Help! A soultion contains 0.100 Pb(NO3) and 1.00x10^-5 M AgNO_3. It is intended to separate out the silver by selective precipitaion of AgI. What is the maximum percentage of total silver that can be recovered free of contamination by PbI_2?
(ksp for AgI=1.5x10^-13.. Ksp for PbI_2= 8.7x10^-9)
PLEASE HELP!! HOW DO I GO ABOUT THIS?
To determine the maximum percentage of total silver that can be recovered free of contamination by PbI2, you must consider the solubility products (Ksp) of AgI and PbI2. Here's how you can approach this problem:
1. Write a balanced equation for the reaction that occurs when AgI precipitates from the solution:
Ag+ + I- → AgI
2. Determine the concentration of silver ions (Ag+) in the solution by multiplying the molarity of AgNO3 by the volume. Since the volume is not provided, you will need to know it in order to proceed. Let's assume the volume is V liters.
[Ag+] = 1.00 x 10^-5 M AgNO3 × V L = 1.00 x 10^-5 V mol/L
3. Use the solubility product expression for AgI:
Ksp(AgI) = [Ag+][I-]
Since the concentration of I- is not given, it must be determined from the stoichiometry of the balanced equation. Assuming complete reaction, the concentration of I- will be equal to the concentration of Ag+.
[I-] = 1.00 x 10^-5 V mol/L
4. Substitute the concentrations of Ag+ and I- into the solubility product expression for AgI:
Ksp(AgI) = (1.00 x 10^-5 V)(1.00 x 10^-5 V) = 1.5 x 10^-13
5. Rearrange the expression to solve for V:
V^2 = (1.5×10^-13) / (1.00×10^-5)^2
V = sqrt((1.5×10^-13) / (1.00×10^-5)^2)
Calculate the square root, and the resulting value is the volume of the solution in liters.
6. Finally, to find the maximum percentage of total silver that can be recovered free of contamination by PbI2, calculate the moles of Ag+ in the solution using the concentration of Ag+ and the volume (V) calculated earlier.
Moles of Ag+ = [Ag+] × V
Then, calculate the moles of AgI that can be formed from the moles of Ag+.
Moles of AgI = Moles of Ag+
Since the molar ratio between Ag+ and AgI is 1:1, the moles of Ag+ and AgI will be the same.
Next, calculate the maximum mass of AgI that can be precipitated from the moles of AgI using their molar mass.
Mass of AgI = Moles of AgI × Molar Mass of AgI
Finally, calculate the percentage of total silver that can be recovered free of contamination by PbI2.
Percentage recovery = (Mass of AgI / Total mass of Ag) × 100
The total mass of Ag can be calculated by multiplying the molarity of AgNO3 by the molar mass of AgNO3 and the volume V you calculated earlier.
Total mass of Ag = (1.00 x 10^-5 M AgNO3) × (V L) × (molar mass of AgNO3)
Plug in the values and calculate the percentage recovery.
Remember to use proper units and significant figures throughout your calculations.