How many mL of 0.15 M magnesium chloride are required to completely precipitate the silver ion in 25.0 mL of .25 M silver nitrate?

Question

Consider the reaction between aqueous magnesium chloride and aqueous silver nitrate. How many mL of 0.15 M magnesium chloride are required to completely precipitate the silver ion in 25.0 mL of .25 M silver nitrate?

Answer 1

Write a correctly balanced equation for the reaction:

MgCl₂(aq) + 2AgNO₃(aq) ==> Mg(NO₃)₂(aq) + 2AgCl(s)

From the net ionic equation:

2Ag⁺(aq) + 2Cl^-(aq) ==> 2AgCl(s)

From the balanced equation, it can be seen that it takes ONE mole MgCl₂ to precipitate TWO moles of AgNO₃ in the form of AgCl.

So, first calculate how many moles of Ag you have  in 25.0 ml of 0.25 M AgNO₃:

0.025 L x 0.25 moles AgNO₃/L x 1 mole Ag/mole AgCl = 0.00625 moles Ag present

Next, calculate moles of MgCl₂ needed to precipitate 0.00625 moles Ag:

0.00625 moles Ag x 1 mole MgCl₂/2 moles Ag = 0.003125 moles MgCl₂

Finally, find the volume of 0.15 M MgCl₂ needed to supply this amount:

(x L)(0.15 moles/L) = 0.03125 moles and x = 0.02083 liters = 20.8 ml = 21 ml (2 significant figures).