How many drop (1drop=0.05ml) of 0.20M potassium iodide must we add to 0.010M Pb(NO3)2 to get precipitation of lead iodide to start if Ksp of lead iodide is 7.1x10^-9.
To determine the number of drops of 0.20M potassium iodide (KI) needed to initiate precipitation of lead iodide (PbI2), we can use the concept of molar ratios and the solubility product constant (Ksp) of lead iodide. Here's how:
1. Write the balanced equation for the precipitation reaction:
Pb(NO3)2 + 2KI -> PbI2 + 2KNO3
2. Determine the molar ratio between lead nitrate (Pb(NO3)2) and lead iodide (PbI2) from the balanced equation:
1 mole of Pb(NO3)2 produces 1 mole of PbI2
3. Calculate the concentration of lead ions (Pb2+) in the lead nitrate solution:
Concentration of Pb(NO3)2 = 0.010 M
4. Use the Ksp expression for lead iodide to calculate the concentration of iodide ions (I-) required for precipitation to occur:
Ksp = [Pb2+][I-]^2
Ksp = 7.1x10^-9
Since 1 mole of PbI2 produces 2 moles of I-, the concentration of I- will be twice that of Pb2+:
[I-] = sqrt(Ksp / 2)
[I-] = sqrt(7.1x10^-9 / 2)
5. Convert the concentration of iodide ions to molarity:
Molarity of I- = [I-] / 0.05 mL
Molarity of I- = (sqrt(7.1x10^-9 / 2)) / 0.05 mL
6. Calculate the volume of 0.20M potassium iodide (KI) solution required to get the desired concentration of iodide ions:
(Moles of I-) = (Molarity of I-) * (Volume of KI solution in L)
Moles of I- = (sqrt(7.1x10^-9 / 2)) * (Volume of KI in L)
Since we know that 1 drop is equal to 0.05 mL, we need to convert the volume to drops:
Volume of KI in drops = (Volume of KI in L) / 0.05 mL
7. Rearrange the equation from Step 6 to solve for the volume of KI in drops:
Volume of KI in drops = (Moles of I-) * (0.05 mL) / (sqrt(7.1x10^-9 / 2))
Now you can plug in the values and calculate the volume of 0.20M potassium iodide needed in drops to initiate precipitation of lead iodide.