What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 195.0 ml of a 0.194 m lead (ii) nitrate solution?

First, we write the reaction equation:

2KI + PbNO₃ → K₂NO₃ + PbI₂

The molar ratio of KI to PbNO₃ is 2 : 1

Moles of PbNO₃ present:

Moles = concentration (M) x volume (dm³)

= 0.194 x 0.195

= 0.038

Moles of KI required = 2 x 0.038 = 0.076 moles

concentration = moles / volume

volume = moles / concentration

= 0.076 / 0.2

= 0.38 L = 380 ml