# Problem: Potassium iodide reacts with lead(II) nitrate in the following 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?

###### FREE Expert Solution
87% (389 ratings)
###### FREE Expert Solution

Step 1: From the problem, the given reaction is:

2 KI (aq) + Pb(NO3)2 (aq) = 2 KNO3 (aq) + Pbl2 (s)          This equation is balanced.

Step 2: We need to do the following:

Molarity of Pb(NO3)2 (volume of Pb(NO3)2) → Moles of Pb(NO3)2 (mole-to-mole comparison)  Moles of KI (molarity of KI)  Volume of KI

We’re given the following:

Molarity of Pb(NO3)2 = 0.112 M

Volume of Pb(NO3)2 = 155.0 mL

Molarity of KI = 0.200 M

Mole-to-mole comparison = 1 mole of Pb(NO3)2 reacts with 2 moles of KI

The volume of KI required is:

87% (389 ratings)
###### Problem Details

Potassium iodide reacts with lead(II) nitrate in the following 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?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems .

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Based on our data, we think this problem is relevant for Professor Ellenberger's class at UGA.