What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a 0.112 m lead (ii) nitrate solution?

First, we write the balanced equation for this reaction:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

Moles = volume (in L) * molarity

We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L

The volume required is 173.6 mL

Anions are negative charged ions