Question

A solution is prepared by mixing 0.12 L of 0.13 M potassium chloride with 0.22 L of a 0.19 M magnesium chloride.

What volume of a 0.22 M silver nitrate solution is required to precipitate all of the chloride ion in the solution as silver chloride?

Answer 1

Moles of KCl = molarity * volume = 0.13 * 0.12 = 1.56 * 10^-2 moles

Moles of MgCl₂ = 0.19 * 0.22 = 4.18 * 10^-2 moles

Moles of chloride = moles of NaCl + 2 * moles of MgCl₂ = 1.56 + 4.18 = 5.74 * 10^-2 moles

Moles of AgNO₃ needed = moles of Chlorine

5.74 * 10^-2 = 0.22 * Volume

Solving, Volume = 0.261 Liters