Well, that's a Cl-ear question! To find out the volume of silver nitrate solution required to precipitate all the Cl- ions, we need to find the limiting reagent in this reaction.
Firstly, let's determine the moles of sodium chloride (NaCl) and magnesium chloride (MgCl2) in the solution:
Moles of NaCl = 0.12 L x 0.12 M = 0.0144 mol
Moles of MgCl2 = 0.22 L x 0.19 M = 0.0418 mol
From the balanced equation, we know that 1 mole of MgCl2 reacts with 2 moles of NaCl to form 2 moles of AgCl. So, the number of moles of AgCl formed would be the same as the number of moles of NaCl. In this case, it's 0.0144 mol.
Now, let's calculate the volume of the silver nitrate (AgNO3) solution required to form 0.0144 moles of AgCl:
Moles of AgNO3 = 0.0144 mol (since it's a 1:1 ratio with AgCl)
Volume of AgNO3 = Moles of AgNO3 / Concentration of AgNO3
The concentration of AgNO3 is given as 0.22 M. So,
Volume of AgNO3 = 0.0144 mol / 0.22 M = 0.0655 L
Therefore, the volume of a 0.22 M silver nitrate solution required to precipitate all the Cl- ions in the solution is 0.0655 liters, or about 65.5 milliliters.
Now, let's hope this answer doesn't precipitate any groans!