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
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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?

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