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:
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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