Potassium iodide reacts with lead (II) nitrate in this precipitation reaction: 2 KI (aq) + Pb (NO3) 2 (aq) → 2 KNO3 (aq) + PbI2 (s) 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? a. 174 mL

b. 86.8 mL

c. 189 mL

d. 94.6 mL

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI (aq) + Pb (NO₃) ₂ (aq) → 2 KNO₃ (aq) + PbI₂ (s)

We have 155.0 mL of a 0.112 M lead (II) nitrate solution. The moles of Pb (NO₃) ₂ are:

0.1550 L * 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb (NO₃) ₂ is 2:1. The moles of KI are:

2 * 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol * (1 L / 0.200 mol) = 0.174 L = 174 mL