Question

A solution is prepared by mixing 0.12 L of 0.14 M sodium chloride with 0.22 L of a 0.17 M MgCl2 solution. What volume of a 0.21 M silver nitrate solution is required to precipitate all the Cl− ion in the solution as AgCl?

Answer 1

Solution- NaCl dissociates as below-

NaCl => Na+ + Cl-

Moles of Cl- = moles of NaCl = volume x concentration

= 0.12 x 0.14 = 0.0168 mol

And MgCl2 dissociates as follow-

MgCl2 => Mg2+ + 2 Cl-

Moles of Cl = 2 x moles of MgCl2 = 2 x volume x concentration

= 2 x 0.22 x 0.17 = 0.0748 mol

Total moles of Cl- = 0.0168 + 0.0748 = 0.0916 mol

Cl- + AgNO3 => AgCl + NO3-

Moles of AgNO3 = Total moles of Cl- = 0.0916 mol

Now,Volume of AgNO3 = moles/concentration

= 0.0916/0.21

= 0.44L