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

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