Question

A solution is prepared by mixing 0.10 L of 0.12 M sodium chloride with 0.23 L of a 0.19 M MgCl2 solution. What volume of a 0.20 M silver nitrate solution is required to precipitate all the Cl− ion in the solution as AgCl?

Answer 1

V = 0.10 L

M = 0.12 NaCl

V2 = 0.23 L

M = 0.19 MgCl2

V3 = ? AgNO3

recipitate all Cl- ions

We must go to equilibrium

Ag+(aq) + Cl-(aq) <---> AgCl(s)

Ksp = [Ag+][Cl-]

[Cl-] = ions of NaCl and ions of MgCl2

find each amount

moles of NaCl = M*V = 0.1*0.12 = 0.012 mol of NaCl present

moles of MgCl2 = M*V = 0.23*0.19 = 0.0437 mol of MgCl2 present (note that Cl is present double!)

Total moles of Cl = 0.012+ 2*0.0437 = 0.0994 mol of Cl-

Ag+(aq) + Cl- <---> AgCl(s)

since Ksp is so small (1-8*10^-10) we can say that there is almost no solubility

therefore, we need 0.0994 mol of Ag+ to react with 0.0994 mol of Cl-

therefore

AgNO3 ---> Ag+ andNO3-

we need 0.0994 mol of AgNO3

Find that in volume of solution

M = mol/V

V = mol/M = 0.0994 mol / 0.20 = 0.497 liters are needed

or 497 ml