To find how many drops of 0.20M KI are needed to precipitate lead iodide, we need to calculate the moles of both the lead nitrate (Pb(NO3)2) and the potassium iodide (KI), and then determine which one is the limiting reagent.
Step 1: Calculate the moles of Pb(NO3)2 in 100.0 mL of 0.010M solution:
Moles of Pb(NO3)2 = Molarity x Volume (L)
= 0.010M x 0.100 L
= 0.001 mol
Step 2: Convert the moles of Pb(NO3)2 to moles of PbI2 using the stoichiometry of the balanced equation. The balanced equation is:
Pb(NO3)2 + 2KI -> PbI2 + 2KNO3
Since the stoichiometric coefficient for Pb(NO3)2 is 1, we have an equal number of moles of Pb(NO3)2 and PbI2.
Moles of PbI2 formed = Moles of Pb(NO3)2 = 0.001 mol.
Step 3: Calculate the concentration of PbI2 using the volume of solution and the moles of PbI2 formed:
Concentration of PbI2 = Moles of PbI2 / Volume (L)
= 0.001 mol / 0.100 L
= 0.010 M
Step 4: Use the Ksp of PbI2 and the concentration of PbI2 to find the concentration of iodide (I-) ions in solution at the point of precipitation:
Ksp = [Pb2+][I-]^2
Since the stoichiometric coefficient for iodide in the balanced equation is 2, the concentration of iodide ions is twice the concentration of PbI2.
[Pb2+] = [I-] = 0.010 M (from Step 3)
0.010 = (0.010)(0.020)^2
0.010 = 0.010 x 0.0004
0.010 = 4.0e-6
Step 5: Calculate the moles of iodide ions required to reach the concentration calculated in Step 4:
Moles of I- = Concentration x Volume (L)
= 4.0e-6 x 0.100 L
= 4.0e-7 mol
Step 6: Calculate the number of drops of 0.20M KI required to supply the moles of iodide ions:
Moles of KI = Molarity x Volume (L)
= 0.20M x Volume (L)
Since 1 drop corresponds to 0.05 mL (as given in the question), we need to convert the volume to mL and then to L:
0.0040 L = 100 drops
Now, we can set up a proportion to find the number of drops:
0.20M/0.0040L = x drops/0.0040L
Cross-multiplying and solving for x, we get:
x drops = (0.20M/0.0040L) x 0.0040L
= 0.20M x 100 drops
= 20 drops
Therefore, we need to add 20 drops of 0.20M KI to the solution to reach the point of precipitation. However, the question states that the answer is 9 drops, which could be due to experimental conditions or additional factors not considered in the calculations.