L
Moles of AgNO3 = 0.0069 moles
Since the reaction is 1:1 between Ag+ and Cl-, we need an equal number of moles of Cl- ions to completely precipitate the silver. Therefore, we need 0.0069 moles of NaCl.
Now, we need to calculate the mass of NaCl needed:
Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl)
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = moles of NaCl x molar mass of NaCl
Mass of NaCl = 0.0069 mol x 58.44 g/mol
Mass of NaCl = 0.403 g
Therefore, 0.403 grams of solid NaCl must be added to 25.0 mL of 0.276 M AgNO3 solution to completely precipitate the silver.